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Python

9 years ago
#!/usr/bin/env python
# -------------------------------------------------------------------------
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# Name: kdetools
# Purpose:
#
# Author: pab
#
# Created: 01.11.2008
# Copyright: (c) pab 2008
# Licence: LGPL
# -------------------------------------------------------------------------
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from __future__ import absolute_import, division
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import copy
import numpy as np
import scipy
import warnings
from itertools import product
from scipy import interpolate, linalg, optimize, sparse, special, stats
from scipy.special import gamma
from numpy import pi, sqrt, atleast_2d, exp, newaxis # @UnresolvedImport
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from wafo.misc import meshgrid, nextpow2, tranproc # , trangood
from wafo.containers import PlotData
from wafo.dctpack import dct, dctn, idctn
from wafo.plotbackend import plotbackend as plt
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try:
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from wafo import fig
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except ImportError:
warnings.warn('fig import only supported on Windows')
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def _invnorm(q):
return special.ndtri(q)
_stats_epan = (1. / 5, 3. / 5, np.inf)
_stats_biwe = (1. / 7, 5. / 7, 45. / 2)
_stats_triw = (1. / 9, 350. / 429, np.inf)
_stats_rect = (1. / 3, 1. / 2, np.inf)
_stats_tria = (1. / 6, 2. / 3, np.inf)
_stats_lapl = (2, 1. / 4, np.inf)
_stats_logi = (pi ** 2 / 3, 1. / 6, 1 / 42)
_stats_gaus = (1, 1. / (2 * sqrt(pi)), 3. / (8 * sqrt(pi)))
__all__ = ['sphere_volume', 'TKDE', 'KDE', 'Kernel', 'accum', 'qlevels',
'iqrange', 'gridcount', 'kde_demo1', 'kde_demo2', 'test_docstrings']
def sphere_volume(d, r=1.0):
"""
Returns volume of d-dimensional sphere with radius r
Parameters
----------
d : scalar or array_like
dimension of sphere
r : scalar or array_like
radius of sphere (default 1)
Example
-------
>>> sphere_volume(2., r=2.)
12.566370614359172
>>> sphere_volume(2., r=1.)
3.1415926535897931
Reference
---------
Wand,M.P. and Jones, M.C. (1995)
'Kernel smoothing'
Chapman and Hall, pp 105
"""
return (r ** d) * 2.0 * pi ** (d / 2.0) / (d * gamma(d / 2.0))
class KDEgauss(object):
""" Kernel-Density Estimator base class.
Parameters
----------
data : (# of dims, # of data)-array
datapoints to estimate from
hs : array-like (optional)
smooting parameter vector/matrix.
(default compute from data using kernel.get_smoothing function)
alpha : real scalar (optional)
sensitivity parameter (default 0 regular KDE)
A good choice might be alpha = 0.5 ( or 1/D)
alpha = 0 Regular KDE (hs is constant)
0 < alpha <= 1 Adaptive KDE (Make hs change)
Members
-------
d : int
number of dimensions
n : int
number of datapoints
Methods
-------
kde.eval_grid_fast(x0, x1,..., xd) : array
evaluate the estimated pdf on meshgrid(x0, x1,..., xd)
kde(x0, x1,..., xd) : array
same as kde.eval_grid_fast(x0, x1,..., xd)
"""
def __init__(self, data, hs=None, kernel=None, alpha=0.0, xmin=None,
xmax=None, inc=512):
self.dataset = atleast_2d(data)
self.hs = hs
self.kernel = kernel if kernel else Kernel('gauss')
self.alpha = alpha
self.xmin = xmin
self.xmax = xmax
self.inc = inc
self.initialize()
def initialize(self):
self.d, self.n = self.dataset.shape
self._set_xlimits()
self._initialize()
def _initialize(self):
self._compute_smoothing()
def _compute_smoothing(self):
"""Computes the smoothing matrix."""
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get_smoothing = self.kernel.get_smoothing
h = self.hs
if h is None:
h = get_smoothing(self.dataset)
h = np.atleast_1d(h)
hsiz = h.shape
if (len(hsiz) == 1) or (self.d == 1):
if max(hsiz) == 1:
h = h * np.ones(self.d)
else:
h.shape = (self.d,) # make sure it has the correct dimension
# If h negative calculate automatic values
ind, = np.where(h <= 0)
for i in ind.tolist():
h[i] = get_smoothing(self.dataset[i])
deth = h.prod()
self.inv_hs = np.diag(1.0 / h)
else: # fully general smoothing matrix
deth = linalg.det(h)
if deth <= 0:
raise ValueError(
'bandwidth matrix h must be positive definit!')
self.inv_hs = linalg.inv(h)
self.hs = h
self._norm_factor = deth * self.n
def _set_xlimits(self):
amin = self.dataset.min(axis=-1)
amax = self.dataset.max(axis=-1)
iqr = iqrange(self.dataset, axis=-1)
sigma = np.minimum(np.std(self.dataset, axis=-1, ddof=1), iqr / 1.34)
# xyzrange = amax - amin
# offset = xyzrange / 4.0
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offset = 2 * sigma
if self.xmin is None:
self.xmin = amin - offset
else:
self.xmin = self.xmin * np.ones((self.d, 1))
if self.xmax is None:
self.xmax = amax + offset
else:
self.xmax = self.xmax * np.ones((self.d, 1))
def eval_grid_fast(self, *args, **kwds):
"""Evaluate the estimated pdf on a grid.
Parameters
----------
arg_0,arg_1,... arg_d-1 : vectors
Alternatively, if no vectors is passed in then
arg_i = linspace(self.xmin[i], self.xmax[i], self.inc)
output : string optional
'value' if value output
'data' if object output
Returns
-------
values : array-like
The values evaluated at meshgrid(*args).
"""
if len(args) == 0:
args = []
for i in range(self.d):
args.append(np.linspace(self.xmin[i], self.xmax[i], self.inc))
self.args = args
return self._eval_grid_fun(self._eval_grid_fast, *args, **kwds)
def _eval_grid_fast(self, *args, **kwds):
X = np.vstack(args)
d, inc = X.shape
# dx = X[:, 1] - X[:, 0]
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R = X.max(axis=-1) - X.min(axis=-1)
t_star = (self.hs / R) ** 2
I = (np.asfarray(np.arange(0, inc)) * pi) ** 2
In = []
for i in range(d):
In.append(I * t_star[i] * 0.5)
Inc = meshgrid(*In) if d > 1 else In
kw = np.zeros((inc,) * d)
for i in range(d):
kw += exp(-Inc[i])
y = kwds.get('y', 1.0)
d, n = self.dataset.shape
# Find the binned kernel weights, c.
c = gridcount(self.dataset, X, y=y) / n
# Perform the convolution.
at = dctn(c) * kw
z = idctn(at) * at.size / np.prod(R)
return z * (z > 0.0)
def _eval_grid_fun(self, eval_grd, *args, **kwds):
output = kwds.pop('output', 'value')
f = eval_grd(*args, **kwds)
if output == 'value':
return f
else:
titlestr = 'Kernel density estimate (%s)' % self.kernel.name
kwds2 = dict(title=titlestr)
kwds2['plot_kwds'] = dict(plotflag=1)
kwds2.update(**kwds)
args = self.args
if self.d == 1:
args = args[0]
wdata = PlotData(f, args, **kwds2)
if self.d > 1:
PL = np.r_[10:90:20, 95, 99, 99.9]
try:
ql = qlevels(f, p=PL)
wdata.clevels = ql
wdata.plevels = PL
except:
pass
return wdata
def _check_shape(self, points):
points = atleast_2d(points)
d, m = points.shape
if d != self.d:
if d == 1 and m == self.d:
# points was passed in as a row vector
points = np.reshape(points, (self.d, 1))
else:
msg = "points have dimension %s, dataset has dimension %s"
raise ValueError(msg % (d, self.d))
return points
def eval_points(self, points, **kwds):
"""Evaluate the estimated pdf on a set of points.
Parameters
----------
points : (# of dimensions, # of points)-array
Alternatively, a (# of dimensions,) vector can be passed in and
treated as a single point.
Returns
-------
values : (# of points,)-array
The values at each point.
Raises
------
ValueError if the dimensionality of the input points is different than
the dimensionality of the KDE.
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"""
points = self._check_shape(points)
return self._eval_points(points, **kwds)
def _eval_points(self, points, **kwds):
pass
__call__ = eval_grid_fast
class _KDE(object):
""" Kernel-Density Estimator base class.
Parameters
----------
data : (# of dims, # of data)-array
datapoints to estimate from
hs : array-like (optional)
smooting parameter vector/matrix.
(default compute from data using kernel.get_smoothing function)
kernel : kernel function object.
kernel must have get_smoothing method
alpha : real scalar (optional)
sensitivity parameter (default 0 regular KDE)
A good choice might be alpha = 0.5 ( or 1/D)
alpha = 0 Regular KDE (hs is constant)
0 < alpha <= 1 Adaptive KDE (Make hs change)
Members
-------
d : int
number of dimensions
n : int
number of datapoints
Methods
-------
kde.eval_grid_fast(x0, x1,..., xd) : array
evaluate the estimated pdf on meshgrid(x0, x1,..., xd)
kde.eval_grid(x0, x1,..., xd) : array
evaluate the estimated pdf on meshgrid(x0, x1,..., xd)
kde.eval_points(points) : array
evaluate the estimated pdf on a provided set of points
kde(x0, x1,..., xd) : array
same as kde.eval_grid(x0, x1,..., xd)
"""
def __init__(self, data, hs=None, kernel=None, alpha=0.0, xmin=None,
xmax=None, inc=512):
self.dataset = atleast_2d(data)
self.hs = hs
self.kernel = kernel if kernel else Kernel('gauss')
self.alpha = alpha
self.xmin = xmin
self.xmax = xmax
self.inc = inc
self.initialize()
def initialize(self):
self.d, self.n = self.dataset.shape
if self.n > 1:
self._set_xlimits()
self._initialize()
def _initialize(self):
pass
def _set_xlimits(self):
amin = self.dataset.min(axis=-1)
amax = self.dataset.max(axis=-1)
iqr = iqrange(self.dataset, axis=-1)
self._sigma = np.minimum(
np.std(self.dataset, axis=-1, ddof=1), iqr / 1.34)
# xyzrange = amax - amin
# offset = xyzrange / 4.0
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offset = self._sigma
if self.xmin is None:
self.xmin = amin - offset
else:
self.xmin = self.xmin * np.ones((self.d, 1))
if self.xmax is None:
self.xmax = amax + offset
else:
self.xmax = self.xmax * np.ones((self.d, 1))
def get_args(self, xmin=None, xmax=None):
if xmin is None:
xmin = self.xmin
else:
xmin = [min(i, j) for i, j in zip(xmin, self.xmin)]
if xmax is None:
xmax = self.xmax
else:
xmax = [max(i, j) for i, j in zip(xmax, self.xmax)]
args = []
for i in range(self.d):
args.append(np.linspace(xmin[i], xmax[i], self.inc))
return args
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def eval_grid_fast(self, *args, **kwds):
"""Evaluate the estimated pdf on a grid.
Parameters
----------
arg_0,arg_1,... arg_d-1 : vectors
Alternatively, if no vectors is passed in then
arg_i = linspace(self.xmin[i], self.xmax[i], self.inc)
output : string optional
'value' if value output
'data' if object output
Returns
-------
values : array-like
The values evaluated at meshgrid(*args).
"""
if len(args) == 0:
args = self.get_args()
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self.args = args
return self._eval_grid_fun(self._eval_grid_fast, *args, **kwds)
def _eval_grid_fast(self, *args, **kwds):
pass
def eval_grid(self, *args, **kwds):
"""Evaluate the estimated pdf on a grid.
Parameters
----------
arg_0,arg_1,... arg_d-1 : vectors
Alternatively, if no vectors is passed in then
arg_i = linspace(self.xmin[i], self.xmax[i], self.inc)
output : string optional
'value' if value output
'data' if object output
Returns
-------
values : array-like
The values evaluated at meshgrid(*args).
"""
if len(args) == 0:
args = []
for i in range(self.d):
args.append(np.linspace(self.xmin[i], self.xmax[i], self.inc))
self.args = args
return self._eval_grid_fun(self._eval_grid, *args, **kwds)
def _eval_grid(self, *args):
pass
def _eval_grid_fun(self, eval_grd, *args, **kwds):
output = kwds.pop('output', 'value')
f = eval_grd(*args, **kwds)
if output == 'value':
return f
else:
titlestr = 'Kernel density estimate (%s)' % self.kernel.name
kwds2 = dict(title=titlestr)
kwds2['plot_kwds'] = kwds.pop('plot_kwds', dict(plotflag=1))
kwds2.update(**kwds)
args = self.args
if self.d == 1:
args = args[0]
wdata = PlotData(f, args, **kwds2)
if self.d > 1:
PL = np.r_[10:90:20, 95, 99, 99.9]
try:
ql = qlevels(f, p=PL)
wdata.clevels = ql
wdata.plevels = PL
except:
pass
return wdata
def _check_shape(self, points):
points = atleast_2d(points)
d, m = points.shape
if d != self.d:
if d == 1 and m == self.d:
# points was passed in as a row vector
points = np.reshape(points, (self.d, 1))
else:
msg = "points have dimension %s, dataset has dimension %s"
raise ValueError(msg % (d, self.d))
return points
def eval_points(self, points, **kwds):
"""Evaluate the estimated pdf on a set of points.
Parameters
----------
points : (# of dimensions, # of points)-array
Alternatively, a (# of dimensions,) vector can be passed in and
treated as a single point.
Returns
-------
values : (# of points,)-array
The values at each point.
Raises
------
ValueError if the dimensionality of the input points is different than
the dimensionality of the KDE.
11 years ago
"""
points = self._check_shape(points)
return self._eval_points(points, **kwds)
def _eval_points(self, points, **kwds):
pass
__call__ = eval_grid
class TKDE(_KDE):
""" Transformation Kernel-Density Estimator.
Parameters
----------
dataset : (# of dims, # of data)-array
datapoints to estimate from
hs : array-like (optional)
smooting parameter vector/matrix.
(default compute from data using kernel.get_smoothing function)
kernel : kernel function object.
kernel must have get_smoothing method
alpha : real scalar (optional)
sensitivity parameter (default 0 regular KDE)
A good choice might be alpha = 0.5 ( or 1/D)
alpha = 0 Regular KDE (hs is constant)
0 < alpha <= 1 Adaptive KDE (Make hs change)
xmin, xmax : vectors
specifying the default argument range for the kde.eval_grid methods.
For the kde.eval_grid_fast methods the values must cover the range of
the data. (default min(data)-range(data)/4, max(data)-range(data)/4)
If a single value of xmin or xmax is given then the boundary is the is
the same for all dimensions.
inc : scalar integer
defining the default dimension of the output from kde.eval_grid methods
(default 512)
(For kde.eval_grid_fast: A value below 50 is very fast to compute but
may give some inaccuracies. Values between 100 and 500 give very
accurate results)
L2 : array-like
vector of transformation parameters (default 1 no transformation)
t(xi;L2) = xi^L2*sign(L2) for L2(i) ~= 0
t(xi;L2) = log(xi) for L2(i) == 0
If single value of L2 is given then the transformation is the same in
all directions.
Members
-------
d : int
number of dimensions
n : int
number of datapoints
Methods
-------
kde.eval_grid_fast(x0, x1,..., xd) : array
evaluate the estimated pdf on meshgrid(x0, x1,..., xd)
kde.eval_grid(x0, x1,..., xd) : array
evaluate the estimated pdf on meshgrid(x0, x1,..., xd)
kde.eval_points(points) : array
evaluate the estimated pdf on a provided set of points
kde(x0, x1,..., xd) : array
same as kde.eval_grid(x0, x1,..., xd)
Example
-------
N = 20
data = np.random.rayleigh(1, size=(N,))
>>> data = np.array([
... 0.75355792, 0.72779194, 0.94149169, 0.07841119,2.32291887,
... 1.10419995, 0.77055114, 0.60288273, 1.36883635, 1.74754326,
... 1.09547561, 1.01671133, 0.73211143, 0.61891719, 0.75903487,
... 1.8919469 , 0.72433808, 1.92973094, 0.44749838, 1.36508452])
>>> import wafo.kdetools as wk
>>> x = np.linspace(0.01, max(data.ravel()) + 1, 10)
>>> kde = wk.TKDE(data, hs=0.5, L2=0.5)
>>> f = kde(x)
>>> f
array([ 1.03982714, 0.45839018, 0.39514782, 0.32860602, 0.26433318,
0.20717946, 0.15907684, 0.1201074 , 0.08941027, 0.06574882])
>>> kde.eval_grid(x)
array([ 1.03982714, 0.45839018, 0.39514782, 0.32860602, 0.26433318,
0.20717946, 0.15907684, 0.1201074 , 0.08941027, 0.06574882])
>>> kde.eval_grid_fast(x)
array([ 1.04018924, 0.45838973, 0.39514689, 0.32860532, 0.26433301,
0.20717976, 0.15907697, 0.1201077 , 0.08941129, 0.06574899])
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import pylab as plb
h1 = plb.plot(x, f) # 1D probability density plot
t = np.trapz(f, x)
"""
def __init__(self, data, hs=None, kernel=None, alpha=0.0, xmin=None,
xmax=None, inc=512, L2=None):
self.L2 = L2
super(TKDE, self).__init__(data, hs, kernel, alpha, xmin, xmax, inc)
def _initialize(self):
self._check_xmin()
tdataset = self._dat2gaus(self.dataset)
xmin = self.xmin
if xmin is not None:
xmin = self._dat2gaus(np.reshape(xmin, (-1, 1)))
xmax = self.xmax
if xmax is not None:
xmax = self._dat2gaus(np.reshape(xmax, (-1, 1)))
self.tkde = KDE(tdataset, self.hs, self.kernel, self.alpha, xmin, xmax,
self.inc)
if self.inc is None:
self.inc = self.tkde.inc
def _check_xmin(self):
if self.L2 is not None:
amin = self.dataset.min(axis=-1)
# default no transformation
L2 = np.atleast_1d(self.L2) * np.ones(self.d)
self.xmin = np.where(L2 != 1, np.maximum(
self.xmin, amin / 100.0), self.xmin).reshape((-1, 1))
def _dat2gaus(self, points):
if self.L2 is None:
return points # default no transformation
# default no transformation
L2 = np.atleast_1d(self.L2) * np.ones(self.d)
tpoints = copy.copy(points)
for i, v2 in enumerate(L2.tolist()):
tpoints[i] = np.log(points[i]) if v2 == 0 else points[i] ** v2
return tpoints
def _gaus2dat(self, tpoints):
if self.L2 is None:
return tpoints # default no transformation
# default no transformation
L2 = np.atleast_1d(self.L2) * np.ones(self.d)
points = copy.copy(tpoints)
for i, v2 in enumerate(L2.tolist()):
points[i] = np.exp(
tpoints[i]) if v2 == 0 else tpoints[i] ** (1.0 / v2)
return points
def _scale_pdf(self, pdf, points):
if self.L2 is None:
return pdf
# default no transformation
L2 = np.atleast_1d(self.L2) * np.ones(self.d)
for i, v2 in enumerate(L2.tolist()):
factor = v2 * np.sign(v2) if v2 else 1
pdf *= np.where(v2 == 1, 1, points[i] ** (v2 - 1) * factor)
if (np.abs(np.diff(pdf)).max() > 10).any():
msg = ''' Numerical problems may have occured due to the power
transformation. Check the KDE for spurious spikes'''
warnings.warn(msg)
return pdf
def eval_grid_fast2(self, *args, **kwds):
"""Evaluate the estimated pdf on a grid.
