import warnings import numpy as np from . import distributions from .._lib._bunch import _make_tuple_bunch 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 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(f"Incompatible lengths ! ({len(y)}<>{len(x)})") # 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] 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(f"Incompatible lengths ! ({len(y)}<>{len(x)})") 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) return SiegelslopesResult(slope=medslope, intercept=medinter)