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Vehicle-Anti-Theft-Face-Rec.../venv/Lib/site-packages/scipy/stats/morestats.py

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import math
import warnings
from collections import namedtuple
import numpy as np
from numpy import (isscalar, r_, log, around, unique, asarray,
zeros, arange, sort, amin, amax, any, atleast_1d,
sqrt, ceil, floor, array, compress,
pi, exp, ravel, count_nonzero, sin, cos, arctan2, hypot)
from scipy import optimize
from scipy import special
from . import statlib
from . import stats
from .stats import find_repeats, _contains_nan
from .contingency import chi2_contingency
from . import distributions
from ._distn_infrastructure import rv_generic
from ._hypotests import _get_wilcoxon_distr
__all__ = ['mvsdist',
'bayes_mvs', 'kstat', 'kstatvar', 'probplot', 'ppcc_max', 'ppcc_plot',
'boxcox_llf', 'boxcox', 'boxcox_normmax', 'boxcox_normplot',
'shapiro', 'anderson', 'ansari', 'bartlett', 'levene', 'binom_test',
'fligner', 'mood', 'wilcoxon', 'median_test',
'circmean', 'circvar', 'circstd', 'anderson_ksamp',
'yeojohnson_llf', 'yeojohnson', 'yeojohnson_normmax',
'yeojohnson_normplot'
]
Mean = namedtuple('Mean', ('statistic', 'minmax'))
Variance = namedtuple('Variance', ('statistic', 'minmax'))
Std_dev = namedtuple('Std_dev', ('statistic', 'minmax'))
def bayes_mvs(data, alpha=0.90):
r"""
Bayesian confidence intervals for the mean, var, and std.
Parameters
----------
data : array_like
Input data, if multi-dimensional it is flattened to 1-D by `bayes_mvs`.
Requires 2 or more data points.
alpha : float, optional
Probability that the returned confidence interval contains
the true parameter.
Returns
-------
mean_cntr, var_cntr, std_cntr : tuple
The three results are for the mean, variance and standard deviation,
respectively. Each result is a tuple of the form::
(center, (lower, upper))
with `center` the mean of the conditional pdf of the value given the
data, and `(lower, upper)` a confidence interval, centered on the
median, containing the estimate to a probability ``alpha``.
See Also
--------
mvsdist
Notes
-----
Each tuple of mean, variance, and standard deviation estimates represent
the (center, (lower, upper)) with center the mean of the conditional pdf
of the value given the data and (lower, upper) is a confidence interval
centered on the median, containing the estimate to a probability
``alpha``.
Converts data to 1-D and assumes all data has the same mean and variance.
Uses Jeffrey's prior for variance and std.
Equivalent to ``tuple((x.mean(), x.interval(alpha)) for x in mvsdist(dat))``
References
----------
T.E. Oliphant, "A Bayesian perspective on estimating mean, variance, and
standard-deviation from data", https://scholarsarchive.byu.edu/facpub/278,
2006.
Examples
--------
First a basic example to demonstrate the outputs:
>>> from scipy import stats
>>> data = [6, 9, 12, 7, 8, 8, 13]
>>> mean, var, std = stats.bayes_mvs(data)
>>> mean
Mean(statistic=9.0, minmax=(7.103650222612533, 10.896349777387467))
>>> var
Variance(statistic=10.0, minmax=(3.176724206..., 24.45910382...))
>>> std
Std_dev(statistic=2.9724954732045084, minmax=(1.7823367265645143, 4.945614605014631))
Now we generate some normally distributed random data, and get estimates of
mean and standard deviation with 95% confidence intervals for those
estimates:
>>> n_samples = 100000
>>> data = stats.norm.rvs(size=n_samples)
>>> res_mean, res_var, res_std = stats.bayes_mvs(data, alpha=0.95)
>>> import matplotlib.pyplot as plt
>>> fig = plt.figure()
>>> ax = fig.add_subplot(111)
>>> ax.hist(data, bins=100, density=True, label='Histogram of data')
>>> ax.vlines(res_mean.statistic, 0, 0.5, colors='r', label='Estimated mean')
>>> ax.axvspan(res_mean.minmax[0],res_mean.minmax[1], facecolor='r',
... alpha=0.2, label=r'Estimated mean (95% limits)')
>>> ax.vlines(res_std.statistic, 0, 0.5, colors='g', label='Estimated scale')
>>> ax.axvspan(res_std.minmax[0],res_std.minmax[1], facecolor='g', alpha=0.2,
... label=r'Estimated scale (95% limits)')
>>> ax.legend(fontsize=10)
>>> ax.set_xlim([-4, 4])
>>> ax.set_ylim([0, 0.5])
>>> plt.show()
"""
m, v, s = mvsdist(data)
if alpha >= 1 or alpha <= 0:
raise ValueError("0 < alpha < 1 is required, but alpha=%s was given."
% alpha)
m_res = Mean(m.mean(), m.interval(alpha))
v_res = Variance(v.mean(), v.interval(alpha))
s_res = Std_dev(s.mean(), s.interval(alpha))
return m_res, v_res, s_res
def mvsdist(data):
"""
'Frozen' distributions for mean, variance, and standard deviation of data.
Parameters
----------
data : array_like
Input array. Converted to 1-D using ravel.
Requires 2 or more data-points.
Returns
-------
mdist : "frozen" distribution object
Distribution object representing the mean of the data.
vdist : "frozen" distribution object
Distribution object representing the variance of the data.
sdist : "frozen" distribution object
Distribution object representing the standard deviation of the data.
See Also
--------
bayes_mvs
Notes
-----
The return values from ``bayes_mvs(data)`` is equivalent to
``tuple((x.mean(), x.interval(0.90)) for x in mvsdist(data))``.
In other words, calling ``<dist>.mean()`` and ``<dist>.interval(0.90)``
on the three distribution objects returned from this function will give
the same results that are returned from `bayes_mvs`.
References
----------
T.E. Oliphant, "A Bayesian perspective on estimating mean, variance, and
standard-deviation from data", https://scholarsarchive.byu.edu/facpub/278,
2006.
Examples
--------
>>> from scipy import stats
>>> data = [6, 9, 12, 7, 8, 8, 13]
>>> mean, var, std = stats.mvsdist(data)
We now have frozen distribution objects "mean", "var" and "std" that we can
examine:
>>> mean.mean()
9.0
>>> mean.interval(0.95)
(6.6120585482655692, 11.387941451734431)
>>> mean.std()
1.1952286093343936
"""
x = ravel(data)
n = len(x)
if n < 2:
raise ValueError("Need at least 2 data-points.")
xbar = x.mean()
C = x.var()
if n > 1000: # gaussian approximations for large n
mdist = distributions.norm(loc=xbar, scale=math.sqrt(C / n))
sdist = distributions.norm(loc=math.sqrt(C), scale=math.sqrt(C / (2. * n)))
vdist = distributions.norm(loc=C, scale=math.sqrt(2.0 / n) * C)
else:
nm1 = n - 1
fac = n * C / 2.
val = nm1 / 2.
mdist = distributions.t(nm1, loc=xbar, scale=math.sqrt(C / nm1))
sdist = distributions.gengamma(val, -2, scale=math.sqrt(fac))
vdist = distributions.invgamma(val, scale=fac)
return mdist, vdist, sdist
def kstat(data, n=2):
r"""
Return the nth k-statistic (1<=n<=4 so far).
The nth k-statistic k_n is the unique symmetric unbiased estimator of the
nth cumulant kappa_n.
Parameters
----------
data : array_like
Input array. Note that n-D input gets flattened.
n : int, {1, 2, 3, 4}, optional
Default is equal to 2.
Returns
-------
kstat : float
The nth k-statistic.
See Also
--------
kstatvar: Returns an unbiased estimator of the variance of the k-statistic.
moment: Returns the n-th central moment about the mean for a sample.
Notes
-----
For a sample size n, the first few k-statistics are given by:
.. math::
k_{1} = \mu
k_{2} = \frac{n}{n-1} m_{2}
k_{3} = \frac{ n^{2} } {(n-1) (n-2)} m_{3}
k_{4} = \frac{ n^{2} [(n + 1)m_{4} - 3(n - 1) m^2_{2}]} {(n-1) (n-2) (n-3)}
where :math:`\mu` is the sample mean, :math:`m_2` is the sample
variance, and :math:`m_i` is the i-th sample central moment.
References
----------
http://mathworld.wolfram.com/k-Statistic.html
http://mathworld.wolfram.com/Cumulant.html
Examples
--------
>>> from scipy import stats
>>> rndm = np.random.RandomState(1234)
As sample size increases, n-th moment and n-th k-statistic converge to the
same number (although they aren't identical). In the case of the normal
distribution, they converge to zero.
>>> for n in [2, 3, 4, 5, 6, 7]:
... x = rndm.normal(size=10**n)
... m, k = stats.moment(x, 3), stats.kstat(x, 3)
... print("%.3g %.3g %.3g" % (m, k, m-k))
-0.631 -0.651 0.0194
0.0282 0.0283 -8.49e-05
-0.0454 -0.0454 1.36e-05
7.53e-05 7.53e-05 -2.26e-09
0.00166 0.00166 -4.99e-09
-2.88e-06 -2.88e-06 8.63e-13
"""
if n > 4 or n < 1:
raise ValueError("k-statistics only supported for 1<=n<=4")
n = int(n)
S = np.zeros(n + 1, np.float64)
data = ravel(data)
N = data.size
# raise ValueError on empty input
if N == 0:
raise ValueError("Data input must not be empty")
# on nan input, return nan without warning
if np.isnan(np.sum(data)):
return np.nan
for k in range(1, n + 1):
S[k] = np.sum(data**k, axis=0)
if n == 1:
return S[1] * 1.0/N
elif n == 2:
return (N*S[2] - S[1]**2.0) / (N*(N - 1.0))
elif n == 3:
return (2*S[1]**3 - 3*N*S[1]*S[2] + N*N*S[3]) / (N*(N - 1.0)*(N - 2.0))
elif n == 4:
return ((-6*S[1]**4 + 12*N*S[1]**2 * S[2] - 3*N*(N-1.0)*S[2]**2 -
4*N*(N+1)*S[1]*S[3] + N*N*(N+1)*S[4]) /
(N*(N-1.0)*(N-2.0)*(N-3.0)))
else:
raise ValueError("Should not be here.")
def kstatvar(data, n=2):
r"""
Return an unbiased estimator of the variance of the k-statistic.
See `kstat` for more details of the k-statistic.
Parameters
----------
data : array_like
Input array. Note that n-D input gets flattened.
n : int, {1, 2}, optional
Default is equal to 2.
Returns
-------
kstatvar : float
The nth k-statistic variance.
See Also
--------
kstat: Returns the n-th k-statistic.
moment: Returns the n-th central moment about the mean for a sample.
Notes
-----
The variances of the first few k-statistics are given by:
.. math::
var(k_{1}) = \frac{\kappa^2}{n}
var(k_{2}) = \frac{\kappa^4}{n} + \frac{2\kappa^2_{2}}{n - 1}
var(k_{3}) = \frac{\kappa^6}{n} + \frac{9 \kappa_2 \kappa_4}{n - 1} +
\frac{9 \kappa^2_{3}}{n - 1} +
\frac{6 n \kappa^3_{2}}{(n-1) (n-2)}
var(k_{4}) = \frac{\kappa^8}{n} + \frac{16 \kappa_2 \kappa_6}{n - 1} +
\frac{48 \kappa_{3} \kappa_5}{n - 1} +
\frac{34 \kappa^2_{4}}{n-1} + \frac{72 n \kappa^2_{2} \kappa_4}{(n - 1) (n - 2)} +
\frac{144 n \kappa_{2} \kappa^2_{3}}{(n - 1) (n - 2)} +
\frac{24 (n + 1) n \kappa^4_{2}}{(n - 1) (n - 2) (n - 3)}
"""
data = ravel(data)
N = len(data)
if n == 1:
return kstat(data, n=2) * 1.0/N
elif n == 2:
k2 = kstat(data, n=2)
k4 = kstat(data, n=4)
return (2*N*k2**2 + (N-1)*k4) / (N*(N+1))
else:
raise ValueError("Only n=1 or n=2 supported.")
def _calc_uniform_order_statistic_medians(n):
"""
Approximations of uniform order statistic medians.
Parameters
----------
n : int
Sample size.
Returns
-------
v : 1d float array
Approximations of the order statistic medians.
References
----------
.. [1] James J. Filliben, "The Probability Plot Correlation Coefficient
Test for Normality", Technometrics, Vol. 17, pp. 111-117, 1975.
Examples
--------
Order statistics of the uniform distribution on the unit interval
are marginally distributed according to beta distributions.
The expectations of these order statistic are evenly spaced across
the interval, but the distributions are skewed in a way that
pushes the medians slightly towards the endpoints of the unit interval:
>>> n = 4
>>> k = np.arange(1, n+1)
>>> from scipy.stats import beta
>>> a = k
>>> b = n-k+1
>>> beta.mean(a, b)
array([ 0.2, 0.4, 0.6, 0.8])
>>> beta.median(a, b)
array([ 0.15910358, 0.38572757, 0.61427243, 0.84089642])
The Filliben approximation uses the exact medians of the smallest
and greatest order statistics, and the remaining medians are approximated
by points spread evenly across a sub-interval of the unit interval:
>>> from scipy.morestats import _calc_uniform_order_statistic_medians
>>> _calc_uniform_order_statistic_medians(n)
array([ 0.15910358, 0.38545246, 0.61454754, 0.84089642])
This plot shows the skewed distributions of the order statistics
of a sample of size four from a uniform distribution on the unit interval:
>>> import matplotlib.pyplot as plt
>>> x = np.linspace(0.0, 1.0, num=50, endpoint=True)
>>> pdfs = [beta.pdf(x, a[i], b[i]) for i in range(n)]
>>> plt.figure()
>>> plt.plot(x, pdfs[0], x, pdfs[1], x, pdfs[2], x, pdfs[3])
"""
v = np.zeros(n, dtype=np.float64)
v[-1] = 0.5**(1.0 / n)
v[0] = 1 - v[-1]
i = np.arange(2, n)
v[1:-1] = (i - 0.3175) / (n + 0.365)
return v
def _parse_dist_kw(dist, enforce_subclass=True):
"""Parse `dist` keyword.