Parameters
----------
arg_0,arg_1,... arg_d-1 : vectors
Alternatively, if no vectors is passed in then
arg_i = gauss2dat(linspace(dat2gauss(self.xmin[i]),
dat2gauss(self.xmax[i]), self.inc))
11 years ago
output : string optional
'value' if value output
'data' if object output
Returns
-------
values : array-like
The values evaluated at meshgrid(*args).
11 years ago
"""
return self._eval_grid_fun(self._eval_grid_fast, *args, **kwds)
def _eval_grid_fast(self, *args, **kwds):
if self.L2 is None:
f = self.tkde.eval_grid_fast(*args, **kwds)
self.args = self.tkde.args
return f
targs = []
if len(args):
targs0 = self._dat2gaus(list(args))
xmin = [min(t) for t in targs0]
xmax = [max(t) for t in targs0]
targs = self.tkde.get_args(xmin, xmax)
tf = self.tkde.eval_grid_fast(*targs)
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self.args = self._gaus2dat(list(self.tkde.args))
points = meshgrid(*self.args) if self.d > 1 else self.args
f = self._scale_pdf(tf, points)
if len(args):
ipoints = meshgrid(*args) if self.d > 1 else args
# shape0 = points[0].shape
# shape0i = ipoints[0].shape
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for i in range(self.d):
points[i].shape = (-1,)
# ipoints[i].shape = (-1,)
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points = np.asarray(points).T
# ipoints = np.asarray(ipoints).T
fi = interpolate.griddata(points, f.ravel(), tuple(ipoints),
method='linear',
fill_value=0.0)
# fi.shape = shape0i
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self.args = args
r = kwds.get('r', 0)
if r == 0:
return fi * (fi > 0)
else:
return fi
return f
def _eval_grid(self, *args, **kwds):
if self.L2 is None:
return self.tkde.eval_grid(*args, **kwds)
targs = self._dat2gaus(list(args))
tf = self.tkde.eval_grid(*targs, **kwds)
points = meshgrid(*args) if self.d > 1 else self.args
f = self._scale_pdf(tf, points)
return f
def _eval_points(self, points):
"""Evaluate the estimated pdf on a set of points.
Parameters
----------
points : (# of dimensions, # of points)-array
Alternatively, a (# of dimensions,) vector can be passed in and
treated as a single point.
Returns
-------
values : (# of points,)-array
The values at each point.
Raises
------
ValueError if the dimensionality of the input points is different than
the dimensionality of the KDE.
11 years ago
"""
if self.L2 is None:
return self.tkde.eval_points(points)
tpoints = self._dat2gaus(points)
tf = self.tkde.eval_points(tpoints)
f = self._scale_pdf(tf, points)
return f
class KDE(_KDE):
""" Kernel-Density Estimator.
Parameters
----------
data : (# of dims, # of data)-array
datapoints to estimate from
hs : array-like (optional)
smooting parameter vector/matrix.
(default compute from data using kernel.get_smoothing function)
kernel : kernel function object.
kernel must have get_smoothing method
alpha : real scalar (optional)
sensitivity parameter (default 0 regular KDE)
A good choice might be alpha = 0.5 ( or 1/D)
alpha = 0 Regular KDE (hs is constant)
0 < alpha <= 1 Adaptive KDE (Make hs change)
xmin, xmax : vectors
specifying the default argument range for the kde.eval_grid methods.
For the kde.eval_grid_fast methods the values must cover the range of
the data.
(default min(data)-range(data)/4, max(data)-range(data)/4)
If a single value of xmin or xmax is given then the boundary is the is
the same for all dimensions.
inc : scalar integer (default 512)
11 years ago
defining the default dimension of the output from kde.eval_grid methods
(For kde.eval_grid_fast: A value below 50 is very fast to compute but
may give some inaccuracies. Values between 100 and 500 give very
accurate results)
Members
-------
d : int
number of dimensions
n : int
number of datapoints
Methods
-------
kde.eval_grid_fast(x0, x1,..., xd) : array
evaluate the estimated pdf on meshgrid(x0, x1,..., xd)
kde.eval_grid(x0, x1,..., xd) : array
evaluate the estimated pdf on meshgrid(x0, x1,..., xd)
kde.eval_points(points) : array
evaluate the estimated pdf on a provided set of points
kde(x0, x1,..., xd) : array
same as kde.eval_grid(x0, x1,..., xd)
Example
-------
N = 20
data = np.random.rayleigh(1, size=(N,))
>>> data = np.array([
... 0.75355792, 0.72779194, 0.94149169, 0.07841119, 2.32291887,
... 1.10419995, 0.77055114, 0.60288273, 1.36883635, 1.74754326,
... 1.09547561, 1.01671133, 0.73211143, 0.61891719, 0.75903487,
... 1.8919469 , 0.72433808, 1.92973094, 0.44749838, 1.36508452])
>>> x = np.linspace(0, max(data.ravel()) + 1, 10)
>>> import wafo.kdetools as wk
>>> kde = wk.KDE(data, hs=0.5, alpha=0.5)
>>> f = kde(x)
>>> f
array([ 0.17252055, 0.41014271, 0.61349072, 0.57023834, 0.37198073,
0.21409279, 0.12738463, 0.07460326, 0.03956191, 0.01887164])
>>> kde.eval_grid(x)
array([ 0.17252055, 0.41014271, 0.61349072, 0.57023834, 0.37198073,
0.21409279, 0.12738463, 0.07460326, 0.03956191, 0.01887164])
>>> kde0 = wk.KDE(data, hs=0.5, alpha=0.0)
>>> kde0.eval_points(x)
array([ 0.2039735 , 0.40252503, 0.54595078, 0.52219649, 0.3906213 ,
0.26381501, 0.16407362, 0.08270612, 0.02991145, 0.00720821])
>>> kde0.eval_grid(x)
array([ 0.2039735 , 0.40252503, 0.54595078, 0.52219649, 0.3906213 ,
0.26381501, 0.16407362, 0.08270612, 0.02991145, 0.00720821])
>>> f = kde0.eval_grid(x, output='plotobj')
>>> f.data
array([ 0.2039735 , 0.40252503, 0.54595078, 0.52219649, 0.3906213 ,
0.26381501, 0.16407362, 0.08270612, 0.02991145, 0.00720821])
>>> f = kde0.eval_grid_fast()
>>> np.allclose(np.interp(x, kde0.args[0], f),
... [ 0.20397743, 0.40252228, 0.54594119, 0.52219025, 0.39062189,
... 0.2638171 , 0.16407487, 0.08270755, 0.04784434, 0.04784434])
True
11 years ago
>>> f1 = kde0.eval_grid_fast(output='plot')
>>> np.allclose(np.interp(x, f1.args, f1.data),
... [ 0.20397743, 0.40252228, 0.54594119, 0.52219025, 0.39062189,
... 0.2638171 , 0.16407487, 0.08270755, 0.04784434, 0.04784434])
True
11 years ago
9 years ago
h = f1.plot()
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import pylab as plb
h1 = plb.plot(x, f) # 1D probability density plot
t = np.trapz(f, x)
"""
def __init__(self, data, hs=None, kernel=None, alpha=0.0, xmin=None,
xmax=None, inc=512):
super(KDE, self).__init__(data, hs, kernel, alpha, xmin, xmax, inc)
def _initialize(self):
self._compute_smoothing()
self._lambda = np.ones(self.n)
if self.alpha > 0:
# pilt = KDE(self.dataset, hs=self.hs, kernel=self.kernel, alpha=0)
# f = pilt.eval_points(self.dataset) # get a pilot estimate by
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# regular KDE (alpha=0)
f = self.eval_points(self.dataset) # pilot estimate
g = np.exp(np.mean(np.log(f)))
self._lambda = (f / g) ** (-self.alpha)
if self.inc is None:
unused_tau, tau = self.kernel.effective_support()
xyzrange = 8 * self._sigma
L1 = 10
self.inc = 2 ** nextpow2(
max(48, (L1 * xyzrange / (tau * self.hs)).max()))
pass
def _compute_smoothing(self):
"""Computes the smoothing matrix."""
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get_smoothing = self.kernel.get_smoothing
h = self.hs
if h is None:
h = get_smoothing(self.dataset)
h = np.atleast_1d(h)
hsiz = h.shape
if (len(hsiz) == 1) or (self.d == 1):
if max(hsiz) == 1:
h = h * np.ones(self.d)
else:
h.shape = (self.d,) # make sure it has the correct dimension
# If h negative calculate automatic values
ind, = np.where(h <= 0)
for i in ind.tolist():
h[i] = get_smoothing(self.dataset[i])
deth = h.prod()
self.inv_hs = np.diag(1.0 / h)
else: # fully general smoothing matrix
deth = linalg.det(h)
if deth <= 0:
raise ValueError(
'bandwidth matrix h must be positive definit!')
self.inv_hs = linalg.inv(h)
self.hs = h
self._norm_factor = deth * self.n
def _eval_grid_fast(self, *args, **kwds):
X = np.vstack(args)
d, inc = X.shape
dx = X[:, 1] - X[:, 0]
Xn = []
nfft0 = 2 * inc
nfft = (nfft0,) * d
x0 = np.linspace(-inc, inc, nfft0 + 1)
for i in range(d):
Xn.append(x0[:-1] * dx[i])
Xnc = meshgrid(*Xn) if d > 1 else Xn
shape0 = Xnc[0].shape
for i in range(d):
Xnc[i].shape = (-1,)
Xn = np.dot(self.inv_hs, np.vstack(Xnc))
# Obtain the kernel weights.
kw = self.kernel(Xn)
# plt.plot(kw)
# plt.draw()
# plt.show()
norm_fact0 = (kw.sum() * dx.prod() * self.n)
norm_fact = (self._norm_factor * self.kernel.norm_factor(d, self.n))
if np.abs(norm_fact0 - norm_fact) > 0.05 * norm_fact:
warnings.warn(
'Numerical inaccuracy due to too low discretization. ' +
'Increase the discretization of the evaluation grid ' +
'(inc=%d)!' % inc)
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norm_fact = norm_fact0
kw = kw / norm_fact
r = kwds.get('r', 0)
if r != 0:
kw *= np.vstack(Xnc) ** r if d > 1 else Xnc[0]
kw.shape = shape0
kw = np.fft.ifftshift(kw)
fftn = np.fft.fftn
ifftn = np.fft.ifftn
y = kwds.get('y', 1.0)
# if self.alpha>0:
# y = y / self._lambda**d
# Find the binned kernel weights, c.
c = gridcount(self.dataset, X, y=y)
# Perform the convolution.
z = np.real(ifftn(fftn(c, s=nfft) * fftn(kw)))
ix = (slice(0, inc),) * d
if r == 0:
return z[ix] * (z[ix] > 0.0)
else:
return z[ix]
def _eval_grid(self, *args, **kwds):
grd = meshgrid(*args) if len(args) > 1 else list(args)
shape0 = grd[0].shape
d = len(grd)
for i in range(d):
grd[i] = grd[i].ravel()
f = self.eval_points(np.vstack(grd), **kwds)
return f.reshape(shape0)
def _eval_points(self, points, **kwds):
"""Evaluate the estimated pdf on a set of points.
Parameters
----------
points : (# of dimensions, # of points)-array
Alternatively, a (# of dimensions,) vector can be passed in and
treated as a single point.
Returns
-------
values : (# of points,)-array
The values at each point.
Raises
------
ValueError if the dimensionality of the input points is different than
the dimensionality of the KDE.
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"""
d, m = points.shape
result = np.zeros((m,))
r = kwds.get('r', 0)
if r == 0:
def fun(xi):
return 1
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else:
def fun(xi):
return (xi ** r).sum(axis=0)
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if m >= self.n:
y = kwds.get('y', np.ones(self.n))
# there are more points than data, so loop over data
for i in range(self.n):
diff = self.dataset[:, i, np.newaxis] - points
tdiff = np.dot(self.inv_hs / self._lambda[i], diff)
result += y[i] * \
fun(diff) * self.kernel(tdiff) / self._lambda[i] ** d
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else:
y = kwds.get('y', 1)
# loop over points
for i in range(m):
diff = self.dataset - points[:, i, np.newaxis]
tdiff = np.dot(self.inv_hs, diff / self._lambda[np.newaxis, :])
tmp = y * fun(diff) * self.kernel(tdiff) / self._lambda ** d
result[i] = tmp.sum(axis=-1)
result /= (self._norm_factor * self.kernel.norm_factor(d, self.n))
return result
class KRegression(_KDE):
""" Kernel-Regression
Parameters
----------
data : (# of dims, # of data)-array
datapoints to estimate from
y : # of data - array
response variable
p : scalar integer (0 or 1)
Nadaraya-Watson estimator if p=0,
local linear estimator if p=1.
hs : array-like (optional)
smooting parameter vector/matrix.
(default compute from data using kernel.get_smoothing function)
kernel : kernel function object.
kernel must have get_smoothing method
alpha : real scalar (optional)
sensitivity parameter (default 0 regular KDE)
A good choice might be alpha = 0.5 ( or 1/D)
alpha = 0 Regular KDE (hs is constant)
0 < alpha <= 1 Adaptive KDE (Make hs change)
xmin, xmax : vectors
specifying the default argument range for the kde.eval_grid methods.
For the kde.eval_grid_fast methods the values must cover the range of
the data. (default min(data)-range(data)/4, max(data)-range(data)/4)
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If a single value of xmin or xmax is given then the boundary is the is
the same for all dimensions.
inc : scalar integer (default 128)
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defining the default dimension of the output from kde.eval_grid methods
(For kde.eval_grid_fast: A value below 50 is very fast to compute but
may give some inaccuracies. Values between 100 and 500 give very
accurate results)
Members
-------
d : int
number of dimensions
n : int
number of datapoints
Methods
-------
kde.eval_grid_fast(x0, x1,..., xd) : array
evaluate the estimated pdf on meshgrid(x0, x1,..., xd)
kde.eval_grid(x0, x1,..., xd) : array
evaluate the estimated pdf on meshgrid(x0, x1,..., xd)
kde.eval_points(points) : array
evaluate the estimated pdf on a provided set of points
kde(x0, x1,..., xd) : array
same as kde.eval_grid(x0, x1,..., xd)
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Example
-------
9 years ago
>>> import wafo.kdetools as wk
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>>> N = 100
>>> x = np.linspace(0, 1, N)
9 years ago
>>> ei = np.random.normal(loc=0, scale=0.075, size=(N,))
>>> ei = np.sqrt(0.075) * np.sin(100*x)
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>>> y = 2*np.exp(-x**2/(2*0.3**2))+3*np.exp(-(x-1)**2/(2*0.7**2)) + ei
>>> kreg = wk.KRegression(x, y)
>>> f = kreg(output='plotobj', title='Kernel regression', plotflag=1)
9 years ago
>>> np.allclose(f.data[:5],
... [ 3.18381052, 3.18362269, 3.18343648, 3.1832536 , 3.1830757 ])
True
h = f.plot(label='p=0')
11 years ago
"""
def __init__(self, data, y, p=0, hs=None, kernel=None, alpha=0.0,
xmin=None, xmax=None, inc=128, L2=None):
self.tkde = TKDE(data, hs=hs, kernel=kernel,
alpha=alpha, xmin=xmin, xmax=xmax, inc=inc, L2=L2)
self.y = y
self.p = p
def eval_grid_fast(self, *args, **kwds):
self._grdfun = self.tkde.eval_grid_fast
return self.tkde._eval_grid_fun(self._eval_gridfun, *args, **kwds)
def eval_grid(self, *args, **kwds):
self._grdfun = self.tkde.eval_grid
return self.tkde._eval_grid_fun(self._eval_gridfun, *args, **kwds)
def _eval_gridfun(self, *args, **kwds):
grdfun = self._grdfun
s0 = grdfun(*args, r=0)
t0 = grdfun(*args, r=0, y=self.y)
if self.p == 0:
return (t0 / (s0 + _TINY)).clip(min=-_REALMAX, max=_REALMAX)
elif self.p == 1:
s1 = grdfun(*args, r=1)
s2 = grdfun(*args, r=2)
t1 = grdfun(*args, r=1, y=self.y)
return ((s2 * t0 - s1 * t1) /
(s2 * s0 - s1 ** 2)).clip(min=-_REALMAX, max=_REALMAX)
__call__ = eval_grid_fast
class BKRegression(object):
'''
Kernel-Regression on binomial data
method : {'beta', 'wilson'}
method is one of the following
'beta', return Bayesian Credible interval using beta-distribution.
