from collections import namedtuple import numpy as np from . import distributions __all__ = ['_find_repeats', 'linregress', 'theilslopes', 'siegelslopes'] LinregressResult = namedtuple('LinregressResult', ('slope', 'intercept', 'rvalue', 'pvalue', 'stderr')) def linregress(x, y=None): """ Calculate a linear least-squares regression for two sets of measurements. Parameters ---------- x, y : array_like Two sets of measurements. Both arrays should have the same length. If only `x` is given (and ``y=None``), then it must be a two-dimensional array where one dimension has length 2. The two sets of measurements are then found by splitting the array along the length-2 dimension. In the case where ``y=None`` and `x` is a 2x2 array, ``linregress(x)`` is equivalent to ``linregress(x[0], x[1])``. Returns ------- slope : float Slope of the regression line. intercept : float Intercept of the regression line. rvalue : float Correlation coefficient. pvalue : float Two-sided p-value for a hypothesis test whose null hypothesis is that the slope is zero, using Wald Test with t-distribution of the test statistic. stderr : float Standard error of the estimated gradient. See also -------- :func:`scipy.optimize.curve_fit` : Use non-linear least squares to fit a function to data. :func:`scipy.optimize.leastsq` : Minimize the sum of squares of a set of equations. Notes ----- Missing values are considered pair-wise: if a value is missing in `x`, the corresponding value in `y` is masked. Examples -------- >>> import matplotlib.pyplot as plt >>> from scipy import stats Generate some data: >>> np.random.seed(12345678) >>> x = np.random.random(10) >>> y = 1.6*x + np.random.random(10) Perform the linear regression: >>> slope, intercept, r_value, p_value, std_err = stats.linregress(x, y) >>> print("slope: %f intercept: %f" % (slope, intercept)) slope: 1.944864 intercept: 0.268578 To get coefficient of determination (R-squared): >>> print("R-squared: %f" % r_value**2) R-squared: 0.735498 Plot the data along with the fitted line: >>> plt.plot(x, y, 'o', label='original data') >>> plt.plot(x, intercept + slope*x, 'r', label='fitted line') >>> plt.legend() >>> plt.show() Example for the case where only x is provided as a 2x2 array: >>> x = np.array([[0, 1], [0, 2]]) >>> r = stats.linregress(x) >>> r.slope, r.intercept (2.0, 0.0) """ TINY = 1.0e-20 if y is None: # x is a (2, N) or (N, 2) shaped array_like x = np.asarray(x) if x.shape[0] == 2: x, y = x elif x.shape[1] == 2: x, y = x.T else: msg = ("If only `x` is given as input, it has to be of shape " "(2, N) or (N, 2), provided shape was %s" % str(x.shape)) raise ValueError(msg) else: x = np.asarray(x) y = np.asarray(y) if x.size == 0 or y.size == 0: raise ValueError("Inputs must not be empty.") n = len(x) xmean = np.mean(x, None) ymean = np.mean(y, None) # average sum of squares: ssxm, ssxym, ssyxm, ssym = np.cov(x, y, bias=1).flat r_num = ssxym r_den = np.sqrt(ssxm * ssym) if r_den == 0.0: r = 0.0 else: r = r_num / r_den # test for numerical error propagation if r > 1.0: r = 1.0 elif r < -1.0: r = -1.0 df = n - 2 slope = r_num / ssxm intercept = ymean - slope*xmean if n == 2: # handle case when only two points are passed in if y[0] == y[1]: prob = 1.0 else: prob = 0.0 sterrest = 0.0 else: t = r * np.sqrt(df / ((1.0 - r + TINY)*(1.0 + r + TINY))) prob = 2 * distributions.t.sf(np.abs(t), df) sterrest = np.sqrt((1 - r**2) * ssym / ssxm / df) return LinregressResult(slope, intercept, r, prob, sterrest) def theilslopes(y, x=None, alpha=0.95): 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". Returns ------- medslope : float Theil slope. medintercept : float Intercept of the Theil line, as ``median(y) - medslope*median(x)``. lo_slope : float Lower bound of the confidence interval on `medslope`. up_slope : float Upper bound of the confidence interval on `medslope`. 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) - medslope*median(x)``, which is given in [3]_. Other definitions of the intercept exist in the literature. A confidence interval for the intercept is not given as this question is not addressed in [1]_. 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. Examples -------- >>> 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) >>> 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() """ # We copy both x and y so we can use _find_repeats. y = np.array(y).flatten() if x is None: x = np.arange(len(y), dtype=float) else: x = np.array(x, dtype=float).flatten() if len(x) != len(y): raise ValueError("Incompatible lengths ! (%s<>%s)" % (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] slopes.sort() medslope = np.median(slopes) 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` 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]] return medslope, medinter, delta[0], 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 ``medslope`` (default option). If 'separate', estimate the intercept independent of the estimated slope. See Notes for details. Returns ------- medslope : float Estimate of the slope of the regression line. medintercept : 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. ``medslope`` 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 ``medslope`` and computes ``medintercept`` as the median of