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@ -16,11 +16,12 @@ from numpy import pi, sqrt, atleast_2d, exp, newaxis #@UnresolvedImport
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from scipy import interpolate, linalg
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from scipy.special import gamma
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from wafo.misc import meshgrid
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from wafo.wafodata import WafoData
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import copy
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import numpy as np
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import scipy
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import warnings
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from wafo.wafodata import WafoData
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import pylab
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_stats_epan = (1. / 5, 3. / 5, np.inf)
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@ -32,7 +33,7 @@ _stats_lapl = (2, 1. / 4, np.inf)
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_stats_logi = (pi ** 2 / 3, 1. / 6, 1 / 42)
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_stats_gaus = (1, 1. / (2 * sqrt(pi)), 3. / (8 * sqrt(pi)))
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__all__ =['sphere_volume','TKDE', 'KDE', 'Kernel', 'accum', 'qlevels',
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__all__ = ['sphere_volume', 'TKDE', 'KDE', 'Kernel', 'accum', 'qlevels',
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'iqrange', 'gridcount', 'kde_demo1', 'kde_demo2']
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def sphere_volume(d, r=1.0):
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"""
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@ -191,7 +192,7 @@ class _KDE(object):
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args = args[0]
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elif self.d > 1:
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PL = np.r_[10:90:20, 95, 99, 99.9]
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ql = qlevels(f,p=PL)
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ql = qlevels(f, p=PL)
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kwds2.setdefault('levels', ql)
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return WafoData(f, args, **kwds2)
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@ -1842,86 +1843,114 @@ def qlevels(pdf, p=(10, 30, 50, 70, 90, 95, 99, 99.9), x1=None, x2=None):
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return ui
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#def qlevels2(r, p=(10,30,50,70,90, 95, 99, 99.9), method=1):
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# '''
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# QLEVELS2 Calculates quantile levels which encloses P% of data
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#
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# CALL: [ql PL] = qlevels2(data,PL,method);
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#
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# ql = the discrete quantile levels, size Np x D
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# data = data matrix, size N x D (D = # of dimensions)
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# PL = percent level vector, length Np (default [10:20:90 95 99 99.9])
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# method = 1 Interpolation so that F(X_(k)) == (k-0.5)/n. (default)
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# 2 Interpolation so that F(X_(k)) == k/(n+1).
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# 3 Based on the empirical distribution.
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#
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# QLEVELS2 sort the columns of data in ascending order and find the
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# quantile levels for each column which encloses P% of the data.
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#
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# Examples : % Finding quantile levels enclosing P% of data:
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# xs = rndnorm(0,1,100000,1);
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# qls = qlevels2(pdfnorm(xs),[10:20:90 95 99 99.9]);
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# % compared with the exact values
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# ql = pdfnorm(invnorm((100-[10:20:90 95 99 99.9])/200));
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#
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# % Finding the median of xs:
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# ql = qlevels2(xs,50);
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#
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# See also qlevels
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# '''
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# [n, d]=size(r);
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# p = np.atleast_1d(p)
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# if np.any((p<0) | (100<p)):
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# ValueError('PL must satisfy 0 <= PL <= 100')
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#
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# if (n==1) && (d>1)
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# r=r(:);
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# n=d;
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# d=1;
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# end
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# if d>1
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# if min(size(p)) > 1
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# error('Not both matrix r and matrix p input')
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# end
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# q = zeros(length(p),d);
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# else
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# q = zeros(size(p));
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# end
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# p = 1-p(:)/100;
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# x = sort(r);
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#
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#
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# if method == 3
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# qq1 = x(ceil(max(1,p*n)),:);
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# qq2 = x(floor(min(p*n+1,n)),:);
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# qq = (qq1+qq2)/2;
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# else
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# x = [x(1,:); x; x(n,:)];
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# if method == 2
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# % This method is from Hjort's "Computer
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# % intensive statistical methods" page 102
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# i = p*(n+1)+1;
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# else % Metod 1
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# i = p*n+1.5;
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# end
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# iu = ceil(i);
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# il = floor(i);
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# d1 = (i-il)*ones(1,d);
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# qq = x(il,:).*(1-d1)+x(iu,:).*d1;
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# end
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#
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# q(:) = qq;
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#
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# return
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def qlevels2(r, p=(10,30,50,70,90, 95, 99, 99.9), method=1):
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'''
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QLEVELS2 Calculates quantile levels which encloses P% of data
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CALL: [ql PL] = qlevels2(data,PL,method);
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ql = the discrete quantile levels, size D X Np
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data = data matrix, size D x N (D = # of dimensions)
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PL = percent level vector, length Np (default [10:20:90 95 99 99.9])
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method = 1 Interpolation so that F(X_(k)) == (k-0.5)/n. (default)
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2 Interpolation so that F(X_(k)) == k/(n+1).