'wilson', return Wilson score interval
a, b : scalars
parameters of the beta distribution defining the apriori distribution
of p, i.e., the Bayes estimator for p: p = (y+a)/(n+a+b).
Setting a=b=0.5 gives Jeffreys interval.
'''
def __init__(self, *args, **kwds):
self.method = kwds.pop('method', 'beta')
self.a = max(kwds.pop('a', 0.5), _TINY)
self.b = max(kwds.pop('b', 0.5), _TINY)
self.kreg = KRegression(*args, **kwds)
# defines bin width (i.e. smoothing) in empirical estimate
self.hs_e = None
# self.x = self.kreg.tkde.dataset
# self.y = self.kreg.y
def _set_smoothing(self, hs):
self.kreg.tkde.hs = hs
self.kreg.tkde.initialize()
x = property(fget=lambda cls: cls.kreg.tkde.dataset.squeeze())
y = property(fget=lambda cls: cls.kreg.y)
kernel = property(fget=lambda cls: cls.kreg.tkde.kernel)
hs = property(fset=_set_smoothing, fget=lambda cls: cls.kreg.tkde.hs)
def _get_max_smoothing(self, fun=None):
"""Return maximum value for smoothing parameter."""
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x = self.x
y = self.y
if fun is None:
get_smoothing = self.kernel.get_smoothing
else:
get_smoothing = getattr(self.kernel, fun)
hs1 = get_smoothing(x)
# hx = np.median(np.abs(x-np.median(x)))/0.6745*(4.0/(3*n))**0.2
if (y == 1).any():
hs2 = get_smoothing(x[y == 1])
# hy = np.median(np.abs(y-np.mean(y)))/0.6745*(4.0/(3*n))**0.2
11 years ago
else:
hs2 = 4 * hs1
# hy = 4*hx
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hopt = sqrt(hs1 * hs2)
return hopt, hs1, hs2
def get_grid(self, hs_e=None):
if hs_e is None:
if self.hs_e is None:
hs1 = self._get_max_smoothing('hste')[0]
hs2 = self._get_max_smoothing('hos')[0]
self.hs_e = sqrt(hs1 * hs2)
hs_e = self.hs_e
x = self.x
xmin, xmax = x.min(), x.max()
ni = max(2 * int((xmax - xmin) / hs_e) + 3, 5)
sml = hs_e # *0.1
xi = np.linspace(xmin - sml, xmax + sml, ni)
return xi
def prb_ci(self, n, p, alpha=0.05, **kwds):
"""Return Confidence Interval for the binomial probability p.
11 years ago
Parameters
----------
n : array-like
number of Bernoulli trials
p : array-like
estimated probability of success in each trial
alpha : scalar
confidence level
method : {'beta', 'wilson'}
method is one of the following
'beta', return Bayesian Credible interval using beta-distribution.
'wilson', return Wilson score interval
a, b : scalars
parameters of the beta distribution defining the apriori
distribution of p, i.e.,
the Bayes estimator for p: p = (y+a)/(n+a+b).
Setting a=b=0.5 gives Jeffreys interval.
"""
11 years ago
if self.method.startswith('w'):
# Wilson score
z0 = -_invnorm(alpha / 2)
den = 1 + (z0 ** 2. / n)
xc = (p + (z0 ** 2) / (2 * n)) / den
halfwidth = (z0 * sqrt((p * (1 - p) / n) +
(z0 ** 2 / (4 * (n ** 2))))) / den
plo = (xc - halfwidth).clip(min=0) # wilson score
pup = (xc + halfwidth).clip(max=1.0) # wilson score
else:
# Jeffreys intervall a=b=0.5
# st.beta.isf(alpha/2, y+a, n-y+b) y = n*p, n-y = n*(1-p)
a = self.a
b = self.b
st = stats
pup = np.where(
p == 1, 1, st.beta.isf(alpha / 2, n * p + a, n * (1 - p) + b))
plo = np.where(p == 0, 0,
st.beta.isf(1 - alpha / 2,
n * p + a, n * (1 - p) + b))
return plo, pup
def prb_empirical(self, xi=None, hs_e=None, alpha=0.05, color='r', **kwds):
"""Returns empirical binomial probabiltity.
11 years ago
Parameters
----------
x : ndarray
position vector
y : ndarray
binomial response variable (zeros and ones)
alpha : scalar
confidence level
color:
used in plot
Returns
-------
P(x) : PlotData object
empirical probability
"""
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if xi is None:
xi = self.get_grid(hs_e)
x = self.x
y = self.y
c = gridcount(x, xi) # + self.a + self.b # count data
if (y == 1).any():
c0 = gridcount(x[y == 1], xi) # + self.a # count success
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else:
c0 = np.zeros(xi.shape)
prb = np.where(c == 0, 0, c0 / (c + _TINY)) # assume prb==0 for c==0
CI = np.vstack(self.prb_ci(c, prb, alpha, **kwds))
prb_e = PlotData(prb, xi, plotmethod='plot', plot_args=['.'],
plot_kwds=dict(markersize=6, color=color, picker=5))
prb_e.dataCI = CI.T
prb_e.count = c
return prb_e
def prb_smoothed(self, prb_e, hs, alpha=0.05, color='r', label=''):
"""Return smoothed binomial probability.
11 years ago
Parameters
----------
prb_e : PlotData object with empirical binomial probabilites
hs : smoothing parameter
alpha : confidence level
color : color of plot object
label : label for plot object
"""
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x_e = prb_e.args
n_e = len(x_e)
dx_e = x_e[1] - x_e[0]
n = self.x.size
x_s = np.linspace(x_e[0], x_e[-1], 10 * n_e + 1)
self.hs = hs
prb_s = self.kreg(x_s, output='plotobj', title='', plot_kwds=dict(
color=color, linewidth=2)) # dict(plotflag=7))
m_nan = np.isnan(prb_s.data)
if m_nan.any(): # assume 0/0 division
prb_s.data[m_nan] = 0.0
# prb_s.data[np.isnan(prb_s.data)] = 0
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# expected number of data in each bin
c_s = self.kreg.tkde.eval_grid_fast(x_s) * dx_e * n
plo, pup = self.prb_ci(c_s, prb_s.data, alpha)
prb_s.dataCI = np.vstack((plo, pup)).T
prb_s.prediction_error_avg = np.trapz(
pup - plo, x_s) / (x_s[-1] - x_s[0])
if label:
prb_s.plot_kwds['label'] = label
prb_s.children = [PlotData([plo, pup], x_s,
plotmethod='fill_between',
plot_kwds=dict(alpha=0.2, color=color)),
prb_e]
# empirical oversmooths the data
# p_s = prb_s.eval_points(self.x)
# dp_s = np.diff(prb_s.data)
# k = (dp_s[:-1]*dp_s[1:]<0).sum() # numpeaks
# p_e = self.y
# n_s = interpolate.interp1d(x_s, c_s)(self.x)
# plo, pup = self.prb_ci(n_s, p_s, alpha)
# sigmai = (pup-plo)
# aicc = (((p_e-p_s)/sigmai)**2).sum()+ 2*k*(k+1)/np.maximum(n-k+1,1)
p_e = prb_e.eval_points(x_s)
p_s = prb_s.data
dp_s = np.sign(np.diff(p_s))
k = (dp_s[:-1] != dp_s[1:]).sum() # numpeaks
# sigmai = (pup-plo)+_EPS
# aicc = (((p_e-p_s)/sigmai)**2).sum()+ 2*k*(k+1)/np.maximum(n_e-k+1,1)
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# + np.abs((p_e-pup).clip(min=0)-(p_e-plo).clip(max=0)).sum()
sigmai = _logit(pup) - _logit(plo) + _EPS
aicc = ((((_logit(p_e) - _logit(p_s)) / sigmai) ** 2).sum() +
2 * k * (k + 1) / np.maximum(n_e - k + 1, 1) +
np.abs((p_e - pup).clip(min=0) -
(p_e - plo).clip(max=0)).sum())
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prb_s.aicc = aicc
# prb_s.labels.title = ''
# prb_s.labels.title='perr=%1.3f,aicc=%1.3f, n=%d, hs=%1.3f' %
# (prb_s.prediction_error_avg,aicc,n,hs)
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return prb_s
def prb_search_best(self, prb_e=None, hsvec=None, hsfun='hste',
alpha=0.05, color='r', label=''):
"""Return best smoothed binomial probability.
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Parameters
----------
prb_e : PlotData object with empirical binomial probabilites
hsvec : arraylike (default np.linspace(hsmax*0.1,hsmax,55))
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vector smoothing parameters
hsfun :
method for calculating hsmax
"""
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if prb_e is None:
prb_e = self.prb_empirical(
hs_e=self.hs_e, alpha=alpha, color=color)
if hsvec is None:
hsmax = self._get_max_smoothing(hsfun)[0] # @UnusedVariable
hsmax = max(hsmax, self.hs_e)
hsvec = np.linspace(hsmax * 0.2, hsmax, 55)
hs_best = hsvec[-1] + 0.1
prb_best = self.prb_smoothed(prb_e, hs_best, alpha, color, label)
aicc = np.zeros(np.size(hsvec))
for i, hi in enumerate(hsvec):
f = self.prb_smoothed(prb_e, hi, alpha, color, label)
aicc[i] = f.aicc
if f.aicc <= prb_best.aicc:
prb_best = f
hs_best = hi
prb_best.score = PlotData(aicc, hsvec)
prb_best.hs = hs_best
self._set_smoothing(hs_best)
return prb_best
class _Kernel(object):
def __init__(self, r=1.0, stats=None):
self.r = r # radius of kernel
self.stats = stats
def norm_factor(self, d=1, n=None):
return 1.0
def norm_kernel(self, x):
X = np.atleast_2d(x)
return self._kernel(X) / self.norm_factor(*X.shape)
def kernel(self, x):
return self._kernel(np.atleast_2d(x))
def deriv4_6_8_10(self, t, numout=4):
raise Exception('Method not implemented for this kernel!')
def effective_support(self):
"""Return the effective support of kernel.
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The kernel must be symmetric and compactly supported on [-tau tau]
if the kernel has infinite support then the kernel must have the
effective support in [-tau tau], i.e., be negligible outside the range
"""
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return self._effective_support()
def _effective_support(self):
return - self.r, self.r
__call__ = kernel
class _KernelMulti(_Kernel):
# p=0; %Sphere = rect for 1D
# p=1; %Multivariate Epanechnikov kernel.
# p=2; %Multivariate Bi-weight Kernel
# p=3; %Multi variate Tri-weight Kernel
# p=4; %Multi variate Four-weight Kernel
def __init__(self, r=1.0, p=1, stats=None):
self.r = r
self.p = p
self.stats = stats
def norm_factor(self, d=1, n=None):
r = self.r
p = self.p
c = 2 ** p * np.prod(np.r_[1:p + 1]) * sphere_volume(d, r) / np.prod(
np.r_[(d + 2):(2 * p + d + 1):2]) # normalizing constant
return c
def _kernel(self, x):
r = self.r
p = self.p
x2 = x ** 2
return ((1.0 - x2.sum(axis=0) / r ** 2).clip(min=0.0)) ** p
mkernel_epanechnikov = _KernelMulti(p=1, stats=_stats_epan)
mkernel_biweight = _KernelMulti(p=2, stats=_stats_biwe)
mkernel_triweight = _KernelMulti(p=3, stats=_stats_triw)
class _KernelProduct(_KernelMulti):
# p=0; %rectangular
# p=1; %1D product Epanechnikov kernel.
# p=2; %1D product Bi-weight Kernel
# p=3; %1D product Tri-weight Kernel
# p=4; %1D product Four-weight Kernel
def norm_factor(self, d=1, n=None):
r = self.r
p = self.p
c = (2 ** p * np.prod(np.r_[1:p + 1]) * sphere_volume(1, r) /
np.prod(np.r_[(1 + 2):(2 * p + 2):2]))
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return c ** d
def _kernel(self, x):
r = self.r # radius
pdf = (1 - (x / r) ** 2).clip(min=0.0)
return pdf.prod(axis=0)
mkernel_p1epanechnikov = _KernelProduct(p=1, stats=_stats_epan)
mkernel_p1biweight = _KernelProduct(p=2, stats=_stats_biwe)
mkernel_p1triweight = _KernelProduct(p=3, stats=_stats_triw)
class _KernelRectangular(_Kernel):
def _kernel(self, x):
return np.where(np.all(np.abs(x) <= self.r, axis=0), 1, 0.0)
def norm_factor(self, d=1, n=None):
r = self.r
return (2 * r) ** d
mkernel_rectangular = _KernelRectangular(stats=_stats_rect)
class _KernelTriangular(_Kernel):
def _kernel(self, x):
pdf = (1 - np.abs(x)).clip(min=0.0)
return pdf.prod(axis=0)
mkernel_triangular = _KernelTriangular(stats=_stats_tria)
class _KernelGaussian(_Kernel):
def _kernel(self, x):
sigma = self.r / 4.0
x2 = (x / sigma) ** 2
return exp(-0.5 * x2.sum(axis=0))
def norm_factor(self, d=1, n=None):
sigma = self.r / 4.0
return (2 * pi * sigma) ** (d / 2.0)
def deriv4_6_8_10(self, t, numout=4):
"""Returns 4th, 6th, 8th and 10th derivatives of the kernel
function."""
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phi0 = exp(-0.5 * t ** 2) / sqrt(2 * pi)
p4 = [1, 0, -6, 0, +3]
p4val = np.polyval(p4, t) * phi0
if numout == 1:
return p4val
out = [p4val]
pn = p4
for unusedix in range(numout - 1):
pnp1 = np.polyadd(-np.r_[pn, 0], np.polyder(pn))
pnp2 = np.polyadd(-np.r_[pnp1, 0], np.polyder(pnp1))
out.append(np.polyval(pnp2, t) * phi0)
pn = pnp2
return out
mkernel_gaussian = _KernelGaussian(r=4.0, stats=_stats_gaus)
# def mkernel_gaussian(X):
# x2 = X ** 2
# d = X.shape[0]
# return (2 * pi) ** (-d / 2) * exp(-0.5 * x2.sum(axis=0))
class _KernelLaplace(_Kernel):
def _kernel(self, x):
absX = np.abs(x)
return exp(-absX.sum(axis=0))
def norm_factor(self, d=1, n=None):
return 2 ** d
mkernel_laplace = _KernelLaplace(r=7.0, stats=_stats_lapl)
class _KernelLogistic(_Kernel):
def _kernel(self, x):
s = exp(-x)
return np.prod(1.0 / (s + 1) ** 2, axis=0)
mkernel_logistic = _KernelLogistic(r=7.0, stats=_stats_logi)
_MKERNEL_DICT = dict(
epan=mkernel_epanechnikov,
biwe=mkernel_biweight,
triw=mkernel_triweight,
p1ep=mkernel_p1epanechnikov,
p1bi=mkernel_p1biweight,
p1tr=mkernel_p1triweight,
rect=mkernel_rectangular,
tria=mkernel_triangular,
lapl=mkernel_laplace,
logi=mkernel_logistic,
gaus=mkernel_gaussian
)
_KERNEL_EXPONENT_DICT = dict(
re=0, sp=0, ep=1, bi=2, tr=3, fo=4, fi=5, si=6, se=7)
class Kernel(object):
"""Multivariate kernel.
11 years ago
Parameters
----------
name : string
defining the kernel. Valid options are:
'epanechnikov' - Epanechnikov kernel.
'biweight' - Bi-weight kernel.
'triweight' - Tri-weight kernel.
'p1epanechnikov' - product of 1D Epanechnikov kernel.
'p1biweight' - product of 1D Bi-weight kernel.
'p1triweight' - product of 1D Tri-weight kernel.
'triangular' - Triangular kernel.
'gaussian' - Gaussian kernel
'rectangular' - Rectangular kernel.
'laplace' - Laplace kernel.
'logistic' - Logistic kernel.
Note that only the first 4 letters of the kernel name is needed.
Examples
--------
N = 20
data = np.random.rayleigh(1, size=(N,))
>>> data = np.array([
... 0.75355792, 0.72779194, 0.94149169, 0.07841119, 2.32291887,
... 1.10419995, 0.77055114, 0.60288273, 1.36883635, 1.74754326,
... 1.09547561, 1.01671133, 0.73211143, 0.61891719, 0.75903487,
... 1.8919469 , 0.72433808, 1.92973094, 0.44749838, 1.36508452])
>>> import wafo.kdetools as wk
>>> gauss = wk.Kernel('gaussian')
>>> gauss.stats()
(1, 0.28209479177387814, 0.21157109383040862)
>>> np.allclose(gauss.hscv(data), 0.21779575)
True
>>> np.allclose(gauss.hstt(data), 0.16341135)
True
>>> np.allclose(gauss.hste(data), 0.19179399)
True
>>> np.allclose(gauss.hldpi(data), 0.22502733)
True
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>>> wk.Kernel('laplace').stats()
(2, 0.25, inf)
>>> triweight = wk.Kernel('triweight')
>>> np.allclose(triweight.stats(),
... (0.1111111111111111, 0.81585081585081587, np.inf))
True
>>> np.allclose(triweight(np.linspace(-1,1,11)),
... [ 0., 0.046656, 0.262144, 0.592704, 0.884736, 1.,
... 0.884736, 0.592704, 0.262144, 0.046656, 0.])