Parameters
----------
dist : str or stats.distributions instance.
Several functions take `dist` as a keyword, hence this utility
function.
enforce_subclass : bool, optional
If True (default), `dist` needs to be a
`_distn_infrastructure.rv_generic` instance.
It can sometimes be useful to set this keyword to False, if a function
wants to accept objects that just look somewhat like such an instance
(for example, they have a ``ppf`` method).
"""
if isinstance(dist, rv_generic):
pass
elif isinstance(dist, str):
try:
dist = getattr(distributions, dist)
except AttributeError:
raise ValueError("%s is not a valid distribution name" % dist)
elif enforce_subclass:
msg = ("`dist` should be a stats.distributions instance or a string "
"with the name of such a distribution.")
raise ValueError(msg)
return dist
def _add_axis_labels_title(plot, xlabel, ylabel, title):
"""Helper function to add axes labels and a title to stats plots"""
try:
if hasattr(plot, 'set_title'):
# Matplotlib Axes instance or something that looks like it
plot.set_title(title)
plot.set_xlabel(xlabel)
plot.set_ylabel(ylabel)
else:
# matplotlib.pyplot module
plot.title(title)
plot.xlabel(xlabel)
plot.ylabel(ylabel)
except Exception:
# Not an MPL object or something that looks (enough) like it.
# Don't crash on adding labels or title
pass
def probplot(x, sparams=(), dist='norm', fit=True, plot=None, rvalue=False):
"""
Calculate quantiles for a probability plot, and optionally show the plot.
Generates a probability plot of sample data against the quantiles of a
specified theoretical distribution (the normal distribution by default).
`probplot` optionally calculates a best-fit line for the data and plots the
results using Matplotlib or a given plot function.
Parameters
----------
x : array_like
Sample/response data from which `probplot` creates the plot.
sparams : tuple, optional
Distribution-specific shape parameters (shape parameters plus location
and scale).
dist : str or stats.distributions instance, optional
Distribution or distribution function name. The default is 'norm' for a
normal probability plot. Objects that look enough like a
stats.distributions instance (i.e. they have a ``ppf`` method) are also
accepted.
fit : bool, optional
Fit a least-squares regression (best-fit) line to the sample data if
True (default).
plot : object, optional
If given, plots the quantiles and least squares fit.
`plot` is an object that has to have methods "plot" and "text".
The `matplotlib.pyplot` module or a Matplotlib Axes object can be used,
or a custom object with the same methods.
Default is None, which means that no plot is created.
Returns
-------
(osm, osr) : tuple of ndarrays
Tuple of theoretical quantiles (osm, or order statistic medians) and
ordered responses (osr). `osr` is simply sorted input `x`.
For details on how `osm` is calculated see the Notes section.
(slope, intercept, r) : tuple of floats, optional
Tuple containing the result of the least-squares fit, if that is
performed by `probplot`. `r` is the square root of the coefficient of
determination. If ``fit=False`` and ``plot=None``, this tuple is not
returned.
Notes
-----
Even if `plot` is given, the figure is not shown or saved by `probplot`;
``plt.show()`` or ``plt.savefig('figname.png')`` should be used after
calling `probplot`.
`probplot` generates a probability plot, which should not be confused with
a Q-Q or a P-P plot. Statsmodels has more extensive functionality of this
type, see ``statsmodels.api.ProbPlot``.
The formula used for the theoretical quantiles (horizontal axis of the
probability plot) is Filliben's estimate::
quantiles = dist.ppf(val), for
0.5**(1/n), for i = n
val = (i - 0.3175) / (n + 0.365), for i = 2, ..., n-1
1 - 0.5**(1/n), for i = 1
where ``i`` indicates the i-th ordered value and ``n`` is the total number
of values.
Examples
--------
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
>>> nsample = 100
>>> np.random.seed(7654321)
A t distribution with small degrees of freedom:
>>> ax1 = plt.subplot(221)
>>> x = stats.t.rvs(3, size=nsample)
>>> res = stats.probplot(x, plot=plt)
A t distribution with larger degrees of freedom:
>>> ax2 = plt.subplot(222)
>>> x = stats.t.rvs(25, size=nsample)
>>> res = stats.probplot(x, plot=plt)
A mixture of two normal distributions with broadcasting:
>>> ax3 = plt.subplot(223)
>>> x = stats.norm.rvs(loc=[0,5], scale=[1,1.5],
... size=(nsample//2,2)).ravel()
>>> res = stats.probplot(x, plot=plt)
A standard normal distribution:
>>> ax4 = plt.subplot(224)
>>> x = stats.norm.rvs(loc=0, scale=1, size=nsample)
>>> res = stats.probplot(x, plot=plt)
Produce a new figure with a loggamma distribution, using the ``dist`` and
``sparams`` keywords:
>>> fig = plt.figure()
>>> ax = fig.add_subplot(111)
>>> x = stats.loggamma.rvs(c=2.5, size=500)
>>> res = stats.probplot(x, dist=stats.loggamma, sparams=(2.5,), plot=ax)
>>> ax.set_title("Probplot for loggamma dist with shape parameter 2.5")
Show the results with Matplotlib:
>>> plt.show()
"""
x = np.asarray(x)
_perform_fit = fit or (plot is not None)
if x.size == 0:
if _perform_fit:
return (x, x), (np.nan, np.nan, 0.0)
else:
return x, x
osm_uniform = _calc_uniform_order_statistic_medians(len(x))
dist = _parse_dist_kw(dist, enforce_subclass=False)
if sparams is None:
sparams = ()
if isscalar(sparams):
sparams = (sparams,)
if not isinstance(sparams, tuple):
sparams = tuple(sparams)
osm = dist.ppf(osm_uniform, *sparams)
osr = sort(x)
if _perform_fit:
# perform a linear least squares fit.
slope, intercept, r, prob, sterrest = stats.linregress(osm, osr)
if plot is not None:
plot.plot(osm, osr, 'bo', osm, slope*osm + intercept, 'r-')
_add_axis_labels_title(plot, xlabel='Theoretical quantiles',
ylabel='Ordered Values',
title='Probability Plot')
# Add R^2 value to the plot as text
if rvalue:
xmin = amin(osm)
xmax = amax(osm)
ymin = amin(x)
ymax = amax(x)
posx = xmin + 0.70 * (xmax - xmin)
posy = ymin + 0.01 * (ymax - ymin)
plot.text(posx, posy, "$R^2=%1.4f$" % r**2)
if fit:
return (osm, osr), (slope, intercept, r)
else:
return osm, osr
def ppcc_max(x, brack=(0.0, 1.0), dist='tukeylambda'):
"""
Calculate the shape parameter that maximizes the PPCC.
The probability plot correlation coefficient (PPCC) plot can be used to
determine the optimal shape parameter for a one-parameter family of
distributions. ppcc_max returns the shape parameter that would maximize the
probability plot correlation coefficient for the given data to a
one-parameter family of distributions.
Parameters
----------
x : array_like
Input array.
brack : tuple, optional
Triple (a,b,c) where (a<b<c). If bracket consists of two numbers (a, c)
then they are assumed to be a starting interval for a downhill bracket
search (see `scipy.optimize.brent`).
dist : str or stats.distributions instance, optional
Distribution or distribution function name. Objects that look enough
like a stats.distributions instance (i.e. they have a ``ppf`` method)
are also accepted. The default is ``'tukeylambda'``.
Returns
-------
shape_value : float
The shape parameter at which the probability plot correlation
coefficient reaches its max value.
See Also
--------
ppcc_plot, probplot, boxcox
Notes
-----
The brack keyword serves as a starting point which is useful in corner
cases. One can use a plot to obtain a rough visual estimate of the location
for the maximum to start the search near it.
References
----------
.. [1] J.J. Filliben, "The Probability Plot Correlation Coefficient Test for
Normality", Technometrics, Vol. 17, pp. 111-117, 1975.
.. [2] https://www.itl.nist.gov/div898/handbook/eda/section3/ppccplot.htm
Examples
--------
First we generate some random data from a Tukey-Lambda distribution,
with shape parameter -0.7:
>>> from scipy import stats
>>> x = stats.tukeylambda.rvs(-0.7, loc=2, scale=0.5, size=10000,
... random_state=1234567) + 1e4
Now we explore this data with a PPCC plot as well as the related
probability plot and Box-Cox normplot. A red line is drawn where we
expect the PPCC value to be maximal (at the shape parameter -0.7 used
above):
>>> import matplotlib.pyplot as plt
>>> fig = plt.figure(figsize=(8, 6))
>>> ax = fig.add_subplot(111)
>>> res = stats.ppcc_plot(x, -5, 5, plot=ax)
We calculate the value where the shape should reach its maximum and a red
line is drawn there. The line should coincide with the highest point in the
ppcc_plot.
>>> max = stats.ppcc_max(x)
>>> ax.vlines(max, 0, 1, colors='r', label='Expected shape value')
>>> plt.show()
"""
dist = _parse_dist_kw(dist)
osm_uniform = _calc_uniform_order_statistic_medians(len(x))
osr = sort(x)
# this function computes the x-axis values of the probability plot
# and computes a linear regression (including the correlation)
# and returns 1-r so that a minimization function maximizes the
# correlation
def tempfunc(shape, mi, yvals, func):
xvals = func(mi, shape)
r, prob = stats.pearsonr(xvals, yvals)
return 1 - r
return optimize.brent(tempfunc, brack=brack, args=(osm_uniform, osr, dist.ppf))
def ppcc_plot(x, a, b, dist='tukeylambda', plot=None, N=80):
"""
Calculate and optionally plot probability plot correlation coefficient.
The probability plot correlation coefficient (PPCC) plot can be used to
determine the optimal shape parameter for a one-parameter family of
distributions. It cannot be used for distributions without shape parameters
(like the normal distribution) or with multiple shape parameters.
By default a Tukey-Lambda distribution (`stats.tukeylambda`) is used. A
Tukey-Lambda PPCC plot interpolates from long-tailed to short-tailed
distributions via an approximately normal one, and is therefore particularly
useful in practice.
Parameters
----------
x : array_like
Input array.
a, b : scalar
Lower and upper bounds of the shape parameter to use.
dist : str or stats.distributions instance, optional
Distribution or distribution function name. Objects that look enough
like a stats.distributions instance (i.e. they have a ``ppf`` method)
are also accepted. The default is ``'tukeylambda'``.
plot : object, optional
If given, plots PPCC against the shape parameter.
`plot` is an object that has to have methods "plot" and "text".
The `matplotlib.pyplot` module or a Matplotlib Axes object can be used,
or a custom object with the same methods.
Default is None, which means that no plot is created.
N : int, optional
Number of points on the horizontal axis (equally distributed from
`a` to `b`).
Returns
-------
svals : ndarray
The shape values for which `ppcc` was calculated.
ppcc : ndarray
The calculated probability plot correlation coefficient values.
See Also
--------
ppcc_max, probplot, boxcox_normplot, tukeylambda
References
----------
J.J. Filliben, "The Probability Plot Correlation Coefficient Test for
Normality", Technometrics, Vol. 17, pp. 111-117, 1975.
Examples
--------
First we generate some random data from a Tukey-Lambda distribution,
with shape parameter -0.7:
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
>>> np.random.seed(1234567)
>>> x = stats.tukeylambda.rvs(-0.7, loc=2, scale=0.5, size=10000) + 1e4
Now we explore this data with a PPCC plot as well as the related
probability plot and Box-Cox normplot. A red line is drawn where we
expect the PPCC value to be maximal (at the shape parameter -0.7 used
above):
>>> fig = plt.figure(figsize=(12, 4))
>>> ax1 = fig.add_subplot(131)
>>> ax2 = fig.add_subplot(132)
>>> ax3 = fig.add_subplot(133)
>>> res = stats.probplot(x, plot=ax1)
>>> res = stats.boxcox_normplot(x, -5, 5, plot=ax2)
>>> res = stats.ppcc_plot(x, -5, 5, plot=ax3)
>>> ax3.vlines(-0.7, 0, 1, colors='r', label='Expected shape value')
>>> plt.show()
"""
if b <= a:
raise ValueError("`b` has to be larger than `a`.")
svals = np.linspace(a, b, num=N)
ppcc = np.empty_like(svals)
for k, sval in enumerate(svals):
_, r2 = probplot(x, sval, dist=dist, fit=True)
ppcc[k] = r2[-1]
if plot is not None:
plot.plot(svals, ppcc, 'x')
_add_axis_labels_title(plot, xlabel='Shape Values',
ylabel='Prob Plot Corr. Coef.',
title='(%s) PPCC Plot' % dist)
return svals, ppcc
def boxcox_llf(lmb, data):
r"""The boxcox log-likelihood function.
Parameters
----------
lmb : scalar
Parameter for Box-Cox transformation. See `boxcox` for details.
data : array_like
Data to calculate Box-Cox log-likelihood for. If `data` is
multi-dimensional, the log-likelihood is calculated along the first
axis.
Returns
-------
llf : float or ndarray
Box-Cox log-likelihood of `data` given `lmb`. A float for 1-D `data`,
an array otherwise.
See Also
--------
boxcox, probplot, boxcox_normplot, boxcox_normmax
Notes
-----
The Box-Cox log-likelihood function is defined here as
.. math::
llf = (\lambda - 1) \sum_i(\log(x_i)) -
N/2 \log(\sum_i (y_i - \bar{y})^2 / N),
where ``y`` is the Box-Cox transformed input data ``x``.