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3 Based on the empirical distribution.
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QLEVELS2 sort the columns of data in ascending order and find the
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quantile levels for each column which encloses P% of the data.
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Examples : % Finding quantile levels enclosing P% of data:
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--------
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>>> import wafo.stats as ws
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>>> PL = np.r_[10:90:20, 90, 95, 99, 99.9]
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>>> xs = ws.norm.rvs(size=2000000)
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>>> np.round(qlevels2(ws.norm.pdf(xs), p=PL), decimals=3)
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array([ 0.396, 0.37 , 0.318, 0.233, 0.103, 0.058, 0.014, 0.002])
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# compared with the exact values
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>>> ws.norm.pdf(ws.norm.ppf((100-PL)/200))
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array([ 0.39580488, 0.370399 , 0.31777657, 0.23315878, 0.10313564,
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0.05844507, 0.01445974, 0.00177719])
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# Finding the median of xs:
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>>> np.round(qlevels2(xs,50), decimals=2)
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array([ 0.])
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See also
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--------
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qlevels
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'''
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q = 100-np.atleast_1d(p)
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return percentile(r, q, axis=-1, method=method)
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_PKDICT = {1: lambda k, w, n: (k - w) / (n - 1),
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2: lambda k, w, n: (k - w / 2) / n,
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3: lambda k, w, n: k / n,
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4: lambda k, w, n: k / (n + 1),
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5: lambda k, w, n: (k - w / 3) / (n + 1 / 3),
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6: lambda k, w, n: (k - w * 3 / 8) / (n + 1 / 4)}
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def _compute_qth_weighted_percentile(a, q, axis, out, method, weights, overwrite_input):
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# normalise weight vector such that sum of the weight vector equals to n
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q = np.atleast_1d(q) / 100.0
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if (q < 0).any() or (q > 1).any():
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raise ValueError, "percentile must be in the range [0,100]"
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shape0 = a.shape
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if axis is None:
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sorted = a.ravel()
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else:
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taxes = range(a.ndim)
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taxes[-1], taxes[axis] = taxes[axis], taxes[-1]
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sorted = np.transpose(a, taxes).reshape(-1, shape0[axis])
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ind = sorted.argsort(axis= -1)
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if overwrite_input:
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sorted.sort(axis= -1)
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else:
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sorted = np.sort(sorted, axis= -1)
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w = np.atleast_1d(weights)
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n = len(w)
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w = w * n / w.sum()
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# Work on each column separately because of weight vector
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m = sorted.shape[0]
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nq = len(q)
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y = np.zeros((m, nq))
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pk_fun = _PKDICT.get(method, 1)
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for i in range(m):
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sortedW = w[ind[i]] # rearrange the weight according to ind
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k = sortedW.cumsum() # cumulative weight
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pk = pk_fun(k, sortedW, n) # different algorithm to compute percentile
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# Interpolation between pk and sorted for given value of q
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y[i] = np.interp(q, pk, sorted[i])
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if axis is None:
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return np.squeeze(y)
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else:
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shape1 = list(shape0)
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shape1[axis], shape1[-1] = shape1[-1], nq
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return np.squeeze(np.transpose(y.reshape(shape1), taxes))
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#method=1: p(k) = k/(n-1)
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#method=2: p(k) = (k+0.5)/n.
