team-10/env/Lib/site-packages/scipy/stats/_stats_mstats_common.py
2025-08-02 07:34:44 +02:00

322 lines
12 KiB
Python

import warnings
import numpy as np
from . import distributions
from .._lib._bunch import _make_tuple_bunch
from ._axis_nan_policy import _axis_nan_policy_factory
from ._stats_pythran import siegelslopes as siegelslopes_pythran
__all__ = ['_find_repeats', 'theilslopes', 'siegelslopes']
# This is not a namedtuple for backwards compatibility. See PR #12983
TheilslopesResult = _make_tuple_bunch('TheilslopesResult',
['slope', 'intercept',
'low_slope', 'high_slope'])
SiegelslopesResult = _make_tuple_bunch('SiegelslopesResult',
['slope', 'intercept'])
def _n_samples_optional_x(kwargs):
return 2 if kwargs.get('x', None) is not None else 1
@_axis_nan_policy_factory(TheilslopesResult, default_axis=None, n_outputs=4,
n_samples=_n_samples_optional_x,
result_to_tuple=lambda x, _: tuple(x), paired=True,
too_small=1)
def theilslopes(y, x=None, alpha=0.95, method='separate'):
r"""
Computes the Theil-Sen estimator for a set of points (x, y).
`theilslopes` implements a method for robust linear regression. It
computes the slope as the median of all slopes between paired values.
Parameters
----------
y : array_like
Dependent variable.
x : array_like or None, optional
Independent variable. If None, use ``arange(len(y))`` instead.
alpha : float, optional
Confidence degree between 0 and 1. Default is 95% confidence.
Note that `alpha` is symmetric around 0.5, i.e. both 0.1 and 0.9 are
interpreted as "find the 90% confidence interval".
method : {'joint', 'separate'}, optional
Method to be used for computing estimate for intercept.
Following methods are supported,
* 'joint': Uses np.median(y - slope * x) as intercept.
* 'separate': Uses np.median(y) - slope * np.median(x)
as intercept.
The default is 'separate'.
.. versionadded:: 1.8.0
Returns
-------
result : ``TheilslopesResult`` instance
The return value is an object with the following attributes:
slope : float
Theil slope.
intercept : float
Intercept of the Theil line.
low_slope : float
Lower bound of the confidence interval on `slope`.
high_slope : float
Upper bound of the confidence interval on `slope`.
See Also
--------
siegelslopes : a similar technique using repeated medians
Notes
-----
The implementation of `theilslopes` follows [1]_. The intercept is
not defined in [1]_, and here it is defined as ``median(y) -
slope*median(x)``, which is given in [3]_. Other definitions of
the intercept exist in the literature such as ``median(y - slope*x)``
in [4]_. The approach to compute the intercept can be determined by the
parameter ``method``. A confidence interval for the intercept is not
given as this question is not addressed in [1]_.
For compatibility with older versions of SciPy, the return value acts
like a ``namedtuple`` of length 4, with fields ``slope``, ``intercept``,
``low_slope``, and ``high_slope``, so one can continue to write::
slope, intercept, low_slope, high_slope = theilslopes(y, x)
References
----------
.. [1] P.K. Sen, "Estimates of the regression coefficient based on
Kendall's tau", J. Am. Stat. Assoc., Vol. 63, pp. 1379-1389, 1968.
.. [2] H. Theil, "A rank-invariant method of linear and polynomial
regression analysis I, II and III", Nederl. Akad. Wetensch., Proc.
53:, pp. 386-392, pp. 521-525, pp. 1397-1412, 1950.
.. [3] W.L. Conover, "Practical nonparametric statistics", 2nd ed.,
John Wiley and Sons, New York, pp. 493.
