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452 lines
14 KiB
Python
452 lines
14 KiB
Python
11 years ago
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"""
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10 years ago
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Additional statistics functions with support for masked arrays.
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11 years ago
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"""
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10 years ago
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# Original author (2007): Pierre GF Gerard-Marchant
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from __future__ import division, print_function, absolute_import
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11 years ago
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__all__ = ['compare_medians_ms',
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'hdquantiles', 'hdmedian', 'hdquantiles_sd',
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'idealfourths',
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'median_cihs','mjci','mquantiles_cimj',
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'rsh',
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'trimmed_mean_ci',]
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10 years ago
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11 years ago
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import numpy as np
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from numpy import float_, int_, ndarray
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import numpy.ma as ma
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from numpy.ma import MaskedArray
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from . import mstats_basic as mstats
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from scipy.stats.distributions import norm, beta, t, binom
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def hdquantiles(data, prob=list([.25,.5,.75]), axis=None, var=False,):
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"""
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Computes quantile estimates with the Harrell-Davis method.
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The quantile estimates are calculated as a weighted linear combination
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of order statistics.
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Parameters
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----------
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data : array_like
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Data array.
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prob : sequence
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Sequence of quantiles to compute.
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axis : int
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Axis along which to compute the quantiles. If None, use a flattened
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array.
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var : boolean
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Whether to return the variance of the estimate.
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Returns
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-------
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hdquantiles : MaskedArray
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A (p,) array of quantiles (if `var` is False), or a (2,p) array of
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quantiles and variances (if `var` is True), where ``p`` is the
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number of quantiles.
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"""
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def _hd_1D(data,prob,var):
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"Computes the HD quantiles for a 1D array. Returns nan for invalid data."
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xsorted = np.squeeze(np.sort(data.compressed().view(ndarray)))
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# Don't use length here, in case we have a numpy scalar
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n = xsorted.size
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10 years ago
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hd = np.empty((2,len(prob)), float_)
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if n < 2:
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hd.flat = np.nan
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if var:
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return hd
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return hd[0]
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10 years ago
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11 years ago
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v = np.arange(n+1) / float(n)
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betacdf = beta.cdf
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for (i,p) in enumerate(prob):
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_w = betacdf(v, (n+1)*p, (n+1)*(1-p))
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w = _w[1:] - _w[:-1]
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hd_mean = np.dot(w, xsorted)
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hd[0,i] = hd_mean
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#
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hd[1,i] = np.dot(w, (xsorted-hd_mean)**2)
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#
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hd[0, prob == 0] = xsorted[0]
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hd[0, prob == 1] = xsorted[-1]
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if var:
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hd[1, prob == 0] = hd[1, prob == 1] = np.nan
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return hd
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return hd[0]
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10 years ago
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# Initialization & checks
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11 years ago
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data = ma.array(data, copy=False, dtype=float_)
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p = np.array(prob, copy=False, ndmin=1)
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# Computes quantiles along axis (or globally)
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if (axis is None) or (data.ndim == 1):
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result = _hd_1D(data, p, var)
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else:
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if data.ndim > 2:
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raise ValueError("Array 'data' must be at most two dimensional, "
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"but got data.ndim = %d" % data.ndim)
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result = ma.apply_along_axis(_hd_1D, axis, data, p, var)
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10 years ago
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return ma.fix_invalid(result, copy=False)
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11 years ago
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def hdmedian(data, axis=-1, var=False):
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"""
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Returns the Harrell-Davis estimate of the median along the given axis.
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Parameters
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----------
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data : ndarray
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Data array.
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axis : int
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Axis along which to compute the quantiles. If None, use a flattened
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array.
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var : boolean
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Whether to return the variance of the estimate.
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"""
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result = hdquantiles(data,[0.5], axis=axis, var=var)
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return result.squeeze()
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def hdquantiles_sd(data, prob=list([.25,.5,.75]), axis=None):
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"""
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The standard error of the Harrell-Davis quantile estimates by jackknife.