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#method=3: p(k) = (k+1)/n
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#method=4: p(k) = (k+1)/(n+1)
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#method=5: p(k) = (k+2/3)/(n+1/3)
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#method=6: p(k) = (k+5/8)/(n+1/4)
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_KDICT = {1:lambda p, n: p * (n - 1),
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2:lambda p, n: p * n - 0.5,
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3:lambda p, n: p * n - 1,
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4:lambda p, n: p * (n + 1) - 1,
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5:lambda p, n: p * (n + 1. / 3) - 2. / 3,
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6:lambda p, n: p * (n + 1. / 4) - 5. / 8}
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def _compute_qth_percentile(sorted, q, axis, out, method):
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if not np.isscalar(q):
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p = [_compute_qth_percentile(sorted, qi, axis, None, method)
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p = [_compute_qth_percentile(sorted, qi, axis, None, method)
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for qi in q]
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if out is not None:
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out.flat = p
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return p
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q = q / 100.0
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@ -1930,38 +1959,29 @@ def _compute_qth_percentile(sorted, q, axis, out, method):
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indexer = [slice(None)] * sorted.ndim
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Nx = sorted.shape[axis]
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if method==1:
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index = q*(Nx-1) # p(k) = k/n
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elif method==2:
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index = q*(Nx-1) + 0.5 # p(k) = (k-0.5)/n.
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elif method==3:
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index = q*Nx # p(k) = k/(n+1)
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elif method==4:
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index = q*(Nx-2) + 1 # p(k) = (k-1)/(n-1)
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elif method==5:
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index = q*(Nx-2./3) + 1./3 #p(k) = (k-1/3)/(n+1/3)
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k_fun = _KDICT.get(method, 1)
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index = np.clip(k_fun(q, Nx), 0, Nx - 1)
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i = int(index)
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if i == index:
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indexer[axis] = slice(i, i+1)
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weights = np.array(1)
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indexer[axis] = slice(i, i + 1)
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weights1 = np.array(1)
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sumval = 1.0
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else:
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indexer[axis] = slice(i, i+2)
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indexer[axis] = slice(i, i + 2)
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j = i + 1
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weights = np.array([(j - index), (index - i)],float)
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wshape = [1]*sorted.ndim
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weights1 = np.array([(j - index), (index - i)], float)
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wshape = [1] * sorted.ndim
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wshape[axis] = 2
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weights.shape = wshape
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sumval = weights.sum()
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weights1.shape = wshape
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sumval = weights1.sum()
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# Use add.reduce in both cases to coerce data type as well as
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# check and use out array.
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return np.add.reduce(sorted[indexer]*weights, axis=axis, out=out)/sumval
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# check and use out array.
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return np.add.reduce(sorted[indexer] * weights1, axis=axis, out=out) / sumval
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def percentile(a, q, axis=None, out=None, overwrite_input=False, method=1):
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def percentile(a, q, axis=None, out=None, overwrite_input=False, method=1, weights=None):
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"""
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Compute the qth percentile of the data along the specified axis.
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@ -1980,23 +2000,6 @@ def percentile(a, q, axis=None, out=None, overwrite_input=False, method=1):
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Alternative output array in which to place the result. It must
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have the same shape and buffer length as the expected output,
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but the type (of the output) will be cast if necessary.
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method : scalar integer
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defining the interpolation method. Valid options are
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1 : p(k) = k/n. That is, linear interpolation of the empirical cdf.
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2 : p(k) = (k-0.5)/n. That is a piecewise linear function where
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the knots are the values midway through the steps of the
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empirical cdf. This is popular amongst hydrologists. (default)
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PRCTILE also uses this formula.
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3 : p(k) = k/(n+1). Thus p(k) = E[F(x[k])].
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This is used by Minitab and by SPSS.
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4 : p(k) = (k-1)/(n-1). In this case, p(k) = mode[F(x[k])].
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This is used by S.
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5 : p(k) = (k-1/3)/(n+1/3). Then p(k) =~ median[F(x[k])].
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The resulting quantile estimates are approximately
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median-unbiased regardless of the distribution of x.