Examples
--------
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
>>> from mpl_toolkits.axes_grid1.inset_locator import inset_axes
>>> np.random.seed(1245)
Generate some random variates and calculate Box-Cox log-likelihood values
for them for a range of ``lmbda`` values:
>>> x = stats.loggamma.rvs(5, loc=10, size=1000)
>>> lmbdas = np.linspace(-2, 10)
>>> llf = np.zeros(lmbdas.shape, dtype=float)
>>> for ii, lmbda in enumerate(lmbdas):
... llf[ii] = stats.boxcox_llf(lmbda, x)
Also find the optimal lmbda value with `boxcox`:
>>> x_most_normal, lmbda_optimal = stats.boxcox(x)
Plot the log-likelihood as function of lmbda. Add the optimal lmbda as a
horizontal line to check that that's really the optimum:
>>> fig = plt.figure()
>>> ax = fig.add_subplot(111)
>>> ax.plot(lmbdas, llf, 'b.-')
>>> ax.axhline(stats.boxcox_llf(lmbda_optimal, x), color='r')
>>> ax.set_xlabel('lmbda parameter')
>>> ax.set_ylabel('Box-Cox log-likelihood')
Now add some probability plots to show that where the log-likelihood is
maximized the data transformed with `boxcox` looks closest to normal:
>>> locs = [3, 10, 4] # 'lower left', 'center', 'lower right'
>>> for lmbda, loc in zip([-1, lmbda_optimal, 9], locs):
... xt = stats.boxcox(x, lmbda=lmbda)
... (osm, osr), (slope, intercept, r_sq) = stats.probplot(xt)
... ax_inset = inset_axes(ax, width="20%", height="20%", loc=loc)
... ax_inset.plot(osm, osr, 'c.', osm, slope*osm + intercept, 'k-')
... ax_inset.set_xticklabels([])
... ax_inset.set_yticklabels([])
... ax_inset.set_title(r'$\lambda=%1.2f$' % lmbda)
>>> plt.show()
"""
data = np.asarray(data)
N = data.shape[0]
if N == 0:
return np.nan
logdata = np.log(data)
# Compute the variance of the transformed data.
if lmb == 0:
variance = np.var(logdata, axis=0)
else:
# Transform without the constant offset 1/lmb. The offset does
# not effect the variance, and the subtraction of the offset can
# lead to loss of precision.
variance = np.var(data**lmb / lmb, axis=0)
return (lmb - 1) * np.sum(logdata, axis=0) - N/2 * np.log(variance)
def _boxcox_conf_interval(x, lmax, alpha):
# Need to find the lambda for which
# f(x,lmbda) >= f(x,lmax) - 0.5*chi^2_alpha;1
fac = 0.5 * distributions.chi2.ppf(1 - alpha, 1)
target = boxcox_llf(lmax, x) - fac
def rootfunc(lmbda, data, target):
return boxcox_llf(lmbda, data) - target
# Find positive endpoint of interval in which answer is to be found
newlm = lmax + 0.5
N = 0
while (rootfunc(newlm, x, target) > 0.0) and (N < 500):
newlm += 0.1
N += 1
if N == 500:
raise RuntimeError("Could not find endpoint.")
lmplus = optimize.brentq(rootfunc, lmax, newlm, args=(x, target))
# Now find negative interval in the same way
newlm = lmax - 0.5
N = 0
while (rootfunc(newlm, x, target) > 0.0) and (N < 500):
newlm -= 0.1
N += 1
if N == 500:
raise RuntimeError("Could not find endpoint.")
lmminus = optimize.brentq(rootfunc, newlm, lmax, args=(x, target))
return lmminus, lmplus
def boxcox(x, lmbda=None, alpha=None):
r"""
Return a dataset transformed by a Box-Cox power transformation.
Parameters
----------
x : ndarray
Input array. Must be positive 1-dimensional. Must not be constant.
lmbda : {None, scalar}, optional
If `lmbda` is not None, do the transformation for that value.
If `lmbda` is None, find the lambda that maximizes the log-likelihood
function and return it as the second output argument.
alpha : {None, float}, optional
If ``alpha`` is not None, return the ``100 * (1-alpha)%`` confidence
interval for `lmbda` as the third output argument.
Must be between 0.0 and 1.0.
Returns
-------
boxcox : ndarray
Box-Cox power transformed array.
maxlog : float, optional
If the `lmbda` parameter is None, the second returned argument is
the lambda that maximizes the log-likelihood function.
(min_ci, max_ci) : tuple of float, optional
If `lmbda` parameter is None and ``alpha`` is not None, this returned
tuple of floats represents the minimum and maximum confidence limits
given ``alpha``.
See Also
--------
probplot, boxcox_normplot, boxcox_normmax, boxcox_llf
Notes
-----
The Box-Cox transform is given by::
y = (x**lmbda - 1) / lmbda, for lmbda > 0
log(x), for lmbda = 0
`boxcox` requires the input data to be positive. Sometimes a Box-Cox
transformation provides a shift parameter to achieve this; `boxcox` does
not. Such a shift parameter is equivalent to adding a positive constant to
`x` before calling `boxcox`.
The confidence limits returned when ``alpha`` is provided give the interval
where:
.. math::
llf(\hat{\lambda}) - llf(\lambda) < \frac{1}{2}\chi^2(1 - \alpha, 1),
with ``llf`` the log-likelihood function and :math:`\chi^2` the chi-squared
function.
References
----------
G.E.P. Box and D.R. Cox, "An Analysis of Transformations", Journal of the
Royal Statistical Society B, 26, 211-252 (1964).
Examples
--------
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
We generate some random variates from a non-normal distribution and make a
probability plot for it, to show it is non-normal in the tails:
>>> fig = plt.figure()
>>> ax1 = fig.add_subplot(211)
>>> x = stats.loggamma.rvs(5, size=500) + 5
>>> prob = stats.probplot(x, dist=stats.norm, plot=ax1)
>>> ax1.set_xlabel('')
>>> ax1.set_title('Probplot against normal distribution')
We now use `boxcox` to transform the data so it's closest to normal:
>>> ax2 = fig.add_subplot(212)
>>> xt, _ = stats.boxcox(x)
>>> prob = stats.probplot(xt, dist=stats.norm, plot=ax2)
>>> ax2.set_title('Probplot after Box-Cox transformation')
>>> plt.show()
"""
x = np.asarray(x)
if x.ndim != 1:
raise ValueError("Data must be 1-dimensional.")
if x.size == 0:
return x
if np.all(x == x[0]):
raise ValueError("Data must not be constant.")
if any(x <= 0):
raise ValueError("Data must be positive.")
if lmbda is not None: # single transformation
return special.boxcox(x, lmbda)
# If lmbda=None, find the lmbda that maximizes the log-likelihood function.
lmax = boxcox_normmax(x, method='mle')
y = boxcox(x, lmax)
if alpha is None:
return y, lmax
else:
# Find confidence interval
interval = _boxcox_conf_interval(x, lmax, alpha)
return y, lmax, interval
def boxcox_normmax(x, brack=(-2.0, 2.0), method='pearsonr'):
"""Compute optimal Box-Cox transform parameter for input data.
Parameters
----------
x : array_like
Input array.
brack : 2-tuple, optional
The starting interval for a downhill bracket search with
`optimize.brent`. Note that this is in most cases not critical; the
final result is allowed to be outside this bracket.
method : str, optional
The method to determine the optimal transform parameter (`boxcox`
``lmbda`` parameter). Options are:
'pearsonr' (default)
Maximizes the Pearson correlation coefficient between
``y = boxcox(x)`` and the expected values for ``y`` if `x` would be
normally-distributed.
'mle'
Minimizes the log-likelihood `boxcox_llf`. This is the method used
in `boxcox`.
'all'
Use all optimization methods available, and return all results.
Useful to compare different methods.
Returns
-------
maxlog : float or ndarray
The optimal transform parameter found. An array instead of a scalar
for ``method='all'``.
See Also
--------
boxcox, boxcox_llf, boxcox_normplot
Examples
--------
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
>>> np.random.seed(1234) # make this example reproducible
Generate some data and determine optimal ``lmbda`` in various ways:
>>> x = stats.loggamma.rvs(5, size=30) + 5
>>> y, lmax_mle = stats.boxcox(x)
>>> lmax_pearsonr = stats.boxcox_normmax(x)
>>> lmax_mle
7.177...
>>> lmax_pearsonr
7.916...
>>> stats.boxcox_normmax(x, method='all')
array([ 7.91667384, 7.17718692])
>>> fig = plt.figure()
>>> ax = fig.add_subplot(111)
>>> prob = stats.boxcox_normplot(x, -10, 10, plot=ax)
>>> ax.axvline(lmax_mle, color='r')
>>> ax.axvline(lmax_pearsonr, color='g', ls='--')
>>> plt.show()
"""
def _pearsonr(x, brack):
osm_uniform = _calc_uniform_order_statistic_medians(len(x))
xvals = distributions.norm.ppf(osm_uniform)
def _eval_pearsonr(lmbda, xvals, samps):
# This function computes the x-axis values of the probability plot
# and computes a linear regression (including the correlation) and
# returns ``1 - r`` so that a minimization function maximizes the
# correlation.
y = boxcox(samps, lmbda)
yvals = np.sort(y)
r, prob = stats.pearsonr(xvals, yvals)
return 1 - r
return optimize.brent(_eval_pearsonr, brack=brack, args=(xvals, x))
def _mle(x, brack):
def _eval_mle(lmb, data):
# function to minimize
return -boxcox_llf(lmb, data)
return optimize.brent(_eval_mle, brack=brack, args=(x,))
def _all(x, brack):
maxlog = np.zeros(2, dtype=float)
maxlog[0] = _pearsonr(x, brack)
maxlog[1] = _mle(x, brack)
return maxlog
methods = {'pearsonr': _pearsonr,
'mle': _mle,
'all': _all}
if method not in methods.keys():
raise ValueError("Method %s not recognized." % method)
optimfunc = methods[method]
return optimfunc(x, brack)
def _normplot(method, x, la, lb, plot=None, N=80):
"""Compute parameters for a Box-Cox or Yeo-Johnson normality plot,
optionally show it. See `boxcox_normplot` or `yeojohnson_normplot` for
details."""
if method == 'boxcox':
title = 'Box-Cox Normality Plot'
transform_func = boxcox
else:
title = 'Yeo-Johnson Normality Plot'
transform_func = yeojohnson
x = np.asarray(x)
if x.size == 0:
return x
if lb <= la:
raise ValueError("`lb` has to be larger than `la`.")
lmbdas = np.linspace(la, lb, num=N)
ppcc = lmbdas * 0.0
for i, val in enumerate(lmbdas):
# Determine for each lmbda the square root of correlation coefficient
# of transformed x
z = transform_func(x, lmbda=val)
_, (_, _, r) = probplot(z, dist='norm', fit=True)
ppcc[i] = r
if plot is not None:
plot.plot(lmbdas, ppcc, 'x')
_add_axis_labels_title(plot, xlabel='$\\lambda$',
ylabel='Prob Plot Corr. Coef.',
title=title)
return lmbdas, ppcc
def boxcox_normplot(x, la, lb, plot=None, N=80):
"""Compute parameters for a Box-Cox normality plot, optionally show it.
A Box-Cox normality plot shows graphically what the best transformation
parameter is to use in `boxcox` to obtain a distribution that is close
to normal.
Parameters
----------
x : array_like
Input array.
la, lb : scalar
The lower and upper bounds for the ``lmbda`` values to pass to `boxcox`
for Box-Cox transformations. These are also the limits of the
horizontal axis of the plot if that is generated.
plot : object, optional
If given, plots the quantiles and least squares fit.
`plot` is an object that has to have methods "plot" and "text".
The `matplotlib.pyplot` module or a Matplotlib Axes object can be used,
or a custom object with the same methods.
Default is None, which means that no plot is created.
N : int, optional
Number of points on the horizontal axis (equally distributed from
`la` to `lb`).
Returns
-------
lmbdas : ndarray
The ``lmbda`` values for which a Box-Cox transform was done.
ppcc : ndarray
Probability Plot Correlelation Coefficient, as obtained from `probplot`
when fitting the Box-Cox transformed input `x` against a normal
distribution.
See Also
--------
probplot, boxcox, boxcox_normmax, boxcox_llf, ppcc_max
Notes
-----
Even if `plot` is given, the figure is not shown or saved by
`boxcox_normplot`; ``plt.show()`` or ``plt.savefig('figname.png')``
should be used after calling `probplot`.
Examples
--------
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
Generate some non-normally distributed data, and create a Box-Cox plot:
>>> x = stats.loggamma.rvs(5, size=500) + 5
>>> fig = plt.figure()
>>> ax = fig.add_subplot(111)
>>> prob = stats.boxcox_normplot(x, -20, 20, plot=ax)
Determine and plot the optimal ``lmbda`` to transform ``x`` and plot it in
the same plot:
>>> _, maxlog = stats.boxcox(x)
>>> ax.axvline(maxlog, color='r')
>>> plt.show()
"""
return _normplot('boxcox', x, la, lb, plot, N)
def yeojohnson(x, lmbda=None):
r"""
Return a dataset transformed by a Yeo-Johnson power transformation.
Parameters
----------
x : ndarray
Input array. Should be 1-dimensional.
lmbda : float, optional
If ``lmbda`` is ``None``, find the lambda that maximizes the
log-likelihood function and return it as the second output argument.
Otherwise the transformation is done for the given value.
Returns
-------
yeojohnson: ndarray
Yeo-Johnson power transformed array.
maxlog : float, optional
If the `lmbda` parameter is None, the second returned argument is
the lambda that maximizes the log-likelihood function.
See Also
--------
probplot, yeojohnson_normplot, yeojohnson_normmax, yeojohnson_llf, boxcox
Notes
-----
The Yeo-Johnson transform is given by::
y = ((x + 1)**lmbda - 1) / lmbda, for x >= 0, lmbda != 0
log(x + 1), for x >= 0, lmbda = 0
-((-x + 1)**(2 - lmbda) - 1) / (2 - lmbda), for x < 0, lmbda != 2
-log(-x + 1), for x < 0, lmbda = 2
Unlike `boxcox`, `yeojohnson` does not require the input data to be
positive.