True
>>> np.allclose(triweight.hns(data), 0.82, rtol=1e-2)
True
>>> np.allclose(triweight.hos(data), 0.88, rtol=1e-2)
True
>>> np.allclose(triweight.hste(data), 0.57, rtol=1e-2)
True
>>> np.allclose(triweight.hscv(data), 0.648, rtol=1e-2)
True
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See also
--------
mkernel
References
----------
B. W. Silverman (1986)
'Density estimation for statistics and data analysis'
Chapman and Hall, pp. 43, 76
Wand, M. P. and Jones, M. C. (1995)
'Density estimation for statistics and data analysis'
Chapman and Hall, pp 31, 103, 175
"""
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def __init__(self, name, fun='hste'): # 'hns'):
self.kernel = _MKERNEL_DICT[name[:4]]
# self.name = self.kernel.__name__.replace('mkernel_', '').title()
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try:
self.get_smoothing = getattr(self, fun)
except:
self.get_smoothing = self.hste
def _get_name(self):
return self.kernel.__class__.__name__.replace('_Kernel', '').title()
name = property(_get_name)
def get_smoothing(self, *args, **kwds):
pass
def stats(self):
"""Return some 1D statistics of the kernel.
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Returns
-------
mu2 : real scalar
2'nd order moment, i.e.,int(x^2*kernel(x))
R : real scalar
integral of squared kernel, i.e., int(kernel(x)^2)
Rdd : real scalar
integral of squared double derivative of kernel,
i.e., int( (kernel''(x))^2 ).
Reference
---------
Wand,M.P. and Jones, M.C. (1995)
'Kernel smoothing'
Chapman and Hall, pp 176.
"""
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return self.kernel.stats
def deriv4_6_8_10(self, t, numout=4):
return self.kernel.deriv4_6_8_10(t, numout)
def effective_support(self):
return self.kernel.effective_support()
def hns(self, data):
"""Returns Normal Scale Estimate of Smoothing Parameter.
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Parameter
---------
data : 2D array
shape d x n (d = # dimensions )
Returns
-------
h : array-like
one dimensional optimal value for smoothing parameter
given the data and kernel. size D
HNS only gives an optimal value with respect to mean integrated
square error, when the true underlying distribution
is Gaussian. This works reasonably well if the data resembles a
Gaussian distribution. However if the distribution is asymmetric,
multimodal or have long tails then HNS may return a to large
smoothing parameter, i.e., the KDE may be oversmoothed and mask
important features of the data. (=> large bias).
One way to remedy this is to reduce H by multiplying with a constant
factor, e.g., 0.85. Another is to try different values for H and make a
visual check by eye.
Example:
data = rndnorm(0, 1,20,1)
h = hns(data,'epan')
See also:
---------
hste, hbcv, hboot, hos, hldpi, hlscv, hscv, hstt, kde
Reference:
---------
B. W. Silverman (1986)
'Density estimation for statistics and data analysis'
Chapman and Hall, pp 43-48
Wand,M.P. and Jones, M.C. (1995)
'Kernel smoothing'
Chapman and Hall, pp 60--63
"""
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A = np.atleast_2d(data)
n = A.shape[1]
# R= int(mkernel(x)^2), mu2= int(x^2*mkernel(x))
mu2, R, unusedRdd = self.stats()
AMISEconstant = (8 * sqrt(pi) * R / (3 * mu2 ** 2 * n)) ** (1. / 5)
iqr = iqrange(A, axis=1) # interquartile range
stdA = np.std(A, axis=1, ddof=1)
# use of interquartile range guards against outliers.
# the use of interquartile range is better if
# the distribution is skew or have heavy tails
# This lessen the chance of oversmoothing.
return np.where(iqr > 0,
np.minimum(stdA, iqr / 1.349), stdA) * AMISEconstant
def hos(self, data):
"""Returns Oversmoothing Parameter.
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Parameter
---------
data = data matrix, size N x D (D = # dimensions )
Returns
-------
h : vector size 1 x D
one dimensional maximum smoothing value for smoothing parameter
given the data and kernel.
The oversmoothing or maximal smoothing principle relies on the fact
that there is a simple upper bound for the AMISE-optimal bandwidth for
estimation of densities with a fixed value of a particular scale
measure. While HOS will give too large bandwidth for optimal estimation
of a general density it provides an excellent starting point for
subjective choice of bandwidth. A sensible strategy is to plot an
estimate with bandwidth HOS and then sucessively look at plots based on
convenient fractions of HOS to see what features are present in the
data for various amount of smoothing. The relation to HNS is given by:
HOS = HNS/0.93
Example:
--------
data = rndnorm(0, 1,20,1)
h = hos(data,'epan');
See also hste, hbcv, hboot, hldpi, hlscv, hscv, hstt, kde, kdefun
Reference
---------
B. W. Silverman (1986)
'Density estimation for statistics and data analysis'
Chapman and Hall, pp 43-48
Wand,M.P. and Jones, M.C. (1986)
'Kernel smoothing'
Chapman and Hall, pp 60--63
"""
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return self.hns(data) / 0.93
def hmns(self, data):
"""Returns Multivariate Normal Scale Estimate of Smoothing Parameter.
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CALL: h = hmns(data,kernel)
h = M dimensional optimal value for smoothing parameter
given the data and kernel. size D x D
data = data matrix, size D x N (D = # dimensions )
kernel = 'epanechnikov' - Epanechnikov kernel.
'biweight' - Bi-weight kernel.
'triweight' - Tri-weight kernel.
'gaussian' - Gaussian kernel
Note that only the first 4 letters of the kernel name is needed.
HMNS only gives a optimal value with respect to mean integrated
square error, when the true underlying distribution is Multivariate
Gaussian. This works reasonably well if the data resembles a
Multivariate Gaussian distribution. However if the distribution is
asymmetric, multimodal or have long tails then HNS is maybe more
appropriate.
Example:
data = rndnorm(0, 1,20,2)
h = hmns(data,'epan')
See also
--------
hns, hste, hbcv, hboot, hos, hldpi, hlscv, hscv, hstt
Reference
----------
B. W. Silverman (1986)
'Density estimation for statistics and data analysis'
Chapman and Hall, pp 43-48, 87
Wand,M.P. and Jones, M.C. (1995)
'Kernel smoothing'
Chapman and Hall, pp 60--63, 86--88
"""
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# TODO: implement more kernels
A = np.atleast_2d(data)
d, n = A.shape
if d == 1:
return self.hns(data)
name = self.name[:4].lower()
if name == 'epan': # Epanechnikov kernel
a = (8.0 * (d + 4.0) * (2 * sqrt(pi)) ** d /
sphere_volume(d)) ** (1. / (4.0 + d))
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elif name == 'biwe': # Bi-weight kernel
a = 2.7779
if d > 2:
raise ValueError('not implemented for d>2')
elif name == 'triw': # Triweight
a = 3.12
if d > 2:
raise ValueError('not implemented for d>2')
elif name == 'gaus': # Gaussian kernel
a = (4.0 / (d + 2.0)) ** (1. / (d + 4.0))
else:
raise ValueError('Unknown kernel.')
covA = scipy.cov(A)
return a * linalg.sqrtm(covA).real * n ** (-1. / (d + 4))
def hste(self, data, h0=None, inc=128, maxit=100, releps=0.01, abseps=0.0):
'''HSTE 2-Stage Solve the Equation estimate of smoothing parameter.
CALL: hs = hste(data,kernel,h0)
hs = one dimensional value for smoothing parameter
given the data and kernel. size 1 x D
data = data matrix, size N x D (D = # dimensions )
kernel = 'gaussian' - Gaussian kernel (default)
( currently the only supported kernel)
h0 = initial starting guess for hs (default h0=hns(A,kernel))
Example:
x = rndnorm(0,1,50,1);
hs = hste(x,'gauss');
See also hbcv, hboot, hos, hldpi, hlscv, hscv, hstt, kde, kdefun
Reference
---------
B. W. Silverman (1986)
'Density estimation for statistics and data analysis'
Chapman and Hall, pp 57--61
Wand,M.P. and Jones, M.C. (1986)
'Kernel smoothing'
Chapman and Hall, pp 74--75
'''
# TODO: NB: this routine can be made faster:
# TODO: replace the iteration in the end with a Newton Raphson scheme
A = np.atleast_2d(data)
d, n = A.shape
# R= int(mkernel(x)^2), mu2= int(x^2*mkernel(x))
mu2, R, unusedRdd = self.stats()
AMISEconstant = (8 * sqrt(pi) * R / (3 * mu2 ** 2 * n)) ** (1. / 5)
STEconstant = R / (mu2 ** (2) * n)
sigmaA = self.hns(A) / AMISEconstant
if h0 is None:
h0 = sigmaA * AMISEconstant
h = np.asarray(h0, dtype=float)
nfft = inc * 2
amin = A.min(axis=1) # Find the minimum value of A.
amax = A.max(axis=1) # Find the maximum value of A.
arange = amax - amin # Find the range of A.
# xa holds the x 'axis' vector, defining a grid of x values where
# the k.d. function will be evaluated.
11 years ago
ax1 = amin - arange / 8.0
bx1 = amax + arange / 8.0
kernel2 = Kernel('gauss')
mu2, R, unusedRdd = kernel2.stats()
STEconstant2 = R / (mu2 ** (2) * n)
fft = np.fft.fft
ifft = np.fft.ifft
for dim in range(d):
s = sigmaA[dim]
ax = ax1[dim]
bx = bx1[dim]
xa = np.linspace(ax, bx, inc)
xn = np.linspace(0, bx - ax, inc)
c = gridcount(A[dim], xa)
# Step 1
psi6NS = -15 / (16 * sqrt(pi) * s ** 7)
psi8NS = 105 / (32 * sqrt(pi) * s ** 9)
# Step 2
k40, k60 = kernel2.deriv4_6_8_10(0, numout=2)
g1 = (-2 * k40 / (mu2 * psi6NS * n)) ** (1.0 / 7)
g2 = (-2 * k60 / (mu2 * psi8NS * n)) ** (1.0 / 9)
# Estimate psi6 given g2.
# kernel weights.
kw4, kw6 = kernel2.deriv4_6_8_10(xn / g2, numout=2)
# Apply fftshift to kw.
kw = np.r_[kw6, 0, kw6[-1:0:-1]]
z = np.real(ifft(fft(c, nfft) * fft(kw))) # convolution.
psi6 = np.sum(c * z[:inc]) / (n * (n - 1) * g2 ** 7)
# Estimate psi4 given g1.
kw4 = kernel2.deriv4_6_8_10(xn / g1, numout=1) # kernel weights.
kw = np.r_[kw4, 0, kw4[-1:0:-1]] # Apply 'fftshift' to kw.
z = np.real(ifft(fft(c, nfft) * fft(kw))) # convolution.
psi4 = np.sum(c * z[:inc]) / (n * (n - 1) * g1 ** 5)
h1 = h[dim]
h_old = 0
count = 0
while ((abs(h_old - h1) > max(releps * h1, abseps)) and
(count < maxit)):
count += 1
h_old = h1
# Step 3
gamma = ((2 * k40 * mu2 * psi4 * h1 ** 5) /
(-psi6 * R)) ** (1.0 / 7)
# Now estimate psi4 given gamma.
# kernel weights.
11 years ago
kw4 = kernel2.deriv4_6_8_10(xn / gamma, numout=1)
kw = np.r_[kw4, 0, kw4[-1:0:-1]] # Apply 'fftshift' to kw.
z = np.real(ifft(fft(c, nfft) * fft(kw))) # convolution.
psi4Gamma = np.sum(c * z[:inc]) / (n * (n - 1) * gamma ** 5)
# Step 4
h1 = (STEconstant2 / psi4Gamma) ** (1.0 / 5)
# Kernel other than Gaussian scale bandwidth
h1 = h1 * (STEconstant / STEconstant2) ** (1.0 / 5)
if count >= maxit:
warnings.warn('The obtained value did not converge.')
h[dim] = h1
# end for dim loop
11 years ago
return h
def hisj(self, data, inc=512, L=7):
'''
HISJ Improved Sheather-Jones estimate of smoothing parameter.
Unlike many other implementations, this one is immune to problems
caused by multimodal densities with widely separated modes. The
estimation does not deteriorate for multimodal densities, because
it do not assume a parametric model for the data.
Parameters
----------
data - a vector of data from which the density estimate is constructed
inc - the number of mesh points used in the uniform discretization
Returns
-------
bandwidth - the optimal bandwidth
Reference
---------
Kernel density estimation via diffusion
Z. I. Botev, J. F. Grotowski, and D. P. Kroese (2010)
Annals of Statistics, Volume 38, Number 5, pages 2916-2957.
'''
A = np.atleast_2d(data)
d, n = A.shape
# R= int(mkernel(x)^2), mu2= int(x^2*mkernel(x))
mu2, R, unusedRdd = self.stats()
STEconstant = R / (n * mu2 ** 2)
amin = A.min(axis=1) # Find the minimum value of A.
amax = A.max(axis=1) # Find the maximum value of A.
arange = amax - amin # Find the range of A.
# xa holds the x 'axis' vector, defining a grid of x values where
# the k.d. function will be evaluated.
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ax1 = amin - arange / 8.0
bx1 = amax + arange / 8.0
kernel2 = Kernel('gauss')
mu2, R, unusedRdd = kernel2.stats()
STEconstant2 = R / (mu2 ** (2) * n)
def fixed_point(t, N, I, a2):
''' this implements the function t-zeta*gamma^[L](t)'''
prod = np.prod
# L = 7
11 years ago
logI = np.log(I)
f = 2 * pi ** (2 * L) * \
(a2 * exp(L * logI - I * pi ** 2 * t)).sum()
11 years ago
for s in range(L - 1, 1, -1):
K0 = prod(np.r_[1:2 * s:2]) / sqrt(2 * pi)
const = (1 + (1. / 2) ** (s + 1. / 2)) / 3
time = (2 * const * K0 / N / f) ** (2. / (3 + 2 * s))
f = 2 * pi ** (2 * s) * \
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(a2 * exp(s * logI - I * pi ** 2 * time)).sum()
return t - (2 * N * sqrt(pi) * f) ** (-2. / 5)
h = np.empty(d)
for dim in range(d):
ax = ax1[dim]
bx = bx1[dim]
xa = np.linspace(ax, bx, inc)
R = bx - ax
c = gridcount(A[dim], xa)
N = len(set(A[dim]))
# a = dct(c/c.sum(), norm=None)
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a = dct(c / len(A[dim]), norm=None)
# now compute the optimal bandwidth^2 using the referenced method
I = np.asfarray(np.arange(1, inc)) ** 2
a2 = (a[1:] / 2) ** 2
def fun(t):
return fixed_point(t, N, I, a2)
11 years ago
x = np.linspace(0, 0.1, 150)
ai = x[0]
f0 = fun(ai)
for bi in x[1:]:
f1 = fun(bi)
if f1 * f0 <= 0:
# print('ai = %g, bi = %g' % (ai,bi))
11 years ago
break
else:
ai = bi
# y = np.asarray([fun(j) for j in x])
11 years ago
# plt.figure(1)
# plt.plot(x,y)
# plt.show()
# use fzero to solve the equation t=zeta*gamma^[5](t)
try:
t_star = optimize.brentq(fun, a=ai, b=bi)
except:
t_star = 0.28 * N ** (-2. / 5)
warnings.warn('Failure in obtaining smoothing parameter')
# smooth the discrete cosine transform of initial data using t_star
# a_t = a*exp(-np.arange(inc)**2*pi**2*t_star/2)
# now apply the inverse discrete cosine transform
# density = idct(a_t)/R;
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# take the rescaling of the data into account
bandwidth = sqrt(t_star) * R
# Kernel other than Gaussian scale bandwidth
h[dim] = bandwidth * (STEconstant / STEconstant2) ** (1.0 / 5)
# end for dim loop
11 years ago
return h
def hstt(self, data, h0=None, inc=128, maxit=100, releps=0.01, abseps=0.0):
'''HSTT Scott-Tapia-Thompson estimate of smoothing parameter.
CALL: hs = hstt(data,kernel)
hs = one dimensional value for smoothing parameter
given the data and kernel. size 1 x D
data = data matrix, size N x D (D = # dimensions )
kernel = 'epanechnikov' - Epanechnikov kernel. (default)
'biweight' - Bi-weight kernel.
'triweight' - Tri-weight kernel.
'triangular' - Triangular kernel.
'gaussian' - Gaussian kernel
'rectangular' - Rectangular kernel.
'laplace' - Laplace kernel.
'logistic' - Logistic kernel.
HSTT returns Scott-Tapia-Thompson (STT) estimate of smoothing
parameter. This is a Solve-The-Equation rule (STE).
Simulation studies shows that the STT estimate of HS
is a good choice under a variety of models. A comparison with
likelihood cross-validation (LCV) indicates that LCV performs slightly
better for short tailed densities.
However, STT method in contrast to LCV is insensitive to outliers.
Example
-------
x = rndnorm(0,1,50,1);
hs = hstt(x,'gauss');
See also
--------
hste, hbcv, hboot, hos, hldpi, hlscv, hscv, kde, kdebin
Reference
---------
B. W. Silverman (1986)
'Density estimation for statistics and data analysis'
Chapman and Hall, pp 57--61
'''
A = np.atleast_2d(data)
d, n = A.shape
# R= int(mkernel(x)^2), mu2= int(x^2*mkernel(x))
mu2, R, unusedRdd = self.stats()
AMISEconstant = (8 * sqrt(pi) * R / (3 * mu2 ** 2 * n)) ** (1. / 5)
STEconstant = R / (mu2 ** (2) * n)
sigmaA = self.hns(A) / AMISEconstant
if h0 is None:
h0 = sigmaA * AMISEconstant
h = np.asarray(h0, dtype=float)
nfft = inc * 2
amin = A.min(axis=1) # Find the minimum value of A.
amax = A.max(axis=1) # Find the maximum value of A.
arange = amax - amin # Find the range of A.