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Parameters
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----------
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data : array_like
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Data array.
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prob : sequence
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Sequence of quantiles to compute.
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axis : int
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Axis along which to compute the quantiles. If None, use a flattened
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array.
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Returns
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-------
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hdquantiles_sd : MaskedArray
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Standard error of the Harrell-Davis quantile estimates.
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"""
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def _hdsd_1D(data,prob):
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"Computes the std error for 1D arrays."
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xsorted = np.sort(data.compressed())
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n = len(xsorted)
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#.........
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hdsd = np.empty(len(prob), float_)
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if n < 2:
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hdsd.flat = np.nan
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10 years ago
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11 years ago
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vv = np.arange(n) / float(n-1)
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betacdf = beta.cdf
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10 years ago
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11 years ago
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for (i,p) in enumerate(prob):
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_w = betacdf(vv, (n+1)*p, (n+1)*(1-p))
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w = _w[1:] - _w[:-1]
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mx_ = np.fromiter([np.dot(w,xsorted[np.r_[list(range(0,k)),
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list(range(k+1,n))].astype(int_)])
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for k in range(n)], dtype=float_)
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mx_var = np.array(mx_.var(), copy=False, ndmin=1) * n / float(n-1)
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hdsd[i] = float(n-1) * np.sqrt(np.diag(mx_var).diagonal() / float(n))
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return hdsd
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10 years ago
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# Initialization & checks
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11 years ago
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data = ma.array(data, copy=False, dtype=float_)
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p = np.array(prob, copy=False, ndmin=1)
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# Computes quantiles along axis (or globally)
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if (axis is None):
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result = _hdsd_1D(data, p)
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else:
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if data.ndim > 2:
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10 years ago
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raise ValueError("Array 'data' must be at most two dimensional, "
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"but got data.ndim = %d" % data.ndim)
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11 years ago
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result = ma.apply_along_axis(_hdsd_1D, axis, data, p)
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10 years ago
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return ma.fix_invalid(result, copy=False).ravel()
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11 years ago
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def trimmed_mean_ci(data, limits=(0.2,0.2), inclusive=(True,True),
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alpha=0.05, axis=None):
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"""
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Selected confidence interval of the trimmed mean along the given axis.
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Parameters
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----------
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data : array_like
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Input data.
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limits : {None, tuple}, optional
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None or a two item tuple.
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Tuple of the percentages to cut on each side of the array, with respect
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to the number of unmasked data, as floats between 0. and 1. If ``n``
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is the number of unmasked data before trimming, then
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10 years ago
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(``n * limits[0]``)th smallest data and (``n * limits[1]``)th
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11 years ago
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largest data are masked. The total number of unmasked data after
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10 years ago
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trimming is ``n * (1. - sum(limits))``.
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11 years ago
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The value of one limit can be set to None to indicate an open interval.
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Defaults to (0.2, 0.2).
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inclusive : (2,) tuple of boolean, optional
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If relative==False, tuple indicating whether values exactly equal to
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the absolute limits are allowed.
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If relative==True, tuple indicating whether the number of data being
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masked on each side should be rounded (True) or truncated (False).
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Defaults to (True, True).
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alpha : float, optional
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Confidence level of the intervals.
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Defaults to 0.05.
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axis : int, optional
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Axis along which to cut. If None, uses a flattened version of `data`.
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Defaults to None.
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Returns
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-------
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trimmed_mean_ci : (2,) ndarray
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The lower and upper confidence intervals of the trimmed data.
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"""
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data = ma.array(data, copy=False)
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trimmed = mstats.trimr(data, limits=limits, inclusive=inclusive, axis=axis)
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tmean = trimmed.mean(axis)
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tstde = mstats.trimmed_stde(data,limits=limits,inclusive=inclusive,axis=axis)
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df = trimmed.count(axis) - 1
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tppf = t.ppf(1-alpha/2.,df)
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return np.array((tmean - tppf*tstde, tmean+tppf*tstde))
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def mjci(data, prob=[0.25,0.5,0.75], axis=None):
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"""
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Returns the Maritz-Jarrett estimators of the standard error of selected
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experimental quantiles of the data.