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6 : p(k) = (k-3/8)/(n+1/4). The resulting quantile estimates are
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approximately unbiased for the expected order statistics
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if x is normally distributed.
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overwrite_input : {False, True}, optional
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If True, then allow use of memory of input array (a) for
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calculations. The input array will be modified by the call to
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@ -2005,6 +2008,23 @@ def percentile(a, q, axis=None, out=None, overwrite_input=False, method=1):
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but it will probably be fully or partially sorted. Default is
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False. Note that, if `overwrite_input` is True and the input
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is not already an ndarray, an error will be raised.
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method : scalar integer
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defining the interpolation method. Valid options are
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1 : p[k] = k/(n-1). In this case, p[k] = mode[F(x[k])].
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This is used by S. (default)
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2 : p[k] = (k+0.5)/n. That is a piecewise linear function where
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the knots are the values midway through the steps of the
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empirical cdf. This is popular amongst hydrologists.
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Matlab also uses this formula.
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3 : p[k] = (k+1)/n. That is, linear interpolation of the empirical cdf.
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4 : p[k] = (k+1)/(n+1). Thus p[k] = E[F(x[k])].
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This is used by Minitab and by SPSS.
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5 : p[k] = (k+2/3)/(n+1/3). Then p[k] =~ median[F(x[k])].
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The resulting quantile estimates are approximately
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median-unbiased regardless of the distribution of x.
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6 : p[k] = (k+5/8)/(n+1/4). The resulting quantile estimates are
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approximately unbiased for the expected order statistics
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if x is normally distributed.
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Returns
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-------
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@ -2029,40 +2049,48 @@ def percentile(a, q, axis=None, out=None, overwrite_input=False, method=1):
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Examples
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--------
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>>> import wafo.kdetools as wk
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>>> a = np.array([[10, 7, 4], [3, 2, 1]])
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>>> a
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array([[10, 7, 4],
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[ 3, 2, 1]])
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>>> np.percentile(a, 50)
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>>> wk.percentile(a, 50)
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3.5
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>>> np.percentile(a, 50, axis=0)
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>>> wk.percentile(a, 50, axis=0)
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array([ 6.5, 4.5, 2.5])
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>>> wk.percentile(a, 50, axis=0, weights=np.ones(2))
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array([ 6.5, 4.5, 2.5])
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>>> np.percentile(a, 50, axis=1)
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>>> wk.percentile(a, 50, axis=1)
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array([ 7., 2.])
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>>> m = np.percentile(a, 50, axis=0)
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>>> wk.percentile(a, 50, axis=1, weights=np.ones(3))
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array([ 7., 2.])
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>>> m = wk.percentile(a, 50, axis=0)
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>>> out = np.zeros_like(m)
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>>> np.percentile(a, 50, axis=0, out=m)
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>>> wk.percentile(a, 50, axis=0, out=m)
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array([ 6.5, 4.5, 2.5])
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>>> m
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array([ 6.5, 4.5, 2.5])
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>>> b = a.copy()
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>>> np.percentile(b, 50, axis=1, overwrite_input=True)
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>>> wk.percentile(b, 50, axis=1, overwrite_input=True)
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array([ 7., 2.])
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>>> assert not np.all(a==b)
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>>> b = a.copy()
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>>> np.percentile(b, 50, axis=None, overwrite_input=True)
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>>> wk.percentile(b, 50, axis=None, overwrite_input=True)
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3.5
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>>> assert not np.all(a==b)
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"""
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a = np.asarray(a)
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if q == 0:
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return a.min(axis=axis, out=out)
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elif q == 100:
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return a.max(axis=axis, out=out)
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if overwrite_input:
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try:
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if q == 0:
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return a.min(axis=axis, out=out)
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elif q == 100:
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return a.max(axis=axis, out=out)
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except:
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pass
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if weights is not None:
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return _compute_qth_weighted_percentile(a, q, axis, out, method, weights, overwrite_input)
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elif overwrite_input:
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if axis is None:
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sorted = a.ravel()
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sorted.sort()
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Per.Andreas.Brodtkorb
14 years ago