# xa holds the x 'axis' vector, defining a grid of x values where
# the k.d. function will be evaluated.
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ax1 = amin - arange / 8.0
bx1 = amax + arange / 8.0
fft = np.fft.fft
ifft = np.fft.ifft
for dim in range(d):
s = sigmaA[dim]
datan = A[dim] / s
ax = ax1[dim] / s
bx = bx1[dim] / s
xa = np.linspace(ax, bx, inc)
xn = np.linspace(0, bx - ax, inc)
c = gridcount(datan, xa)
count = 1
h_old = 0
h1 = h[dim] / s
delta = (bx - ax) / (inc - 1)
while ((abs(h_old - h1) > max(releps * h1, abseps)) and
(count < maxit)):
count += 1
h_old = h1
kw4 = self.kernel(xn / h1) / (n * h1 * self.norm_factor(d=1))
kw = np.r_[kw4, 0, kw4[-1:0:-1]] # Apply 'fftshift' to kw.
f = np.real(ifft(fft(c, nfft) * fft(kw))) # convolution.
# Estimate psi4=R(f'') using simple finite differences and
# quadrature.
ix = np.arange(1, inc - 1)
z = ((f[ix + 1] - 2 * f[ix] + f[ix - 1]) / delta ** 2) ** 2
psi4 = delta * z.sum()
h1 = (STEconstant / psi4) ** (1. / 5)
if count >= maxit:
warnings.warn('The obtained value did not converge.')
h[dim] = h1 * s
# end % for dim loop
return h
def hscv(self, data, hvec=None, inc=128, maxit=100, fulloutput=False):
'''
HSCV Smoothed cross-validation estimate of smoothing parameter.
CALL: [hs,hvec,score] = hscv(data,kernel,hvec)
hs = smoothing parameter
hvec = vector defining possible values of hs
(default linspace(0.25*h0,h0,100), h0=0.62)
score = score vector
data = data vector
kernel = 'gaussian' - Gaussian kernel the only supported
Note that only the first 4 letters of the kernel name is needed.
Example:
data = rndnorm(0,1,20,1)
[hs hvec score] = hscv(data,'epan');
plot(hvec,score)
See also hste, hbcv, hboot, hos, hldpi, hlscv, hstt, kde, kdefun
Wand,M.P. and Jones, M.C. (1986)
'Kernel smoothing'
Chapman and Hall, pp 75--79
'''
# TODO: Add support for other kernels than Gaussian
A = np.atleast_2d(data)
d, n = A.shape
# R= int(mkernel(x)^2), mu2= int(x^2*mkernel(x))
mu2, R, unusedRdd = self.stats()
AMISEconstant = (8 * sqrt(pi) * R / (3 * mu2 ** 2 * n)) ** (1. / 5)
STEconstant = R / (mu2 ** (2) * n)
sigmaA = self.hns(A) / AMISEconstant
if hvec is None:
H = AMISEconstant / 0.93
hvec = np.linspace(0.25 * H, H, maxit)
hvec = np.asarray(hvec, dtype=float)
steps = len(hvec)
score = np.zeros(steps)
nfft = inc * 2
amin = A.min(axis=1) # Find the minimum value of A.
amax = A.max(axis=1) # Find the maximum value of A.
arange = amax - amin # Find the range of A.
# xa holds the x 'axis' vector, defining a grid of x values where
# the k.d. function will be evaluated.
11 years ago
ax1 = amin - arange / 8.0
bx1 = amax + arange / 8.0
kernel2 = Kernel('gauss')
mu2, R, unusedRdd = kernel2.stats()
STEconstant2 = R / (mu2 ** (2) * n)
fft = np.fft.fft
ifft = np.fft.ifft
h = np.zeros(d)
hvec = hvec * (STEconstant2 / STEconstant) ** (1. / 5.)
k40, k60, k80, k100 = kernel2.deriv4_6_8_10(0, numout=4)
psi8 = 105 / (32 * sqrt(pi))
psi12 = 3465. / (512 * sqrt(pi))
g1 = (-2. * k60 / (mu2 * psi8 * n)) ** (1. / 9.)
g2 = (-2. * k100 / (mu2 * psi12 * n)) ** (1. / 13.)
for dim in range(d):
s = sigmaA[dim]
ax = ax1[dim] / s
bx = bx1[dim] / s
datan = A[dim] / s
xa = np.linspace(ax, bx, inc)
xn = np.linspace(0, bx - ax, inc)
c = gridcount(datan, xa)
kw4, kw6 = kernel2.deriv4_6_8_10(xn / g1, numout=2)
kw = np.r_[kw6, 0, kw6[-1:0:-1]]
z = np.real(ifft(fft(c, nfft) * fft(kw)))
psi6 = np.sum(c * z[:inc]) / (n ** 2 * g1 ** 7)
kw4, kw6, kw8, kw10 = kernel2.deriv4_6_8_10(xn / g2, numout=4)
kw = np.r_[kw10, 0, kw10[-1:0:-1]]
z = np.real(ifft(fft(c, nfft) * fft(kw)))
psi10 = np.sum(c * z[:inc]) / (n ** 2 * g2 ** 11)
g3 = (-2. * k40 / (mu2 * psi6 * n)) ** (1. / 7.)
g4 = (-2. * k80 / (mu2 * psi10 * n)) ** (1. / 11.)
kw4 = kernel2.deriv4_6_8_10(xn / g3, numout=1)
kw = np.r_[kw4, 0, kw4[-1:0:-1]]
z = np.real(ifft(fft(c, nfft) * fft(kw)))
psi4 = np.sum(c * z[:inc]) / (n ** 2 * g3 ** 5)
kw4, kw6, kw8 = kernel2.deriv4_6_8_10(xn / g3, numout=3)
kw = np.r_[kw8, 0, kw8[-1:0:-1]]
z = np.real(ifft(fft(c, nfft) * fft(kw)))
psi8 = np.sum(c * z[:inc]) / (n ** 2 * g4 ** 9)
const = (441. / (64 * pi)) ** (1. / 18.) * \
(4 * pi) ** (-1. / 5.) * \
psi4 ** (-2. / 5.) * psi8 ** (-1. / 9.)
M = np.atleast_2d(datan)
Y = (M - M.T).ravel()
for i in range(steps):
g = const * n ** (-23. / 45) * hvec[i] ** (-2)
sig1 = sqrt(2 * hvec[i] ** 2 + 2 * g ** 2)
sig2 = sqrt(hvec[i] ** 2 + 2 * g ** 2)
sig3 = sqrt(2 * g ** 2)
term2 = np.sum(kernel2(Y / sig1) / sig1 - 2 * kernel2(
Y / sig2) / sig2 + kernel2(Y / sig3) / sig3)
score[i] = 1. / (n * hvec[i] * 2. * sqrt(pi)) + term2 / n ** 2
idx = score.argmin()
# Kernel other than Gaussian scale bandwidth
h[dim] = hvec[idx] * (STEconstant / STEconstant2) ** (1 / 5)
if idx == 0:
warnings.warn(
'Optimum is probably lower than hs=%g for dim=%d' %
(h[dim] * s, dim))
elif idx == maxit - 1:
warnings.warn(
'Optimum is probably higher than hs=%g for dim=%d' %
(h[dim] * s, dim))
hvec = hvec * (STEconstant / STEconstant2) ** (1 / 5)
if fulloutput:
return h * sigmaA, score, hvec, sigmaA
else:
return h * sigmaA
def hldpi(self, data, L=2, inc=128):
'''HLDPI L-stage Direct Plug-In estimate of smoothing parameter.
CALL: hs = hldpi(data,kernel,L)
hs = one dimensional value for smoothing parameter
given the data and kernel. size 1 x D
data = data matrix, size N x D (D = # dimensions )
kernel = 'epanechnikov' - Epanechnikov kernel.
'biweight' - Bi-weight kernel.
'triweight' - Tri-weight kernel.
'triangluar' - Triangular kernel.
'gaussian' - Gaussian kernel
'rectangular' - Rectanguler kernel.
'laplace' - Laplace kernel.
'logistic' - Logistic kernel.
L = 0,1,2,3,... (default 2)
Note that only the first 4 letters of the kernel name is needed.
Example:
x = rndnorm(0,1,50,1);
hs = hldpi(x,'gauss',1);
See also hste, hbcv, hboot, hos, hlscv, hscv, hstt, kde, kdefun
Wand,M.P. and Jones, M.C. (1995)
'Kernel smoothing'
Chapman and Hall, pp 67--74
'''
A = np.atleast_2d(data)
d, n = A.shape
# R= int(mkernel(x)^2), mu2= int(x^2*mkernel(x))
mu2, R, unusedRdd = self.stats()
AMISEconstant = (8 * sqrt(pi) * R / (3 * n * mu2 ** 2)) ** (1. / 5)
STEconstant = R / (n * mu2 ** 2)
sigmaA = self.hns(A) / AMISEconstant
nfft = inc * 2
amin = A.min(axis=1) # Find the minimum value of A.
amax = A.max(axis=1) # Find the maximum value of A.
arange = amax - amin # Find the range of A.
# xa holds the x 'axis' vector, defining a grid of x values where
# the k.d. function will be evaluated.
11 years ago
ax1 = amin - arange / 8.0
bx1 = amax + arange / 8.0
kernel2 = Kernel('gauss')
mu2, unusedR, unusedRdd = kernel2.stats()
fft = np.fft.fft
ifft = np.fft.ifft
h = np.zeros(d)
for dim in range(d):
s = sigmaA[dim]
datan = A[dim] # / s
ax = ax1[dim] # / s
bx = bx1[dim] # / s
xa = np.linspace(ax, bx, inc)
xn = np.linspace(0, bx - ax, inc)
c = gridcount(datan, xa)
r = 2 * L + 4
rd2 = L + 2
# Eq. 3.7 in Wand and Jones (1995)
PSI_r = (-1) ** (rd2) * np.prod(
np.r_[rd2 + 1:r + 1]) / (sqrt(pi) * (2 * s) ** (r + 1))
PSI = PSI_r
if L > 0:
# High order derivatives of the Gaussian kernel
Kd = kernel2.deriv4_6_8_10(0, numout=L)
# L-stage iterations to estimate PSI_4
for ix in range(L, 0, -1):
gi = (-2 * Kd[ix - 1] /
(mu2 * PSI * n)) ** (1. / (2 * ix + 5))
# Obtain the kernel weights.
KW0 = kernel2.deriv4_6_8_10(xn / gi, numout=ix)
if ix > 1:
KW0 = KW0[-1]
# Apply 'fftshift' to kw.
kw = np.r_[KW0, 0, KW0[inc - 1:0:-1]]
# Perform the convolution.
z = np.real(ifft(fft(c, nfft) * fft(kw)))
PSI = np.sum(c * z[:inc]) / (n ** 2 * gi ** (2 * ix + 3))
# end
# end
h[dim] = (STEconstant / PSI) ** (1. / 5)
return h
def norm_factor(self, d=1, n=None):
return self.kernel.norm_factor(d, n)
def eval_points(self, points):
return self.kernel(np.atleast_2d(points))
__call__ = eval_points
def mkernel(X, kernel):
"""MKERNEL Multivariate Kernel Function.
11 years ago
Paramaters
----------
X : array-like
matrix size d x n (d = # dimensions, n = # evaluation points)
kernel : string
defining kernel
'epanechnikov' - Epanechnikov kernel.
'biweight' - Bi-weight kernel.
'triweight' - Tri-weight kernel.
'p1epanechnikov' - product of 1D Epanechnikov kernel.
'p1biweight' - product of 1D Bi-weight kernel.
'p1triweight' - product of 1D Tri-weight kernel.
'triangular' - Triangular kernel.
'gaussian' - Gaussian kernel
'rectangular' - Rectangular kernel.
'laplace' - Laplace kernel.
'logistic' - Logistic kernel.
Note that only the first 4 letters of the kernel name is needed.
Returns
-------
z : ndarray
kernel function values evaluated at X
See also
--------
kde, kdefun, kdebin
References
----------
B. W. Silverman (1986)
'Density estimation for statistics and data analysis'
Chapman and Hall, pp. 43, 76
Wand, M. P. and Jones, M. C. (1995)
'Density estimation for statistics and data analysis'
Chapman and Hall, pp 31, 103, 175
"""
11 years ago
fun = _MKERNEL_DICT[kernel[:4]]
return fun(np.atleast_2d(X))
def accumsum(accmap, a, size, dtype=None):
if dtype is None:
dtype = a.dtype
size = np.atleast_1d(size)
if len(size) > 1:
binx = accmap[:, 0]
biny = accmap[:, 1]
out = sparse.coo_matrix(
(a.ravel(), (binx, biny)), shape=size, dtype=dtype).tocsr()
else:
binx = accmap.ravel()
zero = np.zeros(len(binx))
out = sparse.coo_matrix(
(a.ravel(), (binx, zero)), shape=(size, 1), dtype=dtype).tocsr()
return out
def accumsum2(accmap, a, size):
return np.bincount(accmap.ravel(), a.ravel(), np.array(size).max())
def accum(accmap, a, func=None, size=None, fill_value=0, dtype=None):
"""An accumulation function similar to Matlab's `accumarray` function.
11 years ago
Parameters
----------
accmap : ndarray
This is the "accumulation map". It maps input (i.e. indices into
`a`) to their destination in the output array. The first `a.ndim`
dimensions of `accmap` must be the same as `a.shape`. That is,
`accmap.shape[:a.ndim]` must equal `a.shape`. For example, if `a`
has shape (15,4), then `accmap.shape[:2]` must equal (15,4). In this
case `accmap[i,j]` gives the index into the output array where
element (i,j) of `a` is to be accumulated. If the output is, say,
a 2D, then `accmap` must have shape (15,4,2). The value in the
last dimension give indices into the output array. If the output is
1D, then the shape of `accmap` can be either (15,4) or (15,4,1)
a : ndarray
The input data to be accumulated.
func : callable or None
The accumulation function. The function will be passed a list
of values from `a` to be accumulated.
If None, numpy.sum is assumed.
size : ndarray or None
The size of the output array. If None, the size will be determined
from `accmap`.
fill_value : scalar
The default value for elements of the output array.
dtype : numpy data type, or None
The data type of the output array. If None, the data type of
`a` is used.
Returns
-------
out : ndarray
The accumulated results.
The shape of `out` is `size` if `size` is given. Otherwise the
shape is determined by the (lexicographically) largest indices of
the output found in `accmap`.
Examples
--------
>>> from numpy import array, prod
>>> a = array([[1,2,3],[4,-1,6],[-1,8,9]])
>>> a
array([[ 1, 2, 3],
[ 4, -1, 6],
[-1, 8, 9]])
>>> # Sum the diagonals.
>>> accmap = array([[0,1,2],[2,0,1],[1,2,0]])
>>> s = accum(accmap, a)
>>> s
array([ 9, 7, 15])
>>> # A 2D output, from sub-arrays with shapes and positions like this:
>>> # [ (2,2) (2,1)]
>>> # [ (1,2) (1,1)]
>>> accmap = array([
... [[0,0],[0,0],[0,1]],
... [[0,0],[0,0],[0,1]],
... [[1,0],[1,0],[1,1]]])
>>> # Accumulate using a product.
>>> accum(accmap, a, func=prod, dtype=float)
array([[ -8., 18.],
[ -8., 9.]])
>>> # Same accmap, but create an array of lists of values.
>>> accum(accmap, a, func=lambda x: x, dtype='O')
array([[[1, 2, 4, -1], [3, 6]],
[[-1, 8], [9]]], dtype=object)
11 years ago
"""
def create_array_of_python_lists(accmap, a, size):
vals = np.empty(size, dtype='O')
for s in product(*[range(k) for k in size]):
vals[s] = []
for s in product(*[range(k) for k in a.shape]):
indx = tuple(accmap[s])
val = a[s]
vals[indx].append(val)
return vals
11 years ago
# Check for bad arguments and handle the defaults.
if accmap.shape[:a.ndim] != a.shape:
raise ValueError(
"The initial dimensions of accmap must be the same as a.shape")
if func is None:
func = np.sum
if dtype is None:
dtype = a.dtype
if accmap.shape == a.shape:
accmap = np.expand_dims(accmap, -1)
adims = tuple(range(a.ndim))
if size is None:
size = 1 + np.squeeze(np.apply_over_axes(np.max, accmap, axes=adims))
size = np.atleast_1d(size)
# Create an array of python lists of values.
vals = create_array_of_python_lists(accmap, a, size)
11 years ago
# Create the output array.
out = np.empty(size, dtype=dtype)
for s in product(*[range(k) for k in size]):
if vals[s] == []:
out[s] = fill_value
else:
out[s] = func(vals[s])
return out
def qlevels(pdf, p=(10, 30, 50, 70, 90, 95, 99, 99.9), x1=None, x2=None):
"""QLEVELS Calculates quantile levels which encloses P% of PDF.
11 years ago
CALL: [ql PL] = qlevels(pdf,PL,x1,x2);
ql = the discrete quantile levels.
pdf = joint point density function matrix or vector
PL = percent level (default [10:20:90 95 99 99.9])
x1,x2 = vectors of the spacing of the variables
(Default unit spacing)
QLEVELS numerically integrates PDF by decreasing height and find the
quantile levels which encloses P% of the distribution. If X1 and
(or) X2 is unspecified it is assumed that dX1 and dX2 is constant.