.. versionadded:: 1.2.0
References
----------
I. Yeo and R.A. Johnson, "A New Family of Power Transformations to
Improve Normality or Symmetry", Biometrika 87.4 (2000):
Examples
--------
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
We generate some random variates from a non-normal distribution and make a
probability plot for it, to show it is non-normal in the tails:
>>> fig = plt.figure()
>>> ax1 = fig.add_subplot(211)
>>> x = stats.loggamma.rvs(5, size=500) + 5
>>> prob = stats.probplot(x, dist=stats.norm, plot=ax1)
>>> ax1.set_xlabel('')
>>> ax1.set_title('Probplot against normal distribution')
We now use `yeojohnson` to transform the data so it's closest to normal:
>>> ax2 = fig.add_subplot(212)
>>> xt, lmbda = stats.yeojohnson(x)
>>> prob = stats.probplot(xt, dist=stats.norm, plot=ax2)
>>> ax2.set_title('Probplot after Yeo-Johnson transformation')
>>> plt.show()
"""
x = np.asarray(x)
if x.size == 0:
return x
if np.issubdtype(x.dtype, np.complexfloating):
raise ValueError('Yeo-Johnson transformation is not defined for '
'complex numbers.')
if np.issubdtype(x.dtype, np.integer):
x = x.astype(np.float64, copy=False)
if lmbda is not None:
return _yeojohnson_transform(x, lmbda)
# if lmbda=None, find the lmbda that maximizes the log-likelihood function.
lmax = yeojohnson_normmax(x)
y = _yeojohnson_transform(x, lmax)
return y, lmax
def _yeojohnson_transform(x, lmbda):
"""Return x transformed by the Yeo-Johnson power transform with given
parameter lmbda."""
out = np.zeros_like(x)
pos = x >= 0 # binary mask
# when x >= 0
if abs(lmbda) < np.spacing(1.):
out[pos] = np.log1p(x[pos])
else: # lmbda != 0
out[pos] = (np.power(x[pos] + 1, lmbda) - 1) / lmbda
# when x < 0
if abs(lmbda - 2) > np.spacing(1.):
out[~pos] = -(np.power(-x[~pos] + 1, 2 - lmbda) - 1) / (2 - lmbda)
else: # lmbda == 2
out[~pos] = -np.log1p(-x[~pos])
return out
def yeojohnson_llf(lmb, data):
r"""The yeojohnson log-likelihood function.
Parameters
----------
lmb : scalar
Parameter for Yeo-Johnson transformation. See `yeojohnson` for
details.
data : array_like
Data to calculate Yeo-Johnson log-likelihood for. If `data` is
multi-dimensional, the log-likelihood is calculated along the first
axis.
Returns
-------
llf : float
Yeo-Johnson log-likelihood of `data` given `lmb`.
See Also
--------
yeojohnson, probplot, yeojohnson_normplot, yeojohnson_normmax
Notes
-----
The Yeo-Johnson log-likelihood function is defined here as
.. math::
llf = N/2 \log(\hat{\sigma}^2) + (\lambda - 1)
\sum_i \text{ sign }(x_i)\log(|x_i| + 1)
where :math:`\hat{\sigma}^2` is estimated variance of the the Yeo-Johnson
transformed input data ``x``.
.. versionadded:: 1.2.0
Examples
--------
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
>>> from mpl_toolkits.axes_grid1.inset_locator import inset_axes
>>> np.random.seed(1245)
Generate some random variates and calculate Yeo-Johnson log-likelihood
values for them for a range of ``lmbda`` values:
>>> x = stats.loggamma.rvs(5, loc=10, size=1000)
>>> lmbdas = np.linspace(-2, 10)
>>> llf = np.zeros(lmbdas.shape, dtype=float)
>>> for ii, lmbda in enumerate(lmbdas):
... llf[ii] = stats.yeojohnson_llf(lmbda, x)
Also find the optimal lmbda value with `yeojohnson`:
>>> x_most_normal, lmbda_optimal = stats.yeojohnson(x)
Plot the log-likelihood as function of lmbda. Add the optimal lmbda as a
horizontal line to check that that's really the optimum:
>>> fig = plt.figure()
>>> ax = fig.add_subplot(111)
>>> ax.plot(lmbdas, llf, 'b.-')
>>> ax.axhline(stats.yeojohnson_llf(lmbda_optimal, x), color='r')
>>> ax.set_xlabel('lmbda parameter')
>>> ax.set_ylabel('Yeo-Johnson log-likelihood')
Now add some probability plots to show that where the log-likelihood is
maximized the data transformed with `yeojohnson` looks closest to normal:
>>> locs = [3, 10, 4] # 'lower left', 'center', 'lower right'
>>> for lmbda, loc in zip([-1, lmbda_optimal, 9], locs):
... xt = stats.yeojohnson(x, lmbda=lmbda)
... (osm, osr), (slope, intercept, r_sq) = stats.probplot(xt)
... ax_inset = inset_axes(ax, width="20%", height="20%", loc=loc)
... ax_inset.plot(osm, osr, 'c.', osm, slope*osm + intercept, 'k-')
... ax_inset.set_xticklabels([])
... ax_inset.set_yticklabels([])
... ax_inset.set_title(r'$\lambda=%1.2f$' % lmbda)
>>> plt.show()
"""
data = np.asarray(data)
n_samples = data.shape[0]
if n_samples == 0:
return np.nan
trans = _yeojohnson_transform(data, lmb)
loglike = -n_samples / 2 * np.log(trans.var(axis=0))
loglike += (lmb - 1) * (np.sign(data) * np.log(np.abs(data) + 1)).sum(axis=0)
return loglike
def yeojohnson_normmax(x, brack=(-2, 2)):
"""
Compute optimal Yeo-Johnson transform parameter.
Compute optimal Yeo-Johnson transform parameter for input data, using
maximum likelihood estimation.
Parameters
----------
x : array_like
Input array.
brack : 2-tuple, optional
The starting interval for a downhill bracket search with
`optimize.brent`. Note that this is in most cases not critical; the
final result is allowed to be outside this bracket.
Returns
-------
maxlog : float
The optimal transform parameter found.
See Also
--------
yeojohnson, yeojohnson_llf, yeojohnson_normplot
Notes
-----
.. versionadded:: 1.2.0
Examples
--------
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
>>> np.random.seed(1234) # make this example reproducible
Generate some data and determine optimal ``lmbda``
>>> x = stats.loggamma.rvs(5, size=30) + 5
>>> lmax = stats.yeojohnson_normmax(x)
>>> fig = plt.figure()
>>> ax = fig.add_subplot(111)
>>> prob = stats.yeojohnson_normplot(x, -10, 10, plot=ax)
>>> ax.axvline(lmax, color='r')
>>> plt.show()
"""
def _neg_llf(lmbda, data):
return -yeojohnson_llf(lmbda, data)
return optimize.brent(_neg_llf, brack=brack, args=(x,))
def yeojohnson_normplot(x, la, lb, plot=None, N=80):
"""Compute parameters for a Yeo-Johnson normality plot, optionally show it.
A Yeo-Johnson normality plot shows graphically what the best
transformation parameter is to use in `yeojohnson` to obtain a
distribution that is close to normal.
Parameters
----------
x : array_like
Input array.
la, lb : scalar
The lower and upper bounds for the ``lmbda`` values to pass to
`yeojohnson` for Yeo-Johnson transformations. These are also the
limits of the horizontal axis of the plot if that is generated.
plot : object, optional
If given, plots the quantiles and least squares fit.
`plot` is an object that has to have methods "plot" and "text".
The `matplotlib.pyplot` module or a Matplotlib Axes object can be used,
or a custom object with the same methods.
Default is None, which means that no plot is created.
N : int, optional
Number of points on the horizontal axis (equally distributed from
`la` to `lb`).
Returns
-------
lmbdas : ndarray
The ``lmbda`` values for which a Yeo-Johnson transform was done.
ppcc : ndarray
Probability Plot Correlelation Coefficient, as obtained from `probplot`
when fitting the Box-Cox transformed input `x` against a normal
distribution.
See Also
--------
probplot, yeojohnson, yeojohnson_normmax, yeojohnson_llf, ppcc_max
Notes
-----
Even if `plot` is given, the figure is not shown or saved by
`boxcox_normplot`; ``plt.show()`` or ``plt.savefig('figname.png')``
should be used after calling `probplot`.
.. versionadded:: 1.2.0
Examples
--------
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
Generate some non-normally distributed data, and create a Yeo-Johnson plot:
>>> x = stats.loggamma.rvs(5, size=500) + 5
>>> fig = plt.figure()
>>> ax = fig.add_subplot(111)
>>> prob = stats.yeojohnson_normplot(x, -20, 20, plot=ax)
Determine and plot the optimal ``lmbda`` to transform ``x`` and plot it in
the same plot:
>>> _, maxlog = stats.yeojohnson(x)
>>> ax.axvline(maxlog, color='r')
>>> plt.show()
"""
return _normplot('yeojohnson', x, la, lb, plot, N)
ShapiroResult = namedtuple('ShapiroResult', ('statistic', 'pvalue'))
def shapiro(x):
"""
Perform the Shapiro-Wilk test for normality.
The Shapiro-Wilk test tests the null hypothesis that the
data was drawn from a normal distribution.
Parameters
----------
x : array_like
Array of sample data.
Returns
-------
statistic : float
The test statistic.
p-value : float
The p-value for the hypothesis test.
See Also
--------
anderson : The Anderson-Darling test for normality
kstest : The Kolmogorov-Smirnov test for goodness of fit.
Notes
-----
The algorithm used is described in [4]_ but censoring parameters as
described are not implemented. For N > 5000 the W test statistic is accurate
but the p-value may not be.
The chance of rejecting the null hypothesis when it is true is close to 5%
regardless of sample size.
References
----------
.. [1] https://www.itl.nist.gov/div898/handbook/prc/section2/prc213.htm
.. [2] Shapiro, S. S. & Wilk, M.B (1965). An analysis of variance test for
normality (complete samples), Biometrika, Vol. 52, pp. 591-611.
.. [3] Razali, N. M. & Wah, Y. B. (2011) Power comparisons of Shapiro-Wilk,
Kolmogorov-Smirnov, Lilliefors and Anderson-Darling tests, Journal of
Statistical Modeling and Analytics, Vol. 2, pp. 21-33.
.. [4] ALGORITHM AS R94 APPL. STATIST. (1995) VOL. 44, NO. 4.
Examples
--------
>>> from scipy import stats
>>> np.random.seed(12345678)
>>> x = stats.norm.rvs(loc=5, scale=3, size=100)
>>> shapiro_test = stats.shapiro(x)
>>> shapiro_test
ShapiroResult(statistic=0.9772805571556091, pvalue=0.08144091814756393)
>>> shapiro_test.statistic
0.9772805571556091
>>> shapiro_test.pvalue
0.08144091814756393
"""
x = np.ravel(x)
N = len(x)
if N < 3:
raise ValueError("Data must be at least length 3.")
a = zeros(N, 'f')
init = 0
y = sort(x)
a, w, pw, ifault = statlib.swilk(y, a[:N//2], init)
if ifault not in [0, 2]:
warnings.warn("Input data for shapiro has range zero. The results "
"may not be accurate.")
if N > 5000:
warnings.warn("p-value may not be accurate for N > 5000.")
return ShapiroResult(w, pw)
# Values from Stephens, M A, "EDF Statistics for Goodness of Fit and
# Some Comparisons", Journal of the American Statistical
# Association, Vol. 69, Issue 347, Sept. 1974, pp 730-737
_Avals_norm = array([0.576, 0.656, 0.787, 0.918, 1.092])
_Avals_expon = array([0.922, 1.078, 1.341, 1.606, 1.957])
# From Stephens, M A, "Goodness of Fit for the Extreme Value Distribution",
# Biometrika, Vol. 64, Issue 3, Dec. 1977, pp 583-588.
_Avals_gumbel = array([0.474, 0.637, 0.757, 0.877, 1.038])
# From Stephens, M A, "Tests of Fit for the Logistic Distribution Based
# on the Empirical Distribution Function.", Biometrika,
# Vol. 66, Issue 3, Dec. 1979, pp 591-595.
_Avals_logistic = array([0.426, 0.563, 0.660, 0.769, 0.906, 1.010])
AndersonResult = namedtuple('AndersonResult', ('statistic',
'critical_values',
'significance_level'))
def anderson(x, dist='norm'):
"""
Anderson-Darling test for data coming from a particular distribution.
The Anderson-Darling test tests the null hypothesis that a sample is
drawn from a population that follows a particular distribution.
For the Anderson-Darling test, the critical values depend on
which distribution is being tested against. This function works
for normal, exponential, logistic, or Gumbel (Extreme Value
Type I) distributions.
Parameters
----------
x : array_like
Array of sample data.
dist : {'norm', 'expon', 'logistic', 'gumbel', 'gumbel_l', 'gumbel_r', 'extreme1'}, optional
The type of distribution to test against. The default is 'norm'.
The names 'extreme1', 'gumbel_l' and 'gumbel' are synonyms for the
same distribution.
Returns
-------
statistic : float
The Anderson-Darling test statistic.
critical_values : list
The critical values for this distribution.
significance_level : list
The significance levels for the corresponding critical values
in percents. The function returns critical values for a
differing set of significance levels depending on the
distribution that is being tested against.
See Also
--------
kstest : The Kolmogorov-Smirnov test for goodness-of-fit.
Notes
-----
Critical values provided are for the following significance levels:
normal/exponenential
15%, 10%, 5%, 2.5%, 1%
logistic
25%, 10%, 5%, 2.5%, 1%, 0.5%
Gumbel
25%, 10%, 5%, 2.5%, 1%
If the returned statistic is larger than these critical values then
for the corresponding significance level, the null hypothesis that
the data come from the chosen distribution can be rejected.
The returned statistic is referred to as 'A2' in the references.
References
----------
.. [1] https://www.itl.nist.gov/div898/handbook/prc/section2/prc213.htm
.. [2] Stephens, M. A. (1974). EDF Statistics for Goodness of Fit and
Some Comparisons, Journal of the American Statistical Association,
Vol. 69, pp. 730-737.
.. [3] Stephens, M. A. (1976). Asymptotic Results for Goodness-of-Fit
Statistics with Unknown Parameters, Annals of Statistics, Vol. 4,
pp. 357-369.
.. [4] Stephens, M. A. (1977). Goodness of Fit for the Extreme Value
Distribution, Biometrika, Vol. 64, pp. 583-588.
.. [5] Stephens, M. A. (1977). Goodness of Fit with Special Reference
to Tests for Exponentiality , Technical Report No. 262,
Department of Statistics, Stanford University, Stanford, CA.
.. [6] Stephens, M. A. (1979). Tests of Fit for the Logistic Distribution
Based on the Empirical Distribution Function, Biometrika, Vol. 66,
pp. 591-595.