.. [4] https://en.wikipedia.org/wiki/Theil%E2%80%93Sen_estimator
Examples
--------
>>> import numpy as np
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
>>> x = np.linspace(-5, 5, num=150)
>>> y = x + np.random.normal(size=x.size)
>>> y[11:15] += 10 # add outliers
>>> y[-5:] -= 7
Compute the slope, intercept and 90% confidence interval. For comparison,
also compute the least-squares fit with `linregress`:
>>> res = stats.theilslopes(y, x, 0.90, method='separate')
>>> lsq_res = stats.linregress(x, y)
Plot the results. The Theil-Sen regression line is shown in red, with the
dashed red lines illustrating the confidence interval of the slope (note
that the dashed red lines are not the confidence interval of the regression
as the confidence interval of the intercept is not included). The green
line shows the least-squares fit for comparison.
>>> fig = plt.figure()
>>> ax = fig.add_subplot(111)
>>> ax.plot(x, y, 'b.')
>>> ax.plot(x, res[1] + res[0] * x, 'r-')
>>> ax.plot(x, res[1] + res[2] * x, 'r--')
>>> ax.plot(x, res[1] + res[3] * x, 'r--')
>>> ax.plot(x, lsq_res[1] + lsq_res[0] * x, 'g-')
>>> plt.show()
"""
if method not in ['joint', 'separate']:
raise ValueError("method must be either 'joint' or 'separate'."
f"'{method}' is invalid.")
# We copy both x and y so we can use _find_repeats.
y = np.array(y, dtype=float, copy=True).ravel()
if x is None:
x = np.arange(len(y), dtype=float)
else:
x = np.array(x, dtype=float, copy=True).ravel()
if len(x) != len(y):
raise ValueError("Array shapes are incompatible for broadcasting.")
if len(x) < 2:
raise ValueError("`x` and `y` must have length at least 2.")
# Compute sorted slopes only when deltax > 0
deltax = x[:, np.newaxis] - x
deltay = y[:, np.newaxis] - y
slopes = deltay[deltax > 0] / deltax[deltax > 0]
if not slopes.size:
msg = "All `x` coordinates are identical."
warnings.warn(msg, RuntimeWarning, stacklevel=2)
slopes.sort()
medslope = np.median(slopes)
if method == 'joint':
medinter = np.median(y - medslope * x)
else:
medinter = np.median(y) - medslope * np.median(x)
# Now compute confidence intervals
if alpha > 0.5:
alpha = 1. - alpha
z = distributions.norm.ppf(alpha / 2.)
# This implements (2.6) from Sen (1968)
_, nxreps = _find_repeats(x)
_, nyreps = _find_repeats(y)
nt = len(slopes) # N in Sen (1968)
ny = len(y) # n in Sen (1968)
# Equation 2.6 in Sen (1968):
sigsq = 1/18. * (ny * (ny-1) * (2*ny+5) -
sum(k * (k-1) * (2*k + 5) for k in nxreps) -
sum(k * (k-1) * (2*k + 5) for k in nyreps))
# Find the confidence interval indices in `slopes`
try:
sigma = np.sqrt(sigsq)
Ru = min(int(np.round((nt - z*sigma)/2.)), len(slopes)-1)
Rl = max(int(np.round((nt + z*sigma)/2.)) - 1, 0)
delta = slopes[[Rl, Ru]]
except (ValueError, IndexError):
delta = (np.nan, np.nan)
return TheilslopesResult(slope=medslope, intercept=medinter,
low_slope=delta[0], high_slope=delta[1])
def _find_repeats(arr):
# This function assumes it may clobber its input.
if len(arr) == 0:
return np.array(0, np.float64), np.array(0, np.intp)
# XXX This cast was previously needed for the Fortran implementation,
# should we ditch it?
arr = np.asarray(arr, np.float64).ravel()
arr.sort()
# Taken from NumPy 1.9's np.unique.
change = np.concatenate(([True], arr[1:] != arr[:-1]))
unique = arr[change]
change_idx = np.concatenate(np.nonzero(change) + ([arr.size],))
freq = np.diff(change_idx)
atleast2 = freq > 1
return unique[atleast2], freq[atleast2]
@_axis_nan_policy_factory(SiegelslopesResult, default_axis=None, n_outputs=2,
n_samples=_n_samples_optional_x,
result_to_tuple=lambda x, _: tuple(x), paired=True,
too_small=1)
def siegelslopes(y, x=None, method="hierarchical"):
r"""
Computes the Siegel estimator for a set of points (x, y).