NB! QLEVELS normalizes the integral of PDF to N/(N+0.001) before
calculating QL in order to reflect the sampling of PDF is finite.
Currently only able to handle 1D and 2D PDF's if dXi is not constant
(i=1,2).
Example
-------
>>> import wafo.stats as ws
>>> x = np.linspace(-8,8,2001);
>>> PL = np.r_[10:90:20, 90, 95, 99, 99.9]
>>> qlevels(ws.norm.pdf(x),p=PL, x1=x);
array([ 0.39591707, 0.37058719, 0.31830968, 0.23402133, 0.10362052,
0.05862129, 0.01449505, 0.00178806])
# compared with the exact values
>>> ws.norm.pdf(ws.norm.ppf((100-PL)/200))
array([ 0.39580488, 0.370399 , 0.31777657, 0.23315878, 0.10313564,
0.05844507, 0.01445974, 0.00177719])
See also
--------
qlevels2, tranproc
"""
11 years ago
norm = 1 # normalize cdf to unity
pdf = np.atleast_1d(pdf)
if any(pdf.ravel() < 0):
raise ValueError(
'This is not a pdf since one or more values of pdf is negative')
fsiz = pdf.shape
fsizmin = min(fsiz)
if fsizmin == 0:
return []
N = np.prod(fsiz)
d = len(fsiz)
if x1 is None or ((x2 is None) and d > 2):
fdfi = pdf.ravel()
else:
if d == 1: # pdf in one dimension
dx22 = np.ones(1)
else: # % pdf in two dimensions
dx2 = np.diff(x2.ravel()) * 0.5
dx22 = np.r_[0, dx2] + np.r_[dx2, 0]
dx1 = np.diff(x1.ravel()) * 0.5
dx11 = np.r_[0, dx1] + np.r_[dx1, 0]
dx1x2 = dx22[:, None] * dx11
fdfi = (pdf * dx1x2).ravel()
p = np.atleast_1d(p)
if np.any((p < 0) | (100 < p)):
raise ValueError('PL must satisfy 0 <= PL <= 100')
p2 = p / 100.0
ind = np.argsort(pdf.ravel()) # sort by height of pdf
ind = ind[::-1]
fi = pdf.flat[ind]
# integration in the order of decreasing height of pdf
Fi = np.cumsum(fdfi[ind])
if norm: # %normalize Fi to make sure int pdf dx1 dx2 approx 1
Fi = Fi / Fi[-1] * N / (N + 1.5e-8)
maxFi = np.max(Fi)
if maxFi > 1:
warnings.warn('this is not a pdf since cdf>1! normalizing')
Fi = Fi / Fi[-1] * N / (N + 1.5e-8)
elif maxFi < .95:
msg = '''The given pdf is too sparsely sampled since cdf<.95.
Thus QL is questionable'''
warnings.warn(msg)
# make sure Fi is strictly increasing by not considering duplicate values
ind, = np.where(np.diff(np.r_[Fi, 1]) > 0)
# calculating the inverse of Fi to find the index
ui = tranproc(Fi[ind], fi[ind], p2)
# to the desired quantile level
# ui=smooth(Fi(ind),fi(ind),1,p2(:),1) % alternative
# res=ui-ui2
if np.any(ui >= max(pdf.ravel())):
warnings.warn('The lowest percent level is too close to 0%')
if np.any(ui <= min(pdf.ravel())):
msg = '''The given pdf is too sparsely sampled or
the highest percent level is too close to 100%'''
warnings.warn(msg)
ui[ui < 0] = 0.0
return ui
def qlevels2(data, p=(10, 30, 50, 70, 90, 95, 99, 99.9), method=1):
"""QLEVELS2 Calculates quantile levels which encloses P% of data.
11 years ago
CALL: [ql PL] = qlevels2(data,PL,method);
ql = the discrete quantile levels, size D X Np
Parameters
----------
data : data matrix, size D x N (D = # of dimensions)
p : percent level vector, length Np (default [10:20:90 95 99 99.9])
method : integer
1 Interpolation so that F(X_(k)) == (k-0.5)/n. (default)
2 Interpolation so that F(X_(k)) == k/(n+1).
3 Based on the empirical distribution.
Returns
-------
QLEVELS2 sort the columns of data in ascending order and find the
quantile levels for each column which encloses P% of the data.
Examples : Finding quantile levels enclosing P% of data:
--------
>>> import wafo.stats as ws
>>> PL = np.r_[10:90:20, 90, 95, 99, 99.9]
>>> xs = ws.norm.rvs(size=2500000)
>>> np.allclose(qlevels2(ws.norm.pdf(xs), p=PL),
... [0.3958, 0.3704, 0.3179, 0.2331, 0.1031, 0.05841, 0.01451, 0.001751],
... rtol=1e-1)
True
11 years ago
# compared with the exact values
>>> ws.norm.pdf(ws.norm.ppf((100-PL)/200))
array([ 0.39580488, 0.370399 , 0.31777657, 0.23315878, 0.10313564,
0.05844507, 0.01445974, 0.00177719])
# Finding the median of xs:
>>> '%2.2f' % np.abs(qlevels2(xs,50)[0])
'0.00'
See also
--------
qlevels
"""
11 years ago
q = 100 - np.atleast_1d(p)
return percentile(data, q, axis=-1, method=method)
_PKDICT = {1: lambda k, w, n: (k - w) / (n - 1),
2: lambda k, w, n: (k - w / 2) / n,
3: lambda k, w, n: k / n,
4: lambda k, w, n: k / (n + 1),
5: lambda k, w, n: (k - w / 3) / (n + 1 / 3),
6: lambda k, w, n: (k - w * 3 / 8) / (n + 1 / 4)}
def _compute_qth_weighted_percentile(a, q, axis, out, method, weights,
overwrite_input):
# normalise weight vector such that sum of the weight vector equals to n
q = np.atleast_1d(q) / 100.0
if (q < 0).any() or (q > 1).any():
raise ValueError("percentile must be in the range [0,100]")
shape0 = a.shape
if axis is None:
sorted_ = a.ravel()
else:
9 years ago
taxes = [i for i in range(a.ndim)]
11 years ago
taxes[-1], taxes[axis] = taxes[axis], taxes[-1]
sorted_ = np.transpose(a, taxes).reshape(-1, shape0[axis])
ind = sorted_.argsort(axis=-1)
if overwrite_input:
sorted_.sort(axis=-1)
else:
sorted_ = np.sort(sorted_, axis=-1)
w = np.atleast_1d(weights)
n = len(w)
w = w * n / w.sum()
# Work on each column separately because of weight vector
m = sorted_.shape[0]
nq = len(q)
y = np.zeros((m, nq))
pk_fun = _PKDICT.get(method, 1)
for i in range(m):
sortedW = w[ind[i]] # rearrange the weight according to ind
k = sortedW.cumsum() # cumulative weight
# different algorithm to compute percentile
pk = pk_fun(k, sortedW, n)
# Interpolation between pk and sorted_ for given value of q
y[i] = np.interp(q, pk, sorted_[i])
if axis is None:
return np.squeeze(y)
else:
shape1 = list(shape0)
shape1[axis], shape1[-1] = shape1[-1], nq
return np.squeeze(np.transpose(y.reshape(shape1), taxes))
# method=1: p(k) = k/(n-1)
# method=2: p(k) = (k+0.5)/n.
# method=3: p(k) = (k+1)/n
# method=4: p(k) = (k+1)/(n+1)
# method=5: p(k) = (k+2/3)/(n+1/3)
# method=6: p(k) = (k+5/8)/(n+1/4)
_KDICT = {1: lambda p, n: p * (n - 1),
2: lambda p, n: p * n - 0.5,
3: lambda p, n: p * n - 1,
4: lambda p, n: p * (n + 1) - 1,
5: lambda p, n: p * (n + 1. / 3) - 2. / 3,
6: lambda p, n: p * (n + 1. / 4) - 5. / 8}
def _compute_qth_percentile(sorted_, q, axis, out, method):
if not np.isscalar(q):
p = [_compute_qth_percentile(sorted_, qi, axis, None, method)
for qi in q]
if out is not None:
out.flat = p
return p
q = q / 100.0
if (q < 0) or (q > 1):
raise ValueError("percentile must be in the range [0,100]")
indexer = [slice(None)] * sorted_.ndim
Nx = sorted_.shape[axis]
k_fun = _KDICT.get(method, 1)
index = np.clip(k_fun(q, Nx), 0, Nx - 1)
i = int(index)
if i == index:
indexer[axis] = slice(i, i + 1)
weights1 = np.array(1)
sumval = 1.0
else:
indexer[axis] = slice(i, i + 2)
j = i + 1
weights1 = np.array([(j - index), (index - i)], float)
wshape = [1] * sorted_.ndim
wshape[axis] = 2
weights1.shape = wshape
sumval = weights1.sum()
# Use add.reduce in both cases to coerce data type as well as
# check and use out array.
return np.add.reduce(sorted_[indexer] * weights1,
axis=axis, out=out) / sumval
def percentile(a, q, axis=None, out=None, overwrite_input=False, method=1,
weights=None):
"""Compute the qth percentile of the data along the specified axis.
11 years ago
Returns the qth percentile of the array elements.
Parameters
----------
a : array_like
Input array or object that can be converted to an array.
q : float in range of [0,100] (or sequence of floats)
percentile to compute which must be between 0 and 100 inclusive
axis : {None, int}, optional
Axis along which the percentiles are computed. The default (axis=None)
is to compute the median along a flattened version of the array.
out : ndarray, optional
Alternative output array in which to place the result. It must
have the same shape and buffer length as the expected output,
but the type (of the output) will be cast if necessary.
overwrite_input : {False, True}, optional
If True, then allow use of memory of input array (a) for
calculations. The input array will be modified by the call to
median. This will save memory when you do not need to preserve
the contents of the input array. Treat the input as undefined,
but it will probably be fully or partially sorted. Default is
False. Note that, if `overwrite_input` is True and the input
is not already an ndarray, an error will be raised.
method : scalar integer
defining the interpolation method. Valid options are
1 : p[k] = k/(n-1). In this case, p[k] = mode[F(x[k])].
This is used by S. (default)
2 : p[k] = (k+0.5)/n. That is a piecewise linear function where
the knots are the values midway through the steps of the
empirical cdf. This is popular amongst hydrologists.
Matlab also uses this formula.
3 : p[k] = (k+1)/n. That is, linear interpolation of the empirical cdf.
4 : p[k] = (k+1)/(n+1). Thus p[k] = E[F(x[k])].
This is used by Minitab and by SPSS.
5 : p[k] = (k+2/3)/(n+1/3). Then p[k] =~ median[F(x[k])].
The resulting quantile estimates are approximately
median-unbiased regardless of the distribution of x.
6 : p[k] = (k+5/8)/(n+1/4). The resulting quantile estimates are
approximately unbiased for the expected order statistics
if x is normally distributed.
Returns
-------
pcntile : ndarray
A new array holding the result (unless `out` is specified, in
which case that array is returned instead). If the input contains
integers, or floats of smaller precision than 64, then the output
data-type is float64. Otherwise, the output data-type is the same
as that of the input.
See Also
--------
mean, median
Notes
-----
Given a vector V of length N, the qth percentile of V is the qth ranked
value in a sorted copy of V. A weighted average of the two nearest
neighbors is used if the normalized ranking does not match q exactly.
The same as the median if q is 0.5; the same as the min if q is 0;
and the same as the max if q is 1
Examples
--------
>>> import wafo.kdetools as wk
>>> a = np.array([[10, 7, 4], [3, 2, 1]])
>>> a
array([[10, 7, 4],
[ 3, 2, 1]])
>>> wk.percentile(a, 50)
3.5
>>> wk.percentile(a, 50, axis=0)
array([ 6.5, 4.5, 2.5])
>>> wk.percentile(a, 50, axis=0, weights=np.ones(2))
array([ 6.5, 4.5, 2.5])
>>> wk.percentile(a, 50, axis=1)
array([ 7., 2.])
>>> wk.percentile(a, 50, axis=1, weights=np.ones(3))
array([ 7., 2.])
>>> m = wk.percentile(a, 50, axis=0)
>>> out = np.zeros_like(m)
>>> wk.percentile(a, 50, axis=0, out=m)
array([ 6.5, 4.5, 2.5])
>>> m
array([ 6.5, 4.5, 2.5])
>>> b = a.copy()
>>> wk.percentile(b, 50, axis=1, overwrite_input=True)
array([ 7., 2.])
>>> assert not np.all(a==b)
>>> b = a.copy()
>>> wk.percentile(b, 50, axis=None, overwrite_input=True)
3.5
>>> np.all(a==b)
True
11 years ago
"""
a = np.asarray(a)
try:
if q == 0:
return a.min(axis=axis, out=out)
elif q == 100:
return a.max(axis=axis, out=out)
except:
pass
if weights is not None:
return _compute_qth_weighted_percentile(a, q, axis, out, method,
weights, overwrite_input)
elif overwrite_input:
if axis is None:
sorted_ = np.sort(a, axis=axis)
11 years ago
else:
a.sort(axis=axis)
sorted_ = a
else:
sorted_ = np.sort(a, axis=axis)
if axis is None:
axis = 0
return _compute_qth_percentile(sorted_, q, axis, out, method)
def iqrange(data, axis=None):
"""Returns the Inter Quartile Range of data.
11 years ago
Parameters
----------
data : array-like
Input array or object that can be converted to an array.
axis : {None, int}, optional
Axis along which the percentiles are computed. The default (axis=None)
is to compute the median along a flattened version of the array.
Returns
-------
r : array-like
abs(np.percentile(data, 75, axis)-np.percentile(data, 25, axis))
Notes
-----
IQRANGE is a robust measure of spread. The use of interquartile range
guards against outliers if the distribution have heavy tails.
Example
-------
>>> a = np.arange(101)
>>> iqrange(a)
50.0
See also
--------
np.std
"""
11 years ago
return np.abs(np.percentile(data, 75, axis=axis) -
np.percentile(data, 25, axis=axis))
def bitget(int_type, offset):
"""Returns the value of the bit at the offset position in int_type.
11 years ago
Example
-------
>>> bitget(5, np.r_[0:4])
array([1, 0, 1, 0])
"""
11 years ago
return np.bitwise_and(int_type, 1 << offset) >> offset
def gridcount(data, X, y=1):
'''
Returns D-dimensional histogram using linear binning.
Parameters
----------
data = column vectors with D-dimensional data, shape D x Nd
X = row vectors defining discretization, shape D x N
Must include the range of the data.
Returns
-------
c = gridcount, shape N x N x ... x N
GRIDCOUNT obtains the grid counts using linear binning.
There are 2 strategies: simple- or linear- binning.
Suppose that an observation occurs at x and that the nearest point
below and above is y and z, respectively. Then simple binning strategy
assigns a unit weight to either y or z, whichever is closer. Linear
binning, on the other hand, assigns the grid point at y with the weight
of (z-x)/(z-y) and the gridpoint at z a weight of (y-x)/(z-y).
In terms of approximation error of using gridcounts as pdf-estimate,
linear binning is significantly more accurate than simple binning.
NOTE: The interval [min(X);max(X)] must include the range of the data.
The order of C is permuted in the same order as
meshgrid for D==2 or D==3.
Example
-------
>>> import numpy as np
>>> import wafo.kdetools as wk
>>> import pylab as plb
9 years ago
>>> N = 20
11 years ago
>>> data = np.random.rayleigh(1,N)
9 years ago
>>> data = np.array(
... [ 1.07855907, 1.51199717, 1.54382893, 1.54774808, 1.51913566,
... 1.11386486, 1.49146216, 1.51127214, 2.61287913, 0.94793051,
... 2.08532731, 1.35510641, 0.56759888, 1.55766981, 0.77883602,
... 0.9135759 , 0.81177855, 1.02111483, 1.76334202, 0.07571454])
11 years ago
>>> x = np.linspace(0,max(data)+1,50)
>>> dx = x[1]-x[0]
9 years ago
>>> c = wk.gridcount(data, x)
>>> np.allclose(c[:5], [ 0., 0.9731147, 0.0268853, 0., 0.])
True
11 years ago
>>> pdf = c/dx/N
9 years ago
>>> np.allclose(np.trapz(pdf, x), 1)
True
h = plb.plot(x,c,'.') # 1D histogram
h1 = plb.plot(x, pdf) # 1D probability density plot
11 years ago
See also
--------
bincount, accum, kdebin
Reference
----------
Wand,M.P. and Jones, M.C. (1995)
'Kernel smoothing'
Chapman and Hall, pp 182-192
'''
dat = np.atleast_2d(data)
x = np.atleast_2d(X)
y = np.atleast_1d(y).ravel()
d = dat.shape[0]
d1, inc = x.shape
if d != d1:
raise ValueError('Dimension 0 of data and X do not match.')
dx = np.diff(x[:, :2], axis=1)
xlo = x[:, 0]
xup = x[:, -1]
datlo = dat.min(axis=1)
datup = dat.max(axis=1)
if ((datlo < xlo) | (xup < datup)).any():
raise ValueError('X does not include whole range of the data!')
csiz = np.repeat(inc, d)
use_sparse = False
if use_sparse:
acfun = accumsum # faster than accum
else:
acfun = accumsum2 # accum
binx = np.asarray(np.floor((dat - xlo[:, newaxis]) / dx), dtype=int)
w = dx.prod()
abs = np.abs # @ReservedAssignment
if d == 1:
x.shape = (-1,)
c = np.asarray((acfun(binx, (x[binx + 1] - dat) * y, size=(inc, )) +
acfun(binx + 1, (dat - x[binx]) * y, size=(inc, ))) /
w).ravel()
else: # d>2
11 years ago
Nc = csiz.prod()
c = np.zeros((Nc,))
fact2 = np.asarray(np.reshape(inc * np.arange(d), (d, -1)), dtype=int)
fact1 = np.asarray(
np.reshape(csiz.cumprod() / inc, (d, -1)), dtype=int)
# fact1 = fact1(ones(n,1),:);
bt0 = [0, 0]
X1 = X.ravel()
for ir in range(2 ** (d - 1)):
11 years ago
bt0[0] = np.reshape(bitget(ir, np.arange(d)), (d, -1))
bt0[1] = 1 - bt0[0]
for ix in range(2):
11 years ago
one = np.mod(ix, 2)
two = np.mod(ix + 1, 2)
# Convert to linear index
# linear index to c
11 years ago
b1 = np.sum((binx + bt0[one]) * fact1, axis=0)
bt2 = bt0[two] + fact2
b2 = binx + bt2 # linear index to X
c += acfun(
b1, abs(np.prod(X1[b2] - dat, axis=0)) * y, size=(Nc,))
c = np.reshape(c / w, csiz, order='F')
T = [i for i in range(d)]
11 years ago
T[1], T[0] = T[0], T[1]
# make sure c is stored in the same way as meshgrid
c = c.transpose(*T)
return c
def kde_demo1():
"""KDEDEMO1 Demonstrate the smoothing parameter impact on KDE.