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Parameters
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----------
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data: ndarray
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Data array.
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prob: sequence
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Sequence of quantiles to compute.
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axis : int
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Axis along which to compute the quantiles. If None, use a flattened
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array.
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"""
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def _mjci_1D(data, p):
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data = np.sort(data.compressed())
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n = data.size
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prob = (np.array(p) * n + 0.5).astype(int_)
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betacdf = beta.cdf
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10 years ago
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11 years ago
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mj = np.empty(len(prob), float_)
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x = np.arange(1,n+1, dtype=float_) / n
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y = x - 1./n
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for (i,m) in enumerate(prob):
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(m1,m2) = (m-1, n-m)
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W = betacdf(x,m-1,n-m) - betacdf(y,m-1,n-m)
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C1 = np.dot(W,data)
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C2 = np.dot(W,data**2)
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mj[i] = np.sqrt(C2 - C1**2)
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return mj
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10 years ago
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11 years ago
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data = ma.array(data, copy=False)
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if data.ndim > 2:
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10 years ago
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raise ValueError("Array 'data' must be at most two dimensional, "
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"but got data.ndim = %d" % data.ndim)
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11 years ago
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p = np.array(prob, copy=False, ndmin=1)
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# Computes quantiles along axis (or globally)
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if (axis is None):
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return _mjci_1D(data, p)
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else:
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return ma.apply_along_axis(_mjci_1D, axis, data, p)
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def mquantiles_cimj(data, prob=[0.25,0.50,0.75], alpha=0.05, axis=None):
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"""
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Computes the alpha confidence interval for the selected quantiles of the
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data, with Maritz-Jarrett estimators.
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Parameters
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----------
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data : ndarray
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Data array.
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prob : sequence
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Sequence of quantiles to compute.
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alpha : float
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Confidence level of the intervals.
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axis : integer
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Axis along which to compute the quantiles.
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If None, use a flattened array.
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"""
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alpha = min(alpha, 1-alpha)
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z = norm.ppf(1-alpha/2.)
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xq = mstats.mquantiles(data, prob, alphap=0, betap=0, axis=axis)
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smj = mjci(data, prob, axis=axis)
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return (xq - z * smj, xq + z * smj)
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def median_cihs(data, alpha=0.05, axis=None):
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"""
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Computes the alpha-level confidence interval for the median of the data.
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Uses the Hettmasperger-Sheather method.
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Parameters
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----------
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data : array_like
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Input data. Masked values are discarded. The input should be 1D only,
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or `axis` should be set to None.
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alpha : float
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Confidence level of the intervals.
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axis : integer
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Axis along which to compute the quantiles. If None, use a flattened
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array.
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Returns
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-------
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median_cihs :
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Alpha level confidence interval.
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"""
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def _cihs_1D(data, alpha):
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data = np.sort(data.compressed())
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n = len(data)
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alpha = min(alpha, 1-alpha)
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k = int(binom._ppf(alpha/2., n, 0.5))
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gk = binom.cdf(n-k,n,0.5) - binom.cdf(k-1,n,0.5)
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if gk < 1-alpha:
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k -= 1
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gk = binom.cdf(n-k,n,0.5) - binom.cdf(k-1,n,0.5)
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gkk = binom.cdf(n-k-1,n,0.5) - binom.cdf(k,n,0.5)
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I = (gk - 1 + alpha)/(gk - gkk)
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lambd = (n-k) * I / float(k + (n-2*k)*I)
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lims = (lambd*data[k] + (1-lambd)*data[k-1],
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lambd*data[n-k-1] + (1-lambd)*data[n-k])
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return lims
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data = ma.rray(data, copy=False)
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# Computes quantiles along axis (or globally)
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if (axis is None):
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result = _cihs_1D(data.compressed(), alpha)
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else:
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if data.ndim > 2:
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10 years ago
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raise ValueError("Array 'data' must be at most two dimensional, "
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"but got data.ndim = %d" % data.ndim)
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11 years ago
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result = ma.apply_along_axis(_cihs_1D, axis, data, alpha)
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10 years ago
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return result
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11 years ago
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def compare_medians_ms(group_1, group_2, axis=None):
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"""
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Compares the medians from two independent groups along the given axis.