`siegelslopes` implements a method for robust linear regression
using repeated medians (see [1]_) to fit a line to the points (x, y).
The method is robust to outliers with an asymptotic breakdown point
of 50%.
Parameters
----------
y : array_like
Dependent variable.
x : array_like or None, optional
Independent variable. If None, use ``arange(len(y))`` instead.
method : {'hierarchical', 'separate'}
If 'hierarchical', estimate the intercept using the estimated
slope ``slope`` (default option).
If 'separate', estimate the intercept independent of the estimated
slope. See Notes for details.
Returns
-------
result : ``SiegelslopesResult`` instance
The return value is an object with the following attributes:
slope : float
Estimate of the slope of the regression line.
intercept : float
Estimate of the intercept of the regression line.
See Also
--------
theilslopes : a similar technique without repeated medians
Notes
-----
With ``n = len(y)``, compute ``m_j`` as the median of
the slopes from the point ``(x[j], y[j])`` to all other `n-1` points.
``slope`` is then the median of all slopes ``m_j``.
Two ways are given to estimate the intercept in [1]_ which can be chosen
via the parameter ``method``.
The hierarchical approach uses the estimated slope ``slope``
and computes ``intercept`` as the median of ``y - slope*x``.
The other approach estimates the intercept separately as follows: for
each point ``(x[j], y[j])``, compute the intercepts of all the `n-1`
lines through the remaining points and take the median ``i_j``.
``intercept`` is the median of the ``i_j``.
The implementation computes `n` times the median of a vector of size `n`
which can be slow for large vectors. There are more efficient algorithms
(see [2]_) which are not implemented here.
For compatibility with older versions of SciPy, the return value acts
like a ``namedtuple`` of length 2, with fields ``slope`` and
``intercept``, so one can continue to write::
slope, intercept = siegelslopes(y, x)
References
----------
.. [1] A. Siegel, "Robust Regression Using Repeated Medians",
Biometrika, Vol. 69, pp. 242-244, 1982.
.. [2] A. Stein and M. Werman, "Finding the repeated median regression
line", Proceedings of the Third Annual ACM-SIAM Symposium on
Discrete Algorithms, pp. 409-413, 1992.
Examples
--------
>>> import numpy as np
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
>>> x = np.linspace(-5, 5, num=150)
>>> y = x + np.random.normal(size=x.size)
>>> y[11:15] += 10 # add outliers
>>> y[-5:] -= 7
Compute the slope and intercept. For comparison, also compute the
least-squares fit with `linregress`:
>>> res = stats.siegelslopes(y, x)
>>> lsq_res = stats.linregress(x, y)
Plot the results. The Siegel regression line is shown in red. The green
line shows the least-squares fit for comparison.
>>> fig = plt.figure()
>>> ax = fig.add_subplot(111)
>>> ax.plot(x, y, 'b.')
>>> ax.plot(x, res[1] + res[0] * x, 'r-')
>>> ax.plot(x, lsq_res[1] + lsq_res[0] * x, 'g-')
>>> plt.show()
"""
if method not in ['hierarchical', 'separate']:
raise ValueError("method can only be 'hierarchical' or 'separate'")
y = np.asarray(y).ravel()
if x is None:
x = np.arange(len(y), dtype=float)
else:
x = np.asarray(x, dtype=float).ravel()
if len(x) != len(y):
raise ValueError("Array shapes are incompatible for broadcasting.")
if len(x) < 2:
raise ValueError("`x` and `y` must have length at least 2.")
dtype = np.result_type(x, y, np.float32) # use at least float32
y, x = y.astype(dtype), x.astype(dtype)
medslope, medinter = siegelslopes_pythran(y, x, method)
medslope, medinter = np.asarray(medslope)[()], np.asarray(medinter)[()]
return SiegelslopesResult(slope=medslope, intercept=medinter)