11 years ago
KDEDEMO1 shows the true density (dotted) compared to KDE based on 7
observations (solid) and their individual kernels (dashed) for 3
different values of the smoothing parameter, hs.
"""
11 years ago
import scipy.stats as st
x = np.linspace(-4, 4, 101)
x0 = x / 2.0
data = np.random.normal(loc=0, scale=1.0, size=7)
kernel = Kernel('gauss')
hs = kernel.hns(data)
hVec = [hs / 2, hs, 2 * hs]
for ix, h in enumerate(hVec):
plt.figure(ix)
kde = KDE(data, hs=h, kernel=kernel)
f2 = kde(x, output='plot', title='h_s = %2.2f' % h, ylab='Density')
f2.plot('k-')
plt.plot(x, st.norm.pdf(x, 0, 1), 'k:')
n = len(data)
plt.plot(data, np.zeros(data.shape), 'bx')
y = kernel(x0) / (n * h * kernel.norm_factor(d=1, n=n))
for i in range(n):
plt.plot(data[i] + x0 * h, y, 'b--')
plt.plot([data[i], data[i]], [0, np.max(y)], 'b')
plt.axis([x.min(), x.max(), 0, 0.5])
def kde_demo2():
'''Demonstrate the difference between transformation- and ordinary-KDE.
KDEDEMO2 shows that the transformation KDE is a better estimate for
Rayleigh distributed data around 0 than the ordinary KDE.
'''
import scipy.stats as st
data = st.rayleigh.rvs(scale=1, size=300)
x = np.linspace(1.5e-2, 5, 55)
kde = KDE(data)
f = kde(output='plot', title='Ordinary KDE (hs=%g)' % kde.hs)
plt.figure(0)
f.plot()
plt.plot(x, st.rayleigh.pdf(x, scale=1), ':')
# plotnorm((data).^(L2)) % gives a straight line => L2 = 0.5 reasonable
tkde = TKDE(data, L2=0.5)
ft = tkde(x, output='plot', title='Transformation KDE (hs=%g)' %
tkde.tkde.hs)
plt.figure(1)
ft.plot()
plt.plot(x, st.rayleigh.pdf(x, scale=1), ':')
plt.figure(0)
def kde_demo3():
'''Demonstrate the difference between transformation and ordinary-KDE in 2D
KDEDEMO3 shows that the transformation KDE is a better estimate for
Rayleigh distributed data around 0 than the ordinary KDE.
'''
import scipy.stats as st
data = st.rayleigh.rvs(scale=1, size=(2, 300))
# x = np.linspace(1.5e-3, 5, 55)
11 years ago
kde = KDE(data)
f = kde(output='plot', title='Ordinary KDE', plotflag=1)
plt.figure(0)
f.plot()
plt.plot(data[0], data[1], '.')
# plotnorm((data).^(L2)) % gives a straight line => L2 = 0.5 reasonable
tkde = TKDE(data, L2=0.5)
ft = tkde.eval_grid_fast(
output='plot', title='Transformation KDE', plotflag=1)
plt.figure(1)
ft.plot()
plt.plot(data[0], data[1], '.')
plt.figure(0)
def kde_demo4(N=50):
'''Demonstrate that the improved Sheather-Jones plug-in (hisj) is superior
for 1D multimodal distributions
KDEDEMO4 shows that the improved Sheather-Jones plug-in smoothing is a
better compared to normal reference rules (in this case the hns)
'''
import scipy.stats as st
data = np.hstack((st.norm.rvs(loc=5, scale=1, size=(N,)),
st.norm.rvs(loc=-5, scale=1, size=(N,))))
# x = np.linspace(1.5e-3, 5, 55)
11 years ago
kde = KDE(data, kernel=Kernel('gauss', 'hns'))
f = kde(output='plot', title='Ordinary KDE', plotflag=1)
kde1 = KDE(data, kernel=Kernel('gauss', 'hisj'))
f1 = kde1(output='plot', label='Ordinary KDE', plotflag=1)
plt.figure(0)
f.plot('r', label='hns=%g' % kde.hs)
# plt.figure(2)
f1.plot('b', label='hisj=%g' % kde1.hs)
x = np.linspace(-4, 4)
for loc in [-5, 5]:
plt.plot(x + loc, st.norm.pdf(x, 0, scale=1) / 2, 'k:',
label='True density')
11 years ago
plt.legend()
def kde_demo5(N=500):
'''Demonstrate that the improved Sheather-Jones plug-in (hisj) is superior
for 2D multimodal distributions
KDEDEMO5 shows that the improved Sheather-Jones plug-in smoothing is better
compared to normal reference rules (in this case the hns)
'''
import scipy.stats as st
data = np.hstack((st.norm.rvs(loc=5, scale=1, size=(2, N,)),
st.norm.rvs(loc=-5, scale=1, size=(2, N,))))
kde = KDE(data, kernel=Kernel('gauss', 'hns'))
f = kde(output='plot', title='Ordinary KDE (hns=%g %g)' %
tuple(kde.hs.tolist()), plotflag=1)
kde1 = KDE(data, kernel=Kernel('gauss', 'hisj'))
f1 = kde1(output='plot', title='Ordinary KDE (hisj=%g %g)' %
tuple(kde1.hs.tolist()), plotflag=1)
plt.figure(0)
plt.clf()
f.plot()
plt.plot(data[0], data[1], '.')
plt.figure(1)
plt.clf()
f1.plot()
plt.plot(data[0], data[1], '.')
def kreg_demo1(hs=None, fast=False, fun='hisj'):
""""""
11 years ago
N = 100
# ei = np.random.normal(loc=0, scale=0.075, size=(N,))
ei = np.array([
-0.08508516, 0.10462496, 0.07694448, -0.03080661, 0.05777525,
0.06096313, -0.16572389, 0.01838912, -0.06251845, -0.09186784,
-0.04304887, -0.13365788, -0.0185279, -0.07289167, 0.02319097,
0.06887854, -0.08938374, -0.15181813, 0.03307712, 0.08523183,
-0.0378058, -0.06312874, 0.01485772, 0.06307944, -0.0632959,
0.18963205, 0.0369126, -0.01485447, 0.04037722, 0.0085057,
-0.06912903, 0.02073998, 0.1174351, 0.17599277, -0.06842139,
0.12587608, 0.07698113, -0.0032394, -0.12045792, -0.03132877,
0.05047314, 0.02013453, 0.04080741, 0.00158392, 0.10237899,
-0.09069682, 0.09242174, -0.15445323, 0.09190278, 0.07138498,
0.03002497, 0.02495252, 0.01286942, 0.06449978, 0.03031802,
0.11754861, -0.02322272, 0.00455867, -0.02132251, 0.09119446,
-0.03210086, -0.06509545, 0.07306443, 0.04330647, 0.078111,
-0.04146907, 0.05705476, 0.02492201, -0.03200572, -0.02859788,
-0.05893749, 0.00089538, 0.0432551, 0.04001474, 0.04888828,
-0.17708392, 0.16478644, 0.1171006, 0.11664846, 0.01410477,
-0.12458953, -0.11692081, 0.0413047, -0.09292439, -0.07042327,
0.14119701, -0.05114335, 0.04994696, -0.09520663, 0.04829406,
-0.01603065, -0.1933216, 0.19352763, 0.11819496, 0.04567619,
-0.08348306, 0.00812816, -0.00908206, 0.14528945, 0.02901065])
11 years ago
x = np.linspace(0, 1, N)
y0 = 2 * np.exp(-x ** 2 / (2 * 0.3 ** 2)) + \
3 * np.exp(-(x - 1) ** 2 / (2 * 0.7 ** 2))
y = y0 + ei
kernel = Kernel('gauss', fun=fun)
hopt = kernel.hisj(x)
kreg = KRegression(
x, y, p=0, hs=hs, kernel=kernel, xmin=-2 * hopt, xmax=1 + 2 * hopt)
if fast:
kreg.__call__ = kreg.eval_grid_fast
f = kreg(output='plot', title='Kernel regression', plotflag=1)
plt.figure(0)
f.plot(label='p=0')
kreg.p = 1
f1 = kreg(output='plot', title='Kernel regression', plotflag=1)
f1.plot(label='p=1')
# print(f1.data)
plt.plot(x, y, '.', label='data')
plt.plot(x, y0, 'k', label='True model')
plt.legend()
plt.show()
print(kreg.tkde.tkde.inv_hs)
print(kreg.tkde.tkde.hs)
_TINY = np.finfo(float).machar.tiny
11 years ago
_REALMIN = np.finfo(float).machar.xmin
_REALMAX = np.finfo(float).machar.xmax
_EPS = np.finfo(float).eps
def _logit(p):
pc = p.clip(min=0, max=1)
return (np.log(pc) - np.log1p(-pc)).clip(min=-40, max=40)
def _logitinv(x):
return 1.0 / (np.exp(-x) + 1)
def _get_data(n=100, symmetric=False, loc1=1.1, scale1=0.6, scale2=1.0):
import scipy.stats as st
# from sg_filter import SavitzkyGolay
11 years ago
dist = st.norm
norm1 = scale2 * (dist.pdf(-loc1, loc=-loc1, scale=scale1) +
dist.pdf(-loc1, loc=loc1, scale=scale1))
def fun1(x):
return ((dist.pdf(x, loc=-loc1, scale=scale1) +
dist.pdf(x, loc=loc1, scale=scale1)) / norm1).clip(max=1.0)
11 years ago
x = np.sort(6 * np.random.rand(n, 1) - 3, axis=0)
y = (fun1(x) > np.random.rand(n, 1)).ravel()
# y = (np.cos(x)>2*np.random.rand(n, 1)-1).ravel()
11 years ago
x = x.ravel()
if symmetric:
xi = np.hstack((x.ravel(), -x.ravel()))
yi = np.hstack((y, y))
i = np.argsort(xi)
x = xi[i]
y = yi[i]
return x, y, fun1
def kreg_demo2(n=100, hs=None, symmetric=False, fun='hisj', plotlog=False):
x, y, fun1 = _get_data(n, symmetric)
kreg_demo3(x, y, fun1, hs=None, fun='hisj', plotlog=False)
def kreg_demo3(x, y, fun1, hs=None, fun='hisj', plotlog=False):
st = stats
alpha = 0.1
z0 = -_invnorm(alpha / 2)
n = x.size
hopt, hs1, hs2 = _get_regression_smooting(x, y, fun='hos')
if hs is None:
hs = hopt
forward = _logit
reverse = _logitinv
# forward = np.log
# reverse = np.exp
11 years ago
xmin, xmax = x.min(), x.max()
ni = max(2 * int((xmax - xmin) / hopt) + 3, 5)
print(ni)
print(xmin, xmax)
sml = hopt * 0.1
xi = np.linspace(xmin - sml, xmax + sml, ni)
xiii = np.linspace(xmin - sml, xmax + sml, 4 * ni + 1)
c = gridcount(x, xi)
if (y == 1).any():
c0 = gridcount(x[y == 1], xi)
11 years ago
else:
c0 = np.zeros(xi.shape)
yi = np.where(c == 0, 0, c0 / c)
kreg = KRegression(x, y, hs=hs, p=0)
fiii = kreg(xiii)
yiii = interpolate.interp1d(xi, yi)(xiii)
fit = fun1(xiii).clip(max=1.0)
df = np.diff(fiii)
eerr = np.abs((yiii - fiii)).std() + 0.5 * (df[:-1] * df[1:] < 0).sum() / n
err = (fiii - fit).std()
f = kreg(
xiii, output='plotobj',
title='%s err=%1.3f,eerr=%1.3f, n=%d, hs=%1.3f, hs1=%1.3f, hs2=%1.3f' %
11 years ago
(fun, err, eerr, n, hs, hs1, hs2), plotflag=1)
# yi[yi==0] = 1.0/(c[c!=0].min()+4)
# yi[yi==1] = 1-1.0/(c[c!=0].min()+4)
# yi[yi==0] = fi[yi==0]
# yi[yi==0] = np.exp(stineman_interp(xi[yi==0], xi[yi>0],np.log(yi[yi>0])))
# yi[yi==0] = fun1(xi[yi==0])
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try:
yi[yi == 0] = yi[yi > 0].min() / sqrt(n)
except:
yi[yi == 0] = 1. / n
yi[yi == 1] = 1 - (1 - yi[yi < 1].max()) / sqrt(n)
logity = forward(yi)
gkreg = KRegression(xi, logity, hs=hs, xmin=xmin - hopt, xmax=xmax + hopt)
fg = gkreg.eval_grid(
xi, output='plotobj', title='Kernel regression', plotflag=1)
sa = (fg.data - logity).std()
sa2 = iqrange(fg.data - logity) / 1.349
# print('sa=%g %g' % (sa, sa2))
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sa = min(sa, sa2)