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The comparison is performed using the McKean-Schrader estimate of the
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standard error of the medians.
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Parameters
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----------
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group_1 : array_like
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First dataset.
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group_2 : array_like
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Second dataset.
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axis : int, optional
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Axis along which the medians are estimated. If None, the arrays are
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flattened. If `axis` is not None, then `group_1` and `group_2`
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should have the same shape.
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Returns
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-------
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compare_medians_ms : {float, ndarray}
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If `axis` is None, then returns a float, otherwise returns a 1-D
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ndarray of floats with a length equal to the length of `group_1`
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along `axis`.
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"""
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(med_1, med_2) = (ma.median(group_1,axis=axis), ma.median(group_2,axis=axis))
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(std_1, std_2) = (mstats.stde_median(group_1, axis=axis),
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mstats.stde_median(group_2, axis=axis))
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W = np.abs(med_1 - med_2) / ma.sqrt(std_1**2 + std_2**2)
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return 1 - norm.cdf(W)
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def idealfourths(data, axis=None):
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"""
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Returns an estimate of the lower and upper quartiles.
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Uses the ideal fourths algorithm.
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Parameters
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----------
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data : array_like
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Input array.
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axis : int, optional
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Axis along which the quartiles are estimated. If None, the arrays are
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flattened.
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Returns
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-------
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idealfourths : {list of floats, masked array}
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Returns the two internal values that divide `data` into four parts
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using the ideal fourths algorithm either along the flattened array
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(if `axis` is None) or along `axis` of `data`.
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"""
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def _idf(data):
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x = data.compressed()
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n = len(x)
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if n < 3:
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return [np.nan,np.nan]
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(j,h) = divmod(n/4. + 5/12.,1)
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j = int(j)
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qlo = (1-h)*x[j-1] + h*x[j]
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k = n - j
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qup = (1-h)*x[k] + h*x[k-1]
|
||
|
return [qlo, qup]
|
||
|
data = ma.sort(data, axis=axis).view(MaskedArray)
|
||
|
if (axis is None):
|
||
|
return _idf(data)
|
||
|
else:
|
||
|
return ma.apply_along_axis(_idf, axis, data)
|
||
|
|
||
|
|
||
|
def rsh(data, points=None):
|
||
|
"""
|
||
|
Evaluates Rosenblatt's shifted histogram estimators for each point
|
||
|
on the dataset 'data'.
|
||
|
|
||
|
Parameters
|
||
|
----------
|
||
|
data : sequence
|
||
|
Input data. Masked values are ignored.
|
||
|
points : sequence
|
||
|
Sequence of points where to evaluate Rosenblatt shifted histogram.
|
||
|
If None, use the data.
|
||
|
|
||
|
"""
|
||
|
data = ma.array(data, copy=False)
|
||
|
if points is None:
|
||
|
points = data
|
||
|
else:
|
||
|
points = np.array(points, copy=False, ndmin=1)
|
||
10 years ago
|
|
||
11 years ago
|
if data.ndim != 1:
|
||
|
raise AttributeError("The input array should be 1D only !")
|
||
10 years ago
|
|
||
11 years ago
|
n = data.count()
|
||
|
r = idealfourths(data, axis=None)
|
||
|
h = 1.2 * (r[-1]-r[0]) / n**(1./5)
|
||
|
nhi = (data[:,None] <= points[None,:] + h).sum(0)
|
||
|
nlo = (data[:,None] < points[None,:] - h).sum(0)
|
||
|
return (nhi-nlo) / (2.*n*h)
|