"""
if dist not in ['norm', 'expon', 'gumbel', 'gumbel_l',
'gumbel_r', 'extreme1', 'logistic']:
raise ValueError("Invalid distribution; dist must be 'norm', "
"'expon', 'gumbel', 'extreme1' or 'logistic'.")
y = sort(x)
xbar = np.mean(x, axis=0)
N = len(y)
if dist == 'norm':
s = np.std(x, ddof=1, axis=0)
w = (y - xbar) / s
logcdf = distributions.norm.logcdf(w)
logsf = distributions.norm.logsf(w)
sig = array([15, 10, 5, 2.5, 1])
critical = around(_Avals_norm / (1.0 + 4.0/N - 25.0/N/N), 3)
elif dist == 'expon':
w = y / xbar
logcdf = distributions.expon.logcdf(w)
logsf = distributions.expon.logsf(w)
sig = array([15, 10, 5, 2.5, 1])
critical = around(_Avals_expon / (1.0 + 0.6/N), 3)
elif dist == 'logistic':
def rootfunc(ab, xj, N):
a, b = ab
tmp = (xj - a) / b
tmp2 = exp(tmp)
val = [np.sum(1.0/(1+tmp2), axis=0) - 0.5*N,
np.sum(tmp*(1.0-tmp2)/(1+tmp2), axis=0) + N]
return array(val)
sol0 = array([xbar, np.std(x, ddof=1, axis=0)])
sol = optimize.fsolve(rootfunc, sol0, args=(x, N), xtol=1e-5)
w = (y - sol[0]) / sol[1]
logcdf = distributions.logistic.logcdf(w)
logsf = distributions.logistic.logsf(w)
sig = array([25, 10, 5, 2.5, 1, 0.5])
critical = around(_Avals_logistic / (1.0 + 0.25/N), 3)
elif dist == 'gumbel_r':
xbar, s = distributions.gumbel_r.fit(x)
w = (y - xbar) / s
logcdf = distributions.gumbel_r.logcdf(w)
logsf = distributions.gumbel_r.logsf(w)
sig = array([25, 10, 5, 2.5, 1])
critical = around(_Avals_gumbel / (1.0 + 0.2/sqrt(N)), 3)
else: # (dist == 'gumbel') or (dist == 'gumbel_l') or (dist == 'extreme1')
xbar, s = distributions.gumbel_l.fit(x)
w = (y - xbar) / s
logcdf = distributions.gumbel_l.logcdf(w)
logsf = distributions.gumbel_l.logsf(w)
sig = array([25, 10, 5, 2.5, 1])
critical = around(_Avals_gumbel / (1.0 + 0.2/sqrt(N)), 3)
i = arange(1, N + 1)
A2 = -N - np.sum((2*i - 1.0) / N * (logcdf + logsf[::-1]), axis=0)
return AndersonResult(A2, critical, sig)
def _anderson_ksamp_midrank(samples, Z, Zstar, k, n, N):
"""
Compute A2akN equation 7 of Scholz and Stephens.
Parameters
----------
samples : sequence of 1-D array_like
Array of sample arrays.
Z : array_like
Sorted array of all observations.
Zstar : array_like
Sorted array of unique observations.
k : int
Number of samples.
n : array_like
Number of observations in each sample.
N : int
Total number of observations.
Returns
-------
A2aKN : float
The A2aKN statistics of Scholz and Stephens 1987.
"""
A2akN = 0.
Z_ssorted_left = Z.searchsorted(Zstar, 'left')
if N == Zstar.size:
lj = 1.
else:
lj = Z.searchsorted(Zstar, 'right') - Z_ssorted_left
Bj = Z_ssorted_left + lj / 2.
for i in arange(0, k):
s = np.sort(samples[i])
s_ssorted_right = s.searchsorted(Zstar, side='right')
Mij = s_ssorted_right.astype(float)
fij = s_ssorted_right - s.searchsorted(Zstar, 'left')
Mij -= fij / 2.
inner = lj / float(N) * (N*Mij - Bj*n[i])**2 / (Bj*(N - Bj) - N*lj/4.)
A2akN += inner.sum() / n[i]
A2akN *= (N - 1.) / N
return A2akN
def _anderson_ksamp_right(samples, Z, Zstar, k, n, N):
"""
Compute A2akN equation 6 of Scholz & Stephens.
Parameters
----------
samples : sequence of 1-D array_like
Array of sample arrays.
Z : array_like
Sorted array of all observations.
Zstar : array_like
Sorted array of unique observations.
k : int
Number of samples.
n : array_like
Number of observations in each sample.
N : int
Total number of observations.
Returns
-------
A2KN : float
The A2KN statistics of Scholz and Stephens 1987.
"""
A2kN = 0.
lj = Z.searchsorted(Zstar[:-1], 'right') - Z.searchsorted(Zstar[:-1],
'left')
Bj = lj.cumsum()
for i in arange(0, k):
s = np.sort(samples[i])
Mij = s.searchsorted(Zstar[:-1], side='right')
inner = lj / float(N) * (N * Mij - Bj * n[i])**2 / (Bj * (N - Bj))
A2kN += inner.sum() / n[i]
return A2kN
Anderson_ksampResult = namedtuple('Anderson_ksampResult',
('statistic', 'critical_values',
'significance_level'))
def anderson_ksamp(samples, midrank=True):
"""The Anderson-Darling test for k-samples.
The k-sample Anderson-Darling test is a modification of the
one-sample Anderson-Darling test. It tests the null hypothesis
that k-samples are drawn from the same population without having
to specify the distribution function of that population. The
critical values depend on the number of samples.
Parameters
----------
samples : sequence of 1-D array_like
Array of sample data in arrays.
midrank : bool, optional
Type of Anderson-Darling test which is computed. Default
(True) is the midrank test applicable to continuous and
discrete populations. If False, the right side empirical
distribution is used.
Returns
-------
statistic : float
Normalized k-sample Anderson-Darling test statistic.
critical_values : array
The critical values for significance levels 25%, 10%, 5%, 2.5%, 1%,
0.5%, 0.1%.
significance_level : float
An approximate significance level at which the null hypothesis for the
provided samples can be rejected. The value is floored / capped at
0.1% / 25%.
Raises
------
ValueError
If less than 2 samples are provided, a sample is empty, or no
distinct observations are in the samples.
See Also
--------
ks_2samp : 2 sample Kolmogorov-Smirnov test
anderson : 1 sample Anderson-Darling test
Notes
-----
[1]_ defines three versions of the k-sample Anderson-Darling test:
one for continuous distributions and two for discrete
distributions, in which ties between samples may occur. The
default of this routine is to compute the version based on the
midrank empirical distribution function. This test is applicable
to continuous and discrete data. If midrank is set to False, the
right side empirical distribution is used for a test for discrete
data. According to [1]_, the two discrete test statistics differ
only slightly if a few collisions due to round-off errors occur in
the test not adjusted for ties between samples.
The critical values corresponding to the significance levels from 0.01
to 0.25 are taken from [1]_. p-values are floored / capped
at 0.1% / 25%. Since the range of critical values might be extended in
future releases, it is recommended not to test ``p == 0.25``, but rather
``p >= 0.25`` (analogously for the lower bound).
.. versionadded:: 0.14.0
References
----------
.. [1] Scholz, F. W and Stephens, M. A. (1987), K-Sample
Anderson-Darling Tests, Journal of the American Statistical
Association, Vol. 82, pp. 918-924.
Examples
--------
>>> from scipy import stats
>>> np.random.seed(314159)
The null hypothesis that the two random samples come from the same
distribution can be rejected at the 5% level because the returned
test value is greater than the critical value for 5% (1.961) but
not at the 2.5% level. The interpolation gives an approximate
significance level of 3.2%:
>>> stats.anderson_ksamp([np.random.normal(size=50),
... np.random.normal(loc=0.5, size=30)])
(2.4615796189876105,
array([ 0.325, 1.226, 1.961, 2.718, 3.752, 4.592, 6.546]),
0.03176687568842282)
The null hypothesis cannot be rejected for three samples from an
identical distribution. The reported p-value (25%) has been capped and
may not be very accurate (since it corresponds to the value 0.449
whereas the statistic is -0.731):
>>> stats.anderson_ksamp([np.random.normal(size=50),
... np.random.normal(size=30), np.random.normal(size=20)])
(-0.73091722665244196,
array([ 0.44925884, 1.3052767 , 1.9434184 , 2.57696569, 3.41634856,
4.07210043, 5.56419101]),
0.25)
"""
k = len(samples)
if (k < 2):
raise ValueError("anderson_ksamp needs at least two samples")
samples = list(map(np.asarray, samples))
Z = np.sort(np.hstack(samples))
N = Z.size
Zstar = np.unique(Z)
if Zstar.size < 2:
raise ValueError("anderson_ksamp needs more than one distinct "
"observation")
n = np.array([sample.size for sample in samples])
if any(n == 0):
raise ValueError("anderson_ksamp encountered sample without "
"observations")
if midrank:
A2kN = _anderson_ksamp_midrank(samples, Z, Zstar, k, n, N)
else:
A2kN = _anderson_ksamp_right(samples, Z, Zstar, k, n, N)
H = (1. / n).sum()
hs_cs = (1. / arange(N - 1, 1, -1)).cumsum()
h = hs_cs[-1] + 1
g = (hs_cs / arange(2, N)).sum()
a = (4*g - 6) * (k - 1) + (10 - 6*g)*H
b = (2*g - 4)*k**2 + 8*h*k + (2*g - 14*h - 4)*H - 8*h + 4*g - 6
c = (6*h + 2*g - 2)*k**2 + (4*h - 4*g + 6)*k + (2*h - 6)*H + 4*h
d = (2*h + 6)*k**2 - 4*h*k
sigmasq = (a*N**3 + b*N**2 + c*N + d) / ((N - 1.) * (N - 2.) * (N - 3.))
m = k - 1
A2 = (A2kN - m) / math.sqrt(sigmasq)
# The b_i values are the interpolation coefficients from Table 2
# of Scholz and Stephens 1987
b0 = np.array([0.675, 1.281, 1.645, 1.96, 2.326, 2.573, 3.085])
b1 = np.array([-0.245, 0.25, 0.678, 1.149, 1.822, 2.364, 3.615])
b2 = np.array([-0.105, -0.305, -0.362, -0.391, -0.396, -0.345, -0.154])
critical = b0 + b1 / math.sqrt(m) + b2 / m
sig = np.array([0.25, 0.1, 0.05, 0.025, 0.01, 0.005, 0.001])
if A2 < critical.min():
p = sig.max()
warnings.warn("p-value capped: true value larger than {}".format(p),
stacklevel=2)
elif A2 > critical.max():
p = sig.min()
warnings.warn("p-value floored: true value smaller than {}".format(p),
stacklevel=2)
else:
# interpolation of probit of significance level
pf = np.polyfit(critical, log(sig), 2)
p = math.exp(np.polyval(pf, A2))
return Anderson_ksampResult(A2, critical, p)
AnsariResult = namedtuple('AnsariResult', ('statistic', 'pvalue'))
def ansari(x, y):
"""
Perform the Ansari-Bradley test for equal scale parameters.
The Ansari-Bradley test is a non-parametric test for the equality
of the scale parameter of the distributions from which two
samples were drawn.
Parameters
----------
x, y : array_like
Arrays of sample data.
Returns
-------
statistic : float
The Ansari-Bradley test statistic.
pvalue : float
The p-value of the hypothesis test.
See Also
--------
fligner : A non-parametric test for the equality of k variances
mood : A non-parametric test for the equality of two scale parameters
Notes
-----
The p-value given is exact when the sample sizes are both less than
55 and there are no ties, otherwise a normal approximation for the
p-value is used.
References
----------
.. [1] Sprent, Peter and N.C. Smeeton. Applied nonparametric statistical
methods. 3rd ed. Chapman and Hall/CRC. 2001. Section 5.8.2.
"""
x, y = asarray(x), asarray(y)
n = len(x)
m = len(y)
if m < 1:
raise ValueError("Not enough other observations.")
if n < 1:
raise ValueError("Not enough test observations.")
N = m + n
xy = r_[x, y] # combine
rank = stats.rankdata(xy)
symrank = amin(array((rank, N - rank + 1)), 0)
AB = np.sum(symrank[:n], axis=0)
uxy = unique(xy)
repeats = (len(uxy) != len(xy))
exact = ((m < 55) and (n < 55) and not repeats)
if repeats and (m < 55 or n < 55):
warnings.warn("Ties preclude use of exact statistic.")
if exact:
astart, a1, ifault = statlib.gscale(n, m)
ind = AB - astart
total = np.sum(a1, axis=0)
if ind < len(a1)/2.0:
cind = int(ceil(ind))
if ind == cind:
pval = 2.0 * np.sum(a1[:cind+1], axis=0) / total
else:
pval = 2.0 * np.sum(a1[:cind], axis=0) / total
else:
find = int(floor(ind))
if ind == floor(ind):
pval = 2.0 * np.sum(a1[find:], axis=0) / total
else:
pval = 2.0 * np.sum(a1[find+1:], axis=0) / total
return AnsariResult(AB, min(1.0, pval))
# otherwise compute normal approximation
if N % 2: # N odd
mnAB = n * (N+1.0)**2 / 4.0 / N
varAB = n * m * (N+1.0) * (3+N**2) / (48.0 * N**2)
else:
mnAB = n * (N+2.0) / 4.0
varAB = m * n * (N+2) * (N-2.0) / 48 / (N-1.0)
if repeats: # adjust variance estimates
# compute np.sum(tj * rj**2,axis=0)
fac = np.sum(symrank**2, axis=0)
if N % 2: # N odd
varAB = m * n * (16*N*fac - (N+1)**4) / (16.0 * N**2 * (N-1))
else: # N even
varAB = m * n * (16*fac - N*(N+2)**2) / (16.0 * N * (N-1))
z = (AB - mnAB) / sqrt(varAB)
pval = distributions.norm.sf(abs(z)) * 2.0
return AnsariResult(AB, pval)
BartlettResult = namedtuple('BartlettResult', ('statistic', 'pvalue'))
def bartlett(*args):
"""
Perform Bartlett's test for equal variances.
Bartlett's test tests the null hypothesis that all input samples
are from populations with equal variances. For samples
from significantly non-normal populations, Levene's test
`levene` is more robust.
Parameters
----------
sample1, sample2,... : array_like
arrays of sample data. Only 1d arrays are accepted, they may have
different lengths.
Returns
-------
statistic : float
The test statistic.
pvalue : float
The p-value of the test.