# plt.figure(1)
# plt.plot(xi, slogity-logity,'r.')
# plt.plot(xi, logity-,'b.')
# plt.plot(xi, fg.data-logity, 'b.')
# plt.show()
# return
fg = gkreg.eval_grid(
xiii, output='plotobj', title='Kernel regression', plotflag=1)
pi = reverse(fg.data)
dx = xi[1] - xi[0]
ckreg = KDE(x, hs=hs)
# ci = ckreg.eval_grid_fast(xi)*n*dx
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ciii = ckreg.eval_grid_fast(xiii) * dx * x.size # n*(1+symmetric)
# sa1 = np.sqrt(1./(ciii*pi*(1-pi)))
# plo3 = reverse(fg.data-z0*sa)
# pup3 = reverse(fg.data+z0*sa)
fg.data = pi
pi = f.data
# ref Casella and Berger (1990) "Statistical inference" pp444
# a = 2*pi + z0**2/(ciii+1e-16)
# b = 2*(1+z0**2/(ciii+1e-16))
# plo2 = ((a-sqrt(a**2-2*pi**2*b))/b).clip(min=0,max=1)
# pup2 = ((a+sqrt(a**2-2*pi**2*b))/b).clip(min=0,max=1)
# Jeffreys intervall a=b=0.5
# st.beta.isf(alpha/2, x+a, n-x+b)
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ab = 0.07 # 0.055
pi1 = pi # fun1(xiii)
pup2 = np.where(pi == 1,
1,
st.beta.isf(alpha / 2,
ciii * pi1 + ab,
ciii * (1 - pi1) + ab))
plo2 = np.where(pi == 0,
0,
st.beta.isf(1 - alpha / 2,
ciii * pi1 + ab,
ciii * (1 - pi1) + ab))
averr = np.trapz(pup2 - plo2, xiii) / \
(xiii[-1] - xiii[0]) + 0.5 * (df[:-1] * df[1:] < 0).sum()
# f2 = kreg_demo4(x, y, hs, hopt)
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# Wilson score
den = 1 + (z0 ** 2. / ciii)
xc = (pi1 + (z0 ** 2) / (2 * ciii)) / den
halfwidth = (z0 * sqrt((pi1 * (1 - pi1) / ciii) +
(z0 ** 2 / (4 * (ciii ** 2))))) / den
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plo = (xc - halfwidth).clip(min=0) # wilson score
pup = (xc + halfwidth).clip(max=1.0) # wilson score
# pup = (pi + z0*np.sqrt(pi*(1-pi)/ciii)).clip(min=0,max=1) # dont use
# plo = (pi - z0*np.sqrt(pi*(1-pi)/ciii)).clip(min=0,max=1)
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# mi = kreg.eval_grid(x)
# sigma = (stineman_interp(x, xiii, pup)-stineman_interp(x, xiii, plo))/4
# aic = np.abs((y-mi)/sigma).std()+ 0.5*(df[:-1]*df[1:]<0).sum()/n
# aic = np.abs((yiii-fiii)/(pup-plo)).std() + \
# 0.5*(df[:-1]*df[1:]<0).sum() + \
# ((yiii-pup).clip(min=0)-(yiii-plo).clip(max=0)).sum()
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k = (df[:-1] * df[1:] < 0).sum() # numpeaks
sigmai = (pup - plo)
aic = (((yiii - fiii) / sigmai) ** 2).sum() + \
2 * k * (k + 1) / np.maximum(ni - k + 1, 1) + \
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np.abs((yiii - pup).clip(min=0) - (yiii - plo).clip(max=0)).sum()
# aic = (((yiii-fiii)/sigmai)**2).sum()+ 2*k*(k+1)/(ni-k+1) + \
# np.abs((yiii-pup).clip(min=0)-(yiii-plo).clip(max=0)).sum()
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# aic = averr + ((yiii-pup).clip(min=0)-(yiii-plo).clip(max=0)).sum()
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fg.plot(label='KReg grid aic=%2.3f' % (aic))
f.plot(label='KReg averr=%2.3f ' % (averr))
labtxt = '%d CI' % (int(100 * (1 - alpha)))
plt.fill_between(xiii, pup, plo, alpha=0.20,
color='r', linestyle='--', label=labtxt)
plt.fill_between(xiii, pup2, plo2, alpha=0.20, color='b',
linestyle=':', label='%d CI2' % (int(100 * (1 - alpha))))
plt.plot(xiii, fun1(xiii), 'r', label='True model')
plt.scatter(xi, yi, label='data')
print('maxp = %g' % (np.nanmax(f.data)))
print('hs = %g' % (kreg.tkde.tkde.hs))
plt.legend()
h = plt.gca()
if plotlog:
plt.setp(h, yscale='log')
# plt.show()
return hs1, hs2
def kreg_demo4(x, y, hs, hopt, alpha=0.05):
st = stats
n = x.size
xmin, xmax = x.min(), x.max()
ni = max(2 * int((xmax - xmin) / hopt) + 3, 5)
sml = hopt * 0.1
xi = np.linspace(xmin - sml, xmax + sml, ni)
xiii = np.linspace(xmin - sml, xmax + sml, 4 * ni + 1)
kreg = KRegression(x, y, hs=hs, p=0)
dx = xi[1] - xi[0]
ciii = kreg.tkde.eval_grid_fast(xiii) * dx * x.size
# ckreg = KDE(x,hs=hs)
# ciiii = ckreg.eval_grid_fast(xiii)*dx* x.size #n*(1+symmetric)
f = kreg(xiii, output='plotobj') # , plot_kwds=dict(plotflag=7))
pi = f.data
# Jeffreys intervall a=b=0.5
# st.beta.isf(alpha/2, x+a, n-x+b)
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ab = 0.07 # 0.5
pi1 = pi
pup = np.where(pi1 == 1, 1, st.beta.isf(
alpha / 2, ciii * pi1 + ab, ciii * (1 - pi1) + ab))
plo = np.where(pi1 == 0, 0, st.beta.isf(
1 - alpha / 2, ciii * pi1 + ab, ciii * (1 - pi1) + ab))
# Wilson score
# z0 = -_invnorm(alpha/2)
# den = 1+(z0**2./ciii);
# xc=(pi1+(z0**2)/(2*ciii))/den;
# halfwidth=(z0*sqrt((pi1*(1-pi1)/ciii)+(z0**2/(4*(ciii**2)))))/den
# plo2 = (xc-halfwidth).clip(min=0) # wilson score
# pup2 = (xc+halfwidth).clip(max=1.0) # wilson score
# f.dataCI = np.vstack((plo,pup)).T
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f.prediction_error_avg = np.trapz(pup - plo, xiii) / (xiii[-1] - xiii[0])
fiii = f.data
c = gridcount(x, xi)
if (y == 1).any():
c0 = gridcount(x[y == 1], xi)
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else:
c0 = np.zeros(xi.shape)
yi = np.where(c == 0, 0, c0 / c)
f.children = [PlotData([plo, pup], xiii, plotmethod='fill_between',
plot_kwds=dict(alpha=0.2, color='r')),
PlotData(yi, xi, plotmethod='scatter',
plot_kwds=dict(color='r', s=5))]
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yiii = interpolate.interp1d(xi, yi)(xiii)
df = np.diff(fiii)
k = (df[:-1] * df[1:] < 0).sum() # numpeaks
sigmai = (pup - plo)
aicc = (((yiii - fiii) / sigmai) ** 2).sum() + \
2 * k * (k + 1) / np.maximum(ni - k + 1, 1) + \
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np.abs((yiii - pup).clip(min=0) - (yiii - plo).clip(max=0)).sum()
f.aicc = aicc
f.labels.title = 'perr=%1.3f,aicc=%1.3f, n=%d, hs=%1.3f' % (
f.prediction_error_avg, aicc, n, hs)
return f
def check_kreg_demo3():
plt.ion()
k = 0
for n in [50, 100, 300, 600, 4000]:
x, y, fun1 = _get_data(
n, symmetric=True, loc1=1.0, scale1=0.6, scale2=1.25)
k0 = k
for fun in ['hste', ]:
hsmax, _hs1, _hs2 = _get_regression_smooting(x, y, fun=fun)
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for hi in np.linspace(hsmax * 0.25, hsmax, 9):
plt.figure(k)
k += 1
unused = kreg_demo3(x, y, fun1, hs=hi, fun=fun, plotlog=False)
# kreg_demo2(n=n,symmetric=True,fun='hste', plotlog=False)
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fig.tile(range(k0, k))
plt.ioff()
plt.show()
def check_kreg_demo4():
plt.ion()
# test_docstrings()
# kde_demo2()
# kreg_demo1(fast=True)
# kde_gauss_demo()
# kreg_demo2(n=120,symmetric=True,fun='hste', plotlog=True)
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k = 0
for _i, n in enumerate([100, 300, 600, 4000]):
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x, y, fun1 = _get_data(
n, symmetric=True, loc1=0.1, scale1=0.6, scale2=0.75)
# k0 = k
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hopt1, _h1, _h2 = _get_regression_smooting(x, y, fun='hos')
hopt2, _h1, _h2 = _get_regression_smooting(x, y, fun='hste')
hopt = sqrt(hopt1 * hopt2)
# hopt = _get_regression_smooting(x,y,fun='hos')[0]
for _j, fun in enumerate(['hste']): # , 'hisj', 'hns', 'hstt'
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hsmax, _hs1, _hs2 = _get_regression_smooting(x, y, fun=fun)
fmax = kreg_demo4(x, y, hsmax + 0.1, hopt)
for hi in np.linspace(hsmax * 0.1, hsmax, 55):
f = kreg_demo4(x, y, hi, hopt)
if f.aicc <= fmax.aicc:
fmax = f
plt.figure(k)
k += 1
fmax.plot()
plt.plot(x, fun1(x), 'r')
# kreg_demo2(n=n,symmetric=True,fun='hste', plotlog=False)
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fig.tile(range(0, k))
plt.ioff()
plt.show()
def check_regression_bin():
plt.ion()
# test_docstrings()
# kde_demo2()
# kreg_demo1(fast=True)
# kde_gauss_demo()
# kreg_demo2(n=120,symmetric=True,fun='hste', plotlog=True)
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k = 0
for _i, n in enumerate([100, 300, 600, 4000]):
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x, y, fun1 = _get_data(
n, symmetric=True, loc1=0.1, scale1=0.6, scale2=0.75)
fbest = regressionbin(x, y, alpha=0.05, color='g', label='Transit_D')
figk = plt.figure(k)
ax = figk.gca()
k += 1
fbest.labels.title = 'N = %d' % n
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fbest.plot(axis=ax)
ax.plot(x, fun1(x), 'r')
ax.legend(frameon=False, markerscale=4)
# ax = plt.gca()
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ax.set_yticklabels(ax.get_yticks() * 100.0)
ax.grid(True)
fig.tile(range(0, k))
plt.ioff()
plt.show()
def check_bkregression():
plt.ion()
k = 0
for _i, n in enumerate([50, 100, 300, 600]):
11 years ago
x, y, fun1 = _get_data(
n, symmetric=True, loc1=0.1, scale1=0.6, scale2=0.75)
bkreg = BKRegression(x, y)
fbest = bkreg.prb_search_best(
hsfun='hste', alpha=0.05, color='g', label='Transit_D')
figk = plt.figure(k)
ax = figk.gca()
k += 1
# fbest.score.plot(axis=ax)
# axsize = ax.axis()
# ax.vlines(fbest.hs,axsize[2]+1,axsize[3])
# ax.set(yscale='log')
fbest.labels.title = 'N = %d' % n
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fbest.plot(axis=ax)
ax.plot(x, fun1(x), 'r')
ax.legend(frameon=False, markerscale=4)
# ax = plt.gca()
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ax.set_yticklabels(ax.get_yticks() * 100.0)
ax.grid(True)
fig.tile(range(0, k))
plt.ioff()
plt.show()
def _get_regression_smooting(x, y, fun='hste'):
hs1 = Kernel('gauss', fun=fun).get_smoothing(x)
# hx = np.median(np.abs(x-np.median(x)))/0.6745*(4.0/(3*n))**0.2
if (y == 1).any():
hs2 = Kernel('gauss', fun=fun).get_smoothing(x[y == 1])
# hy = np.median(np.abs(y-np.mean(y)))/0.6745*(4.0/(3*n))**0.2
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else:
hs2 = 4 * hs1
# hy = 4*hx
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# hy2 = Kernel('gauss', fun=fun).get_smoothing(y)
# kernel = Kernel('gauss',fun=fun)
# hopt = (hs1+2*hs2)/3
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# hopt = (hs1+4*hs2)/5 #kernel.get_smoothing(x)
# hopt = hs2
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hopt = sqrt(hs1 * hs2)
return hopt, hs1, hs2
def empirical_bin_prb(x, y, hopt, color='r'):
"""Returns empirical binomial probabiltity.
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Parameters
----------
x : ndarray
position ve
y : ndarray
binomial response variable (zeros and ones)
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Returns
-------
P(x) : PlotData object
empirical probability
"""
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xmin, xmax = x.min(), x.max()
ni = max(2 * int((xmax - xmin) / hopt) + 3, 5)
sml = hopt # *0.1
xi = np.linspace(xmin - sml, xmax + sml, ni)
c = gridcount(x, xi)
if (y == 1).any():
c0 = gridcount(x[y == 1], xi)
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else:
c0 = np.zeros(xi.shape)
yi = np.where(c == 0, 0, c0 / c)
return PlotData(yi, xi, plotmethod='scatter',
plot_kwds=dict(color=color, s=5))
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def smoothed_bin_prb(x, y, hs, hopt, alpha=0.05, color='r', label='',
bin_prb=None):
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'''
Parameters
----------
x,y
hs : smoothing parameter
hopt : spacing in empirical_bin_prb
alpha : confidence level
color : color of plot object
bin_prb : PlotData object with empirical bin prb
'''
if bin_prb is None:
bin_prb = empirical_bin_prb(x, y, hopt, color)
xi = bin_prb.args
yi = bin_prb.data
ni = len(xi)
dxi = xi[1] - xi[0]
n = x.size
xiii = np.linspace(xi[0], xi[-1], 10 * ni + 1)
kreg = KRegression(x, y, hs=hs, p=0)
# expected number of data in each bin
ciii = kreg.tkde.eval_grid_fast(xiii) * dxi * n
f = kreg(xiii, output='plotobj') # , plot_kwds=dict(plotflag=7))
pi = f.data
st = stats
# Jeffreys intervall a=b=0.5
# st.beta.isf(alpha/2, x+a, n-x+b)
11 years ago
ab = 0.07 # 0.5
pi1 = pi
pup = np.where(pi1 == 1, 1, st.beta.isf(
alpha / 2, ciii * pi1 + ab, ciii * (1 - pi1) + ab))
plo = np.where(pi1 == 0, 0, st.beta.isf(
1 - alpha / 2, ciii * pi1 + ab, ciii * (1 - pi1) + ab))
# Wilson score
# z0 = -_invnorm(alpha/2)
# den = 1+(z0**2./ciii);
# xc=(pi1+(z0**2)/(2*ciii))/den;
# halfwidth=(z0*sqrt((pi1*(1-pi1)/ciii)+(z0**2/(4*(ciii**2)))))/den
# plo2 = (xc-halfwidth).clip(min=0) # wilson score
# pup2 = (xc+halfwidth).clip(max=1.0) # wilson score
# f.dataCI = np.vstack((plo,pup)).T
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f.prediction_error_avg = np.trapz(pup - plo, xiii) / (xiii[-1] - xiii[0])
fiii = f.data
f.plot_kwds['color'] = color
f.plot_kwds['linewidth'] = 2
if label:
f.plot_kwds['label'] = label
f.children = [PlotData([plo, pup], xiii, plotmethod='fill_between',
plot_kwds=dict(alpha=0.2, color=color)),
bin_prb]
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yiii = interpolate.interp1d(xi, yi)(xiii)
df = np.diff(fiii)
k = (df[:-1] * df[1:] < 0).sum() # numpeaks
sigmai = (pup - plo)
aicc = (((yiii - fiii) / sigmai) ** 2).sum() + \
2 * k * (k + 1) / np.maximum(ni - k + 1, 1) + \
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np.abs((yiii - pup).clip(min=0) - (yiii - plo).clip(max=0)).sum()
f.aicc = aicc
f.fun = kreg
f.labels.title = 'perr=%1.3f,aicc=%1.3f, n=%d, hs=%1.3f' % (
f.prediction_error_avg, aicc, n, hs)
return f
def regressionbin(x, y, alpha=0.05, color='r', label=''):
"""Return kernel regression estimate for binomial data.
11 years ago
Parameters
----------
x : arraylike
positions
y : arraylike
of 0 and 1
"""
hopt1, _h1, _h2 = _get_regression_smooting(x, y, fun='hos')
hopt2, _h1, _h2 = _get_regression_smooting(x, y, fun='hste')
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hopt = sqrt(hopt1 * hopt2)
fbest = smoothed_bin_prb(x, y, hopt2 + 0.1, hopt, alpha, color, label)
bin_prb = fbest.children[-1]
for fun in ['hste']: # , 'hisj', 'hns', 'hstt'
hsmax, _hs1, _hs2 = _get_regression_smooting(x, y, fun=fun)
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for hi in np.linspace(hsmax * 0.1, hsmax, 55):
f = smoothed_bin_prb(x, y, hi, hopt, alpha, color, label, bin_prb)
if f.aicc <= fbest.aicc:
fbest = f
# hbest = hi
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return fbest
def kde_gauss_demo(n=50):
"""KDEDEMO Demonstrate the KDEgauss.
11 years ago
KDEDEMO1 shows the true density (dotted) compared to KDE based on 7
observations (solid) and their individual kernels (dashed) for 3
different values of the smoothing parameter, hs.
"""
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st = stats
# x = np.linspace(-4, 4, 101)
# data = np.random.normal(loc=0, scale=1.0, size=n)
# data = np.random.exponential(scale=1.0, size=n)
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# n1 = 128
# I = (np.arange(n1)*pi)**2 *0.01*0.5
# kw = exp(-I)
# plt.plot(idctn(kw))
# return
# dist = st.norm
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dist = st.expon
data = dist.rvs(loc=0, scale=1.0, size=n)
d, _N = np.atleast_2d(data).shape
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if d == 1:
plot_options = [dict(color='red'), dict(
color='green'), dict(color='black')]
else:
plot_options = [dict(colors='red'), dict(colors='green'),
dict(colors='black')]
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plt.figure(1)
kde0 = KDE(data, kernel=Kernel('gauss', 'hste'))
f0 = kde0.eval_grid_fast(output='plot', ylab='Density')
f0.plot(**plot_options[0])
kde1 = TKDE(data, kernel=Kernel('gauss', 'hisj'), L2=.5)
f1 = kde1.eval_grid_fast(output='plot', ylab='Density')
f1.plot(**plot_options[1])
kde2 = KDEgauss(data)
f2 = kde2(output='plot', ylab='Density')
x = f2.args
f2.plot(**plot_options[2])
fmax = dist.pdf(x, 0, 1).max()
if d == 1:
plt.plot(x, dist.pdf(x, 0, 1), 'k:')
plt.axis([x.min(), x.max(), 0, fmax])
plt.show()
print(fmax / f2.data.max())
format_ = ''.join(('%g, ') * d)
format_ = 'hs0=%s hs1=%s hs2=%s' % (format_, format_, format_)
print(format_ % tuple(kde0.hs.tolist() +
kde1.tkde.hs.tolist() + kde2.hs.tolist()))
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print('inc0 = %d, inc1 = %d, inc2 = %d' % (kde0.inc, kde1.inc, kde2.inc))
def test_kde():
data = np.array([
0.75355792, 0.72779194, 0.94149169, 0.07841119, 2.32291887,
1.10419995, 0.77055114, 0.60288273, 1.36883635, 1.74754326,
1.09547561, 1.01671133, 0.73211143, 0.61891719, 0.75903487,
1.8919469, 0.72433808, 1.92973094, 0.44749838, 1.36508452])
x = np.linspace(0.01, max(data.ravel()) + 1, 10)
kde = TKDE(data, hs=0.5, L2=0.5)
_f = kde(x)
# f = array([1.03982714, 0.45839018, 0.39514782, 0.32860602, 0.26433318,
# 0.20717946, 0.15907684, 0.1201074 , 0.08941027, 0.06574882])
_f1 = kde.eval_grid(x)
# array([ 1.03982714, 0.45839018, 0.39514782, 0.32860602, 0.26433318,
# 0.20717946, 0.15907684, 0.1201074 , 0.08941027, 0.06574882])
_f2 = kde.eval_grid_fast(x)
# array([ 1.06437223, 0.46203314, 0.39593137, 0.32781899, 0.26276433,
# 0.20532206, 0.15723498, 0.11843998, 0.08797755, 0. ])
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def test_docstrings():
import doctest
print('Testing docstrings in %s' % __file__)
doctest.testmod(optionflags=doctest.NORMALIZE_WHITESPACE)
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if __name__ == '__main__':
test_docstrings()
# test_kde()
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# check_bkregression()
# check_regression_bin()
# check_kreg_demo3()
# check_kreg_demo4()
# test_smoothn_1d()
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# test_smoothn_2d()
# kde_demo2()
# kreg_demo1(fast=True)
# kde_gauss_demo()
# kreg_demo2(n=120,symmetric=True,fun='hste', plotlog=True)
# plt.show('hold')