See Also
--------
fligner : A non-parametric test for the equality of k variances
levene : A robust parametric test for equality of k variances
Notes
-----
Conover et al. (1981) examine many of the existing parametric and
nonparametric tests by extensive simulations and they conclude that the
tests proposed by Fligner and Killeen (1976) and Levene (1960) appear to be
superior in terms of robustness of departures from normality and power
([3]_).
References
----------
.. [1] https://www.itl.nist.gov/div898/handbook/eda/section3/eda357.htm
.. [2] Snedecor, George W. and Cochran, William G. (1989), Statistical
Methods, Eighth Edition, Iowa State University Press.
.. [3] Park, C. and Lindsay, B. G. (1999). Robust Scale Estimation and
Hypothesis Testing based on Quadratic Inference Function. Technical
Report #99-03, Center for Likelihood Studies, Pennsylvania State
University.
.. [4] Bartlett, M. S. (1937). Properties of Sufficiency and Statistical
Tests. Proceedings of the Royal Society of London. Series A,
Mathematical and Physical Sciences, Vol. 160, No.901, pp. 268-282.
Examples
--------
Test whether or not the lists `a`, `b` and `c` come from populations
with equal variances.
>>> from scipy.stats import bartlett
>>> a = [8.88, 9.12, 9.04, 8.98, 9.00, 9.08, 9.01, 8.85, 9.06, 8.99]
>>> b = [8.88, 8.95, 9.29, 9.44, 9.15, 9.58, 8.36, 9.18, 8.67, 9.05]
>>> c = [8.95, 9.12, 8.95, 8.85, 9.03, 8.84, 9.07, 8.98, 8.86, 8.98]
>>> stat, p = bartlett(a, b, c)
>>> p
1.1254782518834628e-05
The very small p-value suggests that the populations do not have equal
variances.
This is not surprising, given that the sample variance of `b` is much
larger than that of `a` and `c`:
>>> [np.var(x, ddof=1) for x in [a, b, c]]
[0.007054444444444413, 0.13073888888888888, 0.008890000000000002]
"""
# Handle empty input and input that is not 1d
for a in args:
if np.asanyarray(a).size == 0:
return BartlettResult(np.nan, np.nan)
if np.asanyarray(a).ndim > 1:
raise ValueError('Samples must be one-dimensional.')
k = len(args)
if k < 2:
raise ValueError("Must enter at least two input sample vectors.")
Ni = zeros(k)
ssq = zeros(k, 'd')
for j in range(k):
Ni[j] = len(args[j])
ssq[j] = np.var(args[j], ddof=1)
Ntot = np.sum(Ni, axis=0)
spsq = np.sum((Ni - 1)*ssq, axis=0) / (1.0*(Ntot - k))
numer = (Ntot*1.0 - k) * log(spsq) - np.sum((Ni - 1.0)*log(ssq), axis=0)
denom = 1.0 + 1.0/(3*(k - 1)) * ((np.sum(1.0/(Ni - 1.0), axis=0)) -
1.0/(Ntot - k))
T = numer / denom
pval = distributions.chi2.sf(T, k - 1) # 1 - cdf
return BartlettResult(T, pval)
LeveneResult = namedtuple('LeveneResult', ('statistic', 'pvalue'))
def levene(*args, **kwds):
"""
Perform Levene test for equal variances.
The Levene test tests the null hypothesis that all input samples
are from populations with equal variances. Levene's test is an
alternative to Bartlett's test `bartlett` in the case where
there are significant deviations from normality.
Parameters
----------
sample1, sample2, ... : array_like
The sample data, possibly with different lengths. Only one-dimensional
samples are accepted.
center : {'mean', 'median', 'trimmed'}, optional
Which function of the data to use in the test. The default
is 'median'.
proportiontocut : float, optional
When `center` is 'trimmed', this gives the proportion of data points
to cut from each end. (See `scipy.stats.trim_mean`.)
Default is 0.05.
Returns
-------
statistic : float
The test statistic.
pvalue : float
The p-value for the test.
Notes
-----
Three variations of Levene's test are possible. The possibilities
and their recommended usages are:
* 'median' : Recommended for skewed (non-normal) distributions>
* 'mean' : Recommended for symmetric, moderate-tailed distributions.
* 'trimmed' : Recommended for heavy-tailed distributions.
The test version using the mean was proposed in the original article
of Levene ([2]_) while the median and trimmed mean have been studied by
Brown and Forsythe ([3]_), sometimes also referred to as Brown-Forsythe
test.
References
----------
.. [1] https://www.itl.nist.gov/div898/handbook/eda/section3/eda35a.htm
.. [2] Levene, H. (1960). In Contributions to Probability and Statistics:
Essays in Honor of Harold Hotelling, I. Olkin et al. eds.,
Stanford University Press, pp. 278-292.
.. [3] Brown, M. B. and Forsythe, A. B. (1974), Journal of the American
Statistical Association, 69, 364-367
Examples
--------
Test whether or not the lists `a`, `b` and `c` come from populations
with equal variances.
>>> from scipy.stats import levene
>>> a = [8.88, 9.12, 9.04, 8.98, 9.00, 9.08, 9.01, 8.85, 9.06, 8.99]
>>> b = [8.88, 8.95, 9.29, 9.44, 9.15, 9.58, 8.36, 9.18, 8.67, 9.05]
>>> c = [8.95, 9.12, 8.95, 8.85, 9.03, 8.84, 9.07, 8.98, 8.86, 8.98]
>>> stat, p = levene(a, b, c)
>>> p
0.002431505967249681
The small p-value suggests that the populations do not have equal
variances.
This is not surprising, given that the sample variance of `b` is much
larger than that of `a` and `c`:
>>> [np.var(x, ddof=1) for x in [a, b, c]]
[0.007054444444444413, 0.13073888888888888, 0.008890000000000002]
"""
# Handle keyword arguments.
center = 'median'
proportiontocut = 0.05
for kw, value in kwds.items():
if kw not in ['center', 'proportiontocut']:
raise TypeError("levene() got an unexpected keyword "
"argument '%s'" % kw)
if kw == 'center':
center = value
else:
proportiontocut = value
k = len(args)
if k < 2:
raise ValueError("Must enter at least two input sample vectors.")
# check for 1d input
for j in range(k):
if np.asanyarray(args[j]).ndim > 1:
raise ValueError('Samples must be one-dimensional.')
Ni = zeros(k)
Yci = zeros(k, 'd')
if center not in ['mean', 'median', 'trimmed']:
raise ValueError("Keyword argument <center> must be 'mean', 'median'"
" or 'trimmed'.")
if center == 'median':
func = lambda x: np.median(x, axis=0)
elif center == 'mean':
func = lambda x: np.mean(x, axis=0)
else: # center == 'trimmed'
args = tuple(stats.trimboth(np.sort(arg), proportiontocut)
for arg in args)
func = lambda x: np.mean(x, axis=0)
for j in range(k):
Ni[j] = len(args[j])
Yci[j] = func(args[j])
Ntot = np.sum(Ni, axis=0)
# compute Zij's
Zij = [None] * k
for i in range(k):
Zij[i] = abs(asarray(args[i]) - Yci[i])
# compute Zbari
Zbari = zeros(k, 'd')
Zbar = 0.0
for i in range(k):
Zbari[i] = np.mean(Zij[i], axis=0)
Zbar += Zbari[i] * Ni[i]
Zbar /= Ntot
numer = (Ntot - k) * np.sum(Ni * (Zbari - Zbar)**2, axis=0)
# compute denom_variance
dvar = 0.0
for i in range(k):
dvar += np.sum((Zij[i] - Zbari[i])**2, axis=0)
denom = (k - 1.0) * dvar
W = numer / denom
pval = distributions.f.sf(W, k-1, Ntot-k) # 1 - cdf
return LeveneResult(W, pval)
def binom_test(x, n=None, p=0.5, alternative='two-sided'):
"""
Perform a test that the probability of success is p.
This is an exact, two-sided test of the null hypothesis
that the probability of success in a Bernoulli experiment
is `p`.
Parameters
----------
x : int or array_like
The number of successes, or if x has length 2, it is the
number of successes and the number of failures.
n : int
The number of trials. This is ignored if x gives both the
number of successes and failures.
p : float, optional
The hypothesized probability of success. ``0 <= p <= 1``. The
default value is ``p = 0.5``.
alternative : {'two-sided', 'greater', 'less'}, optional
Indicates the alternative hypothesis. The default value is
'two-sided'.
Returns
-------
p-value : float
The p-value of the hypothesis test.
References
----------
.. [1] https://en.wikipedia.org/wiki/Binomial_test
Examples
--------
>>> from scipy import stats
A car manufacturer claims that no more than 10% of their cars are unsafe.
15 cars are inspected for safety, 3 were found to be unsafe. Test the
manufacturer's claim:
>>> stats.binom_test(3, n=15, p=0.1, alternative='greater')
0.18406106910639114
The null hypothesis cannot be rejected at the 5% level of significance
because the returned p-value is greater than the critical value of 5%.
"""
x = atleast_1d(x).astype(np.int_)
if len(x) == 2:
n = x[1] + x[0]
x = x[0]
elif len(x) == 1:
x = x[0]
if n is None or n < x:
raise ValueError("n must be >= x")
n = np.int_(n)
else:
raise ValueError("Incorrect length for x.")
if (p > 1.0) or (p < 0.0):
raise ValueError("p must be in range [0,1]")
if alternative not in ('two-sided', 'less', 'greater'):
raise ValueError("alternative not recognized\n"
"should be 'two-sided', 'less' or 'greater'")
if alternative == 'less':
pval = distributions.binom.cdf(x, n, p)
return pval
if alternative == 'greater':
pval = distributions.binom.sf(x-1, n, p)
return pval
# if alternative was neither 'less' nor 'greater', then it's 'two-sided'
d = distributions.binom.pmf(x, n, p)
rerr = 1 + 1e-7
if x == p * n:
# special case as shortcut, would also be handled by `else` below
pval = 1.
elif x < p * n:
i = np.arange(np.ceil(p * n), n+1)
y = np.sum(distributions.binom.pmf(i, n, p) <= d*rerr, axis=0)
pval = (distributions.binom.cdf(x, n, p) +
distributions.binom.sf(n - y, n, p))
else:
i = np.arange(np.floor(p*n) + 1)
y = np.sum(distributions.binom.pmf(i, n, p) <= d*rerr, axis=0)
pval = (distributions.binom.cdf(y-1, n, p) +
distributions.binom.sf(x-1, n, p))
return min(1.0, pval)
def _apply_func(x, g, func):
# g is list of indices into x
# separating x into different groups
# func should be applied over the groups
g = unique(r_[0, g, len(x)])
output = [func(x[g[k]:g[k+1]]) for k in range(len(g) - 1)]
return asarray(output)
FlignerResult = namedtuple('FlignerResult', ('statistic', 'pvalue'))
def fligner(*args, **kwds):
"""
Perform Fligner-Killeen test for equality of variance.
Fligner's test tests the null hypothesis that all input samples
are from populations with equal variances. Fligner-Killeen's test is
distribution free when populations are identical [2]_.
Parameters
----------
sample1, sample2, ... : array_like
Arrays of sample data. Need not be the same length.
center : {'mean', 'median', 'trimmed'}, optional
Keyword argument controlling which function of the data is used in
computing the test statistic. The default is 'median'.
proportiontocut : float, optional
When `center` is 'trimmed', this gives the proportion of data points
to cut from each end. (See `scipy.stats.trim_mean`.)
Default is 0.05.
Returns
-------
statistic : float
The test statistic.
pvalue : float
The p-value for the hypothesis test.
See Also
--------
bartlett : A parametric test for equality of k variances in normal samples
levene : A robust parametric test for equality of k variances
Notes
-----
As with Levene's test there are three variants of Fligner's test that
differ by the measure of central tendency used in the test. See `levene`
for more information.
Conover et al. (1981) examine many of the existing parametric and
nonparametric tests by extensive simulations and they conclude that the
tests proposed by Fligner and Killeen (1976) and Levene (1960) appear to be
superior in terms of robustness of departures from normality and power [3]_.
References
----------
.. [1] Park, C. and Lindsay, B. G. (1999). Robust Scale Estimation and
Hypothesis Testing based on Quadratic Inference Function. Technical
Report #99-03, Center for Likelihood Studies, Pennsylvania State
University.
https://cecas.clemson.edu/~cspark/cv/paper/qif/draftqif2.pdf
.. [2] Fligner, M.A. and Killeen, T.J. (1976). Distribution-free two-sample
tests for scale. 'Journal of the American Statistical Association.'
71(353), 210-213.
.. [3] Park, C. and Lindsay, B. G. (1999). Robust Scale Estimation and
Hypothesis Testing based on Quadratic Inference Function. Technical
Report #99-03, Center for Likelihood Studies, Pennsylvania State
University.
.. [4] Conover, W. J., Johnson, M. E. and Johnson M. M. (1981). A
comparative study of tests for homogeneity of variances, with
applications to the outer continental shelf biding data.
Technometrics, 23(4), 351-361.
Examples
--------
Test whether or not the lists `a`, `b` and `c` come from populations
with equal variances.
>>> from scipy.stats import fligner
>>> a = [8.88, 9.12, 9.04, 8.98, 9.00, 9.08, 9.01, 8.85, 9.06, 8.99]
>>> b = [8.88, 8.95, 9.29, 9.44, 9.15, 9.58, 8.36, 9.18, 8.67, 9.05]
>>> c = [8.95, 9.12, 8.95, 8.85, 9.03, 8.84, 9.07, 8.98, 8.86, 8.98]
>>> stat, p = fligner(a, b, c)
>>> p
0.00450826080004775
The small p-value suggests that the populations do not have equal
variances.
This is not surprising, given that the sample variance of `b` is much
larger than that of `a` and `c`:
>>> [np.var(x, ddof=1) for x in [a, b, c]]
[0.007054444444444413, 0.13073888888888888, 0.008890000000000002]
"""
# Handle empty input
for a in args:
if np.asanyarray(a).size == 0:
return FlignerResult(np.nan, np.nan)
# Handle keyword arguments.
center = 'median'
proportiontocut = 0.05
for kw, value in kwds.items():
if kw not in ['center', 'proportiontocut']:
raise TypeError("fligner() got an unexpected keyword "
"argument '%s'" % kw)
if kw == 'center':
center = value
else:
proportiontocut = value
k = len(args)
if k < 2:
raise ValueError("Must enter at least two input sample vectors.")
if center not in ['mean', 'median', 'trimmed']:
raise ValueError("Keyword argument <center> must be 'mean', 'median'"
" or 'trimmed'.")
if center == 'median':
func = lambda x: np.median(x, axis=0)
elif center == 'mean':
func = lambda x: np.mean(x, axis=0)
else: # center == 'trimmed'
args = tuple(stats.trimboth(arg, proportiontocut) for arg in args)
func = lambda x: np.mean(x, axis=0)
Ni = asarray([len(args[j]) for j in range(k)])
Yci = asarray([func(args[j]) for j in range(k)])
Ntot = np.sum(Ni, axis=0)
# compute Zij's
Zij = [abs(asarray(args[i]) - Yci[i]) for i in range(k)]
allZij = []
g = [0]
for i in range(k):
allZij.extend(list(Zij[i]))
g.append(len(allZij))
ranks = stats.rankdata(allZij)
a = distributions.norm.ppf(ranks / (2*(Ntot + 1.0)) + 0.5)
# compute Aibar
Aibar = _apply_func(a, g, np.sum) / Ni
anbar = np.mean(a, axis=0)
varsq = np.var(a, axis=0, ddof=1)
Xsq = np.sum(Ni * (asarray(Aibar) - anbar)**2.0, axis=0) / varsq
pval = distributions.chi2.sf(Xsq, k - 1) # 1 - cdf
return FlignerResult(Xsq, pval)
def mood(x, y, axis=0):
"""
Perform Mood's test for equal scale parameters.
Mood's two-sample test for scale parameters is a non-parametric
test for the null hypothesis that two samples are drawn from the
same distribution with the same scale parameter.
Parameters
----------
x, y : array_like
Arrays of sample data.
axis : int, optional
The axis along which the samples are tested. `x` and `y` can be of
different length along `axis`.
If `axis` is None, `x` and `y` are flattened and the test is done on
all values in the flattened arrays.
Returns
-------
z : scalar or ndarray
The z-score for the hypothesis test. For 1-D inputs a scalar is
returned.
p-value : scalar ndarray
The p-value for the hypothesis test.
See Also
--------
fligner : A non-parametric test for the equality of k variances
ansari : A non-parametric test for the equality of 2 variances
bartlett : A parametric test for equality of k variances in normal samples
levene : A parametric test for equality of k variances
Notes
-----
The data are assumed to be drawn from probability distributions ``f(x)``
and ``f(x/s) / s`` respectively, for some probability density function f.
The null hypothesis is that ``s == 1``.
For multi-dimensional arrays, if the inputs are of shapes
``(n0, n1, n2, n3)`` and ``(n0, m1, n2, n3)``, then if ``axis=1``, the
resulting z and p values will have shape ``(n0, n2, n3)``. Note that
``n1`` and ``m1`` don't have to be equal, but the other dimensions do.
Examples
--------
>>> from scipy import stats
>>> np.random.seed(1234)
>>> x2 = np.random.randn(2, 45, 6, 7)
>>> x1 = np.random.randn(2, 30, 6, 7)
>>> z, p = stats.mood(x1, x2, axis=1)
>>> p.shape
(2, 6, 7)
Find the number of points where the difference in scale is not significant:
>>> (p > 0.1).sum()
74
Perform the test with different scales:
>>> x1 = np.random.randn(2, 30)
>>> x2 = np.random.randn(2, 35) * 10.0
>>> stats.mood(x1, x2, axis=1)
(array([-5.7178125 , -5.25342163]), array([ 1.07904114e-08, 1.49299218e-07]))
"""
x = np.asarray(x, dtype=float)
y = np.asarray(y, dtype=float)
if axis is None:
x = x.flatten()
y = y.flatten()
axis = 0
# Determine shape of the result arrays
res_shape = tuple([x.shape[ax] for ax in range(len(x.shape)) if ax != axis])
if not (res_shape == tuple([y.shape[ax] for ax in range(len(y.shape)) if
ax != axis])):
raise ValueError("Dimensions of x and y on all axes except `axis` "
"should match")
n = x.shape[axis]
m = y.shape[axis]
N = m + n
if N < 3:
raise ValueError("Not enough observations.")
xy = np.concatenate((x, y), axis=axis)
if axis != 0:
xy = np.rollaxis(xy, axis)
xy = xy.reshape(xy.shape[0], -1)
# Generalized to the n-dimensional case by adding the axis argument, and
# using for loops, since rankdata is not vectorized. For improving
# performance consider vectorizing rankdata function.
all_ranks = np.zeros_like(xy)
for j in range(xy.shape[1]):
all_ranks[:, j] = stats.rankdata(xy[:, j])
Ri = all_ranks[:n]
M = np.sum((Ri - (N + 1.0) / 2)**2, axis=0)
# Approx stat.
mnM = n * (N * N - 1.0) / 12
varM = m * n * (N + 1.0) * (N + 2) * (N - 2) / 180
z = (M - mnM) / sqrt(varM)
# sf for right tail, cdf for left tail. Factor 2 for two-sidedness
z_pos = z > 0
pval = np.zeros_like(z)
pval[z_pos] = 2 * distributions.norm.sf(z[z_pos])
pval[~z_pos] = 2 * distributions.norm.cdf(z[~z_pos])
if res_shape == ():
# Return scalars, not 0-D arrays
z = z[0]
pval = pval[0]
else:
z.shape = res_shape
pval.shape = res_shape
return z, pval
WilcoxonResult = namedtuple('WilcoxonResult', ('statistic', 'pvalue'))
def wilcoxon(x, y=None, zero_method="wilcox", correction=False,
alternative="two-sided", mode='auto'):
"""
Calculate the Wilcoxon signed-rank test.
The Wilcoxon signed-rank test tests the null hypothesis that two
related paired samples come from the same distribution. In particular,
it tests whether the distribution of the differences x - y is symmetric
about zero. It is a non-parametric version of the paired T-test.
Parameters
----------
x : array_like
Either the first set of measurements (in which case `y` is the second
set of measurements), or the differences between two sets of
measurements (in which case `y` is not to be specified.) Must be
one-dimensional.
y : array_like, optional
Either the second set of measurements (if `x` is the first set of
measurements), or not specified (if `x` is the differences between
two sets of measurements.) Must be one-dimensional.
zero_method : {'pratt', 'wilcox', 'zsplit'}, optional
The following options are available (default is 'wilcox'):
* 'pratt': Includes zero-differences in the ranking process,
but drops the ranks of the zeros, see [4]_, (more conservative).
* 'wilcox': Discards all zero-differences, the default.
* 'zsplit': Includes zero-differences in the ranking process and
split the zero rank between positive and negative ones.
correction : bool, optional
If True, apply continuity correction by adjusting the Wilcoxon rank
statistic by 0.5 towards the mean value when computing the
z-statistic if a normal approximation is used. Default is False.
alternative : {"two-sided", "greater", "less"}, optional
The alternative hypothesis to be tested, see Notes. Default is
"two-sided".
mode : {"auto", "exact", "approx"}
Method to calculate the p-value, see Notes. Default is "auto".
Returns
-------
statistic : float
If `alternative` is "two-sided", the sum of the ranks of the
differences above or below zero, whichever is smaller.
Otherwise the sum of the ranks of the differences above zero.
pvalue : float
The p-value for the test depending on `alternative` and `mode`.
See Also
--------
kruskal, mannwhitneyu
Notes
-----
The test has been introduced in [4]_. Given n independent samples
(xi, yi) from a bivariate distribution (i.e. paired samples),
it computes the differences di = xi - yi. One assumption of the test
is that the differences are symmetric, see [2]_.
The two-sided test has the null hypothesis that the median of the
differences is zero against the alternative that it is different from
zero. The one-sided test has the null hypothesis that the median is
positive against the alternative that it is negative
(``alternative == 'less'``), or vice versa (``alternative == 'greater.'``).
To derive the p-value, the exact distribution (``mode == 'exact'``)
can be used for sample sizes of up to 25. The default ``mode == 'auto'``
uses the exact distribution if there are at most 25 observations and no
ties, otherwise a normal approximation is used (``mode == 'approx'``).
The treatment of ties can be controlled by the parameter `zero_method`.
If ``zero_method == 'pratt'``, the normal approximation is adjusted as in
[5]_. A typical rule is to require that n > 20 ([2]_, p. 383).
References
----------
.. [1] https://en.wikipedia.org/wiki/Wilcoxon_signed-rank_test
.. [2] Conover, W.J., Practical Nonparametric Statistics, 1971.
.. [3] Pratt, J.W., Remarks on Zeros and Ties in the Wilcoxon Signed
Rank Procedures, Journal of the American Statistical Association,
Vol. 54, 1959, pp. 655-667. :doi:`10.1080/01621459.1959.10501526`
.. [4] Wilcoxon, F., Individual Comparisons by Ranking Methods,
Biometrics Bulletin, Vol. 1, 1945, pp. 80-83. :doi:`10.2307/3001968`
.. [5] Cureton, E.E., The Normal Approximation to the Signed-Rank
Sampling Distribution When Zero Differences are Present,
Journal of the American Statistical Association, Vol. 62, 1967,
pp. 1068-1069. :doi:`10.1080/01621459.1967.10500917`
Examples
--------
In [4]_, the differences in height between cross- and self-fertilized
corn plants is given as follows:
>>> d = [6, 8, 14, 16, 23, 24, 28, 29, 41, -48, 49, 56, 60, -67, 75]
Cross-fertilized plants appear to be be higher. To test the null
hypothesis that there is no height difference, we can apply the
two-sided test:
>>> from scipy.stats import wilcoxon
>>> w, p = wilcoxon(d)
>>> w, p
(24.0, 0.041259765625)
Hence, we would reject the null hypothesis at a confidence level of 5%,
concluding that there is a difference in height between the groups.
To confirm that the median of the differences can be assumed to be
positive, we use:
>>> w, p = wilcoxon(d, alternative='greater')
>>> w, p
(96.0, 0.0206298828125)
This shows that the null hypothesis that the median is negative can be
rejected at a confidence level of 5% in favor of the alternative that
the median is greater than zero. The p-values above are exact. Using the
normal approximation gives very similar values:
>>> w, p = wilcoxon(d, mode='approx')
>>> w, p
(24.0, 0.04088813291185591)
Note that the statistic changed to 96 in the one-sided case (the sum
of ranks of positive differences) whereas it is 24 in the two-sided
case (the minimum of sum of ranks above and below zero).
"""
if mode not in ["auto", "approx", "exact"]:
raise ValueError("mode must be either 'auto', 'approx' or 'exact'")
if zero_method not in ["wilcox", "pratt", "zsplit"]:
raise ValueError("Zero method must be either 'wilcox' "
"or 'pratt' or 'zsplit'")
if alternative not in ["two-sided", "less", "greater"]:
raise ValueError("Alternative must be either 'two-sided', "
"'greater' or 'less'")
if y is None:
d = asarray(x)
if d.ndim > 1:
raise ValueError('Sample x must be one-dimensional.')
else:
x, y = map(asarray, (x, y))
if x.ndim > 1 or y.ndim > 1:
raise ValueError('Samples x and y must be one-dimensional.')
if len(x) != len(y):
raise ValueError('The samples x and y must have the same length.')
d = x - y
if mode == "auto":
if len(d) <= 25:
mode = "exact"
else:
mode = "approx"
n_zero = np.sum(d == 0)
if n_zero > 0 and mode == "exact":
mode = "approx"
warnings.warn("Exact p-value calculation does not work if there are "
"ties. Switching to normal approximation.")
if mode == "approx":
if zero_method in ["wilcox", "pratt"]:
if n_zero == len(d):
raise ValueError("zero_method 'wilcox' and 'pratt' do not "
"work if x - y is zero for all elements.")
if zero_method == "wilcox":
# Keep all non-zero differences
d = compress(np.not_equal(d, 0), d)
count = len(d)
if count < 10 and mode == "approx":
warnings.warn("Sample size too small for normal approximation.")
r = stats.rankdata(abs(d))
r_plus = np.sum((d > 0) * r)
r_minus = np.sum((d < 0) * r)
if zero_method == "zsplit":
r_zero = np.sum((d == 0) * r)
r_plus += r_zero / 2.
r_minus += r_zero / 2.
# return min for two-sided test, but r_plus for one-sided test
# the literature is not consistent here
# r_plus is more informative since r_plus + r_minus = count*(count+1)/2,
# i.e. the sum of the ranks, so r_minus and the min can be inferred
# (If alternative='pratt', r_plus + r_minus = count*(count+1)/2 - r_zero.)
# [3] uses the r_plus for the one-sided test, keep min for two-sided test
# to keep backwards compatibility
if alternative == "two-sided":
T = min(r_plus, r_minus)
else:
T = r_plus
if mode == "approx":
mn = count * (count + 1.) * 0.25
se = count * (count + 1.) * (2. * count + 1.)
if zero_method == "pratt":
r = r[d != 0]
# normal approximation needs to be adjusted, see Cureton (1967)
mn -= n_zero * (n_zero + 1.) * 0.25
se -= n_zero * (n_zero + 1.) * (2. * n_zero + 1.)
replist, repnum = find_repeats(r)
if repnum.size != 0:
# Correction for repeated elements.
se -= 0.5 * (repnum * (repnum * repnum - 1)).sum()
se = sqrt(se / 24)
# apply continuity correction if applicable
d = 0
if correction:
if alternative == "two-sided":
d = 0.5 * np.sign(T - mn)
elif alternative == "less":
d = -0.5
else:
d = 0.5
# compute statistic and p-value using normal approximation
z = (T - mn - d) / se
if alternative == "two-sided":
prob = 2. * distributions.norm.sf(abs(z))
elif alternative == "greater":
# large T = r_plus indicates x is greater than y; i.e.
# accept alternative in that case and return small p-value (sf)
prob = distributions.norm.sf(z)
else:
prob = distributions.norm.cdf(z)
elif mode == "exact":
# get frequencies cnt of the possible positive ranksums r_plus
cnt = _get_wilcoxon_distr(count)
# note: r_plus is int (ties not allowed), need int for slices below
r_plus = int(r_plus)
if alternative == "two-sided":
if r_plus == (len(cnt) - 1) // 2:
# r_plus is the center of the distribution.
prob = 1.0
else:
p_less = np.sum(cnt[:r_plus + 1]) / 2**count
p_greater = np.sum(cnt[r_plus:]) / 2**count
prob = 2*min(p_greater, p_less)
elif alternative == "greater":
prob = np.sum(cnt[r_plus:]) / 2**count
else:
prob = np.sum(cnt[:r_plus + 1]) / 2**count
return WilcoxonResult(T, prob)
def median_test(*args, **kwds):
"""
Perform a Mood's median test.
Test that two or more samples come from populations with the same median.
Let ``n = len(args)`` be the number of samples. The "grand median" of
all the data is computed, and a contingency table is formed by
classifying the values in each sample as being above or below the grand
median. The contingency table, along with `correction` and `lambda_`,
are passed to `scipy.stats.chi2_contingency` to compute the test statistic
and p-value.
Parameters
----------
sample1, sample2, ... : array_like
The set of samples. There must be at least two samples.
Each sample must be a one-dimensional sequence containing at least
one value. The samples are not required to have the same length.
ties : str, optional
Determines how values equal to the grand median are classified in
the contingency table. The string must be one of::
"below":
Values equal to the grand median are counted as "below".
"above":
Values equal to the grand median are counted as "above".
"ignore":
Values equal to the grand median are not counted.
The default is "below".
correction : bool, optional
If True, *and* there are just two samples, apply Yates' correction
for continuity when computing the test statistic associated with
the contingency table. Default is True.
lambda_ : float or str, optional
By default, the statistic computed in this test is Pearson's
chi-squared statistic. `lambda_` allows a statistic from the
Cressie-Read power divergence family to be used instead. See
`power_divergence` for details.
Default is 1 (Pearson's chi-squared statistic).
nan_policy : {'propagate', 'raise', 'omit'}, optional
Defines how to handle when input contains nan. 'propagate' returns nan,
'raise' throws an error, 'omit' performs the calculations ignoring nan
values. Default is 'propagate'.
Returns
-------
stat : float
The test statistic. The statistic that is returned is determined by
`lambda_`. The default is Pearson's chi-squared statistic.
p : float
The p-value of the test.
m : float
The grand median.
table : ndarray
The contingency table. The shape of the table is (2, n), where
n is the number of samples. The first row holds the counts of the
values above the grand median, and the second row holds the counts
of the values below the grand median. The table allows further
analysis with, for example, `scipy.stats.chi2_contingency`, or with
`scipy.stats.fisher_exact` if there are two samples, without having
to recompute the table. If ``nan_policy`` is "propagate" and there
are nans in the input, the return value for ``table`` is ``None``.
See Also
--------
kruskal : Compute the Kruskal-Wallis H-test for independent samples.
mannwhitneyu : Computes the Mann-Whitney rank test on samples x and y.
Notes
-----
.. versionadded:: 0.15.0
References
----------
.. [1] Mood, A. M., Introduction to the Theory of Statistics. McGraw-Hill
(1950), pp. 394-399.
.. [2] Zar, J. H., Biostatistical Analysis, 5th ed. Prentice Hall (2010).
See Sections 8.12 and 10.15.
Examples
--------
A biologist runs an experiment in which there are three groups of plants.
Group 1 has 16 plants, group 2 has 15 plants, and group 3 has 17 plants.
Each plant produces a number of seeds. The seed counts for each group
are::
Group 1: 10 14 14 18 20 22 24 25 31 31 32 39 43 43 48 49
Group 2: 28 30 31 33 34 35 36 40 44 55 57 61 91 92 99
Group 3: 0 3 9 22 23 25 25 33 34 34 40 45 46 48 62 67 84
The following code applies Mood's median test to these samples.
>>> g1 = [10, 14, 14, 18, 20, 22, 24, 25, 31, 31, 32, 39, 43, 43, 48, 49]
>>> g2 = [28, 30, 31, 33, 34, 35, 36, 40, 44, 55, 57, 61, 91, 92, 99]
>>> g3 = [0, 3, 9, 22, 23, 25, 25, 33, 34, 34, 40, 45, 46, 48, 62, 67, 84]
>>> from scipy.stats import median_test
>>> stat, p, med, tbl = median_test(g1, g2, g3)
The median is
>>> med
34.0
and the contingency table is
>>> tbl
array([[ 5, 10, 7],
[11, 5, 10]])
`p` is too large to conclude that the medians are not the same:
>>> p
0.12609082774093244
The "G-test" can be performed by passing ``lambda_="log-likelihood"`` to
`median_test`.
>>> g, p, med, tbl = median_test(g1, g2, g3, lambda_="log-likelihood")
>>> p
0.12224779737117837
The median occurs several times in the data, so we'll get a different
result if, for example, ``ties="above"`` is used:
>>> stat, p, med, tbl = median_test(g1, g2, g3, ties="above")
>>> p
0.063873276069553273
>>> tbl
array([[ 5, 11, 9],
[11, 4, 8]])
This example demonstrates that if the data set is not large and there
are values equal to the median, the p-value can be sensitive to the
choice of `ties`.
"""
ties = kwds.pop('ties', 'below')
correction = kwds.pop('correction', True)
lambda_ = kwds.pop('lambda_', None)
nan_policy = kwds.pop('nan_policy', 'propagate')
if len(kwds) > 0:
bad_kwd = kwds.keys()[0]
raise TypeError("median_test() got an unexpected keyword "
"argument %r" % bad_kwd)
if len(args) < 2:
raise ValueError('median_test requires two or more samples.')
ties_options = ['below', 'above', 'ignore']
if ties not in ties_options:
raise ValueError("invalid 'ties' option '%s'; 'ties' must be one "
"of: %s" % (ties, str(ties_options)[1:-1]))
data = [np.asarray(arg) for arg in args]
# Validate the sizes and shapes of the arguments.
for k, d in enumerate(data):
if d.size == 0:
raise ValueError("Sample %d is empty. All samples must "
"contain at least one value." % (k + 1))
if d.ndim != 1:
raise ValueError("Sample %d has %d dimensions. All "
"samples must be one-dimensional sequences." %
(k + 1, d.ndim))
cdata = np.concatenate(data)
contains_nan, nan_policy = _contains_nan(cdata, nan_policy)
if contains_nan and nan_policy == 'propagate':
return np.nan, np.nan, np.nan, None
if contains_nan:
grand_median = np.median(cdata[~np.isnan(cdata)])
else:
grand_median = np.median(cdata)
# When the minimum version of numpy supported by scipy is 1.9.0,
# the above if/else statement can be replaced by the single line:
# grand_median = np.nanmedian(cdata)
# Create the contingency table.
table = np.zeros((2, len(data)), dtype=np.int64)
for k, sample in enumerate(data):
sample = sample[~np.isnan(sample)]
nabove = count_nonzero(sample > grand_median)
nbelow = count_nonzero(sample < grand_median)
nequal = sample.size - (nabove + nbelow)
table[0, k] += nabove
table[1, k] += nbelow
if ties == "below":
table[1, k] += nequal
elif ties == "above":
table[0, k] += nequal
# Check that no row or column of the table is all zero.
# Such a table can not be given to chi2_contingency, because it would have
# a zero in the table of expected frequencies.
rowsums = table.sum(axis=1)
if rowsums[0] == 0:
raise ValueError("All values are below the grand median (%r)." %
grand_median)
if rowsums[1] == 0:
raise ValueError("All values are above the grand median (%r)." %
grand_median)
if ties == "ignore":
# We already checked that each sample has at least one value, but it
# is possible that all those values equal the grand median. If `ties`
# is "ignore", that would result in a column of zeros in `table`. We
# check for that case here.
zero_cols = np.nonzero((table == 0).all(axis=0))[0]
if len(zero_cols) > 0:
msg = ("All values in sample %d are equal to the grand "
"median (%r), so they are ignored, resulting in an "
"empty sample." % (zero_cols[0] + 1, grand_median))
raise ValueError(msg)
stat, p, dof, expected = chi2_contingency(table, lambda_=lambda_,
correction=correction)
return stat, p, grand_median, table
def _circfuncs_common(samples, high, low, nan_policy='propagate'):
# Ensure samples are array-like and size is not zero
samples = np.asarray(samples)
if samples.size == 0:
return np.nan, np.asarray(np.nan), np.asarray(np.nan), None
# Recast samples as radians that range between 0 and 2 pi and calculate
# the sine and cosine
sin_samp = sin((samples - low)*2.*pi / (high - low))
cos_samp = cos((samples - low)*2.*pi / (high - low))
# Apply the NaN policy
contains_nan, nan_policy = _contains_nan(samples, nan_policy)
if contains_nan and nan_policy == 'omit':
mask = np.isnan(samples)
# Set the sines and cosines that are NaN to zero
sin_samp[mask] = 0.0
cos_samp[mask] = 0.0
else:
mask = None
return samples, sin_samp, cos_samp, mask
def circmean(samples, high=2*pi, low=0, axis=None, nan_policy='propagate'):
"""
Compute the circular mean for samples in a range.
Parameters
----------
samples : array_like
Input array.
high : float or int, optional
High boundary for circular mean range. Default is ``2*pi``.
low : float or int, optional
Low boundary for circular mean range. Default is 0.
axis : int, optional
Axis along which means are computed. The default is to compute
the mean of the flattened array.
nan_policy : {'propagate', 'raise', 'omit'}, optional
Defines how to handle when input contains nan. 'propagate' returns nan,
'raise' throws an error, 'omit' performs the calculations ignoring nan
values. Default is 'propagate'.
Returns
-------
circmean : float
Circular mean.
Examples
--------
>>> from scipy.stats import circmean
>>> circmean([0.1, 2*np.pi+0.2, 6*np.pi+0.3])
0.2
>>> from scipy.stats import circmean
>>> circmean([0.2, 1.4, 2.6], high = 1, low = 0)
0.4
"""
samples, sin_samp, cos_samp, nmask = _circfuncs_common(samples, high, low,
nan_policy=nan_policy)
sin_sum = sin_samp.sum(axis=axis)
cos_sum = cos_samp.sum(axis=axis)
res = arctan2(sin_sum, cos_sum)
mask_nan = ~np.isnan(res)
if mask_nan.ndim > 0:
mask = res[mask_nan] < 0
else:
mask = res < 0
if mask.ndim > 0:
mask_nan[mask_nan] = mask
res[mask_nan] += 2*pi
elif mask:
res += 2*pi
# Set output to NaN if no samples went into the mean
if nmask is not None:
if nmask.all():
res = np.full(shape=res.shape, fill_value=np.nan)
else:
# Find out if any of the axis that are being averaged consist
# entirely of NaN. If one exists, set the result (res) to NaN
nshape = 0 if axis is None else axis
smask = nmask.shape[nshape] == nmask.sum(axis=axis)
if smask.any():
res[smask] = np.nan
return res*(high - low)/2.0/pi + low
def circvar(samples, high=2*pi, low=0, axis=None, nan_policy='propagate'):
"""
Compute the circular variance for samples assumed to be in a range.
Parameters
----------
samples : array_like
Input array.
high : float or int, optional
High boundary for circular variance range. Default is ``2*pi``.
low : float or int, optional
Low boundary for circular variance range. Default is 0.
axis : int, optional
Axis along which variances are computed. The default is to compute
the variance of the flattened array.
nan_policy : {'propagate', 'raise', 'omit'}, optional
Defines how to handle when input contains nan. 'propagate' returns nan,
'raise' throws an error, 'omit' performs the calculations ignoring nan
values. Default is 'propagate'.
Returns
-------
circvar : float
Circular variance.
Notes
-----
This uses a definition of circular variance that in the limit of small
angles returns a number close to the 'linear' variance.
Examples
--------
>>> from scipy.stats import circvar
>>> circvar([0, 2*np.pi/3, 5*np.pi/3])
2.19722457734
"""
samples, sin_samp, cos_samp, mask = _circfuncs_common(samples, high, low,
nan_policy=nan_policy)
if mask is None:
sin_mean = sin_samp.mean(axis=axis)
cos_mean = cos_samp.mean(axis=axis)
else:
nsum = np.asarray(np.sum(~mask, axis=axis).astype(float))
nsum[nsum == 0] = np.nan
sin_mean = sin_samp.sum(axis=axis) / nsum
cos_mean = cos_samp.sum(axis=axis) / nsum
R = hypot(sin_mean, cos_mean)
return ((high - low)/2.0/pi)**2 * 2 * log(1/R)
def circstd(samples, high=2*pi, low=0, axis=None, nan_policy='propagate'):
"""
Compute the circular standard deviation for samples assumed to be in the
range [low to high].
Parameters
----------
samples : array_like
Input array.
high : float or int, optional
High boundary for circular standard deviation range.
Default is ``2*pi``.
low : float or int, optional
Low boundary for circular standard deviation range. Default is 0.
axis : int, optional
Axis along which standard deviations are computed. The default is
to compute the standard deviation of the flattened array.
nan_policy : {'propagate', 'raise', 'omit'}, optional
Defines how to handle when input contains nan. 'propagate' returns nan,
'raise' throws an error, 'omit' performs the calculations ignoring nan
values. Default is 'propagate'.
Returns
-------
circstd : float
Circular standard deviation.
Notes
-----
This uses a definition of circular standard deviation that in the limit of
small angles returns a number close to the 'linear' standard deviation.
Examples
--------
>>> from scipy.stats import circstd
>>> circstd([0, 0.1*np.pi/2, 0.001*np.pi, 0.03*np.pi/2])
0.063564063306
"""
samples, sin_samp, cos_samp, mask = _circfuncs_common(samples, high, low,
nan_policy=nan_policy)
if mask is None:
sin_mean = sin_samp.mean(axis=axis)
cos_mean = cos_samp.mean(axis=axis)
else:
nsum = np.asarray(np.sum(~mask, axis=axis).astype(float))
nsum[nsum == 0] = np.nan
sin_mean = sin_samp.sum(axis=axis) / nsum
cos_mean = cos_samp.sum(axis=axis) / nsum
R = hypot(sin_mean, cos_mean)
return ((high - low)/2.0/pi) * sqrt(-2*log(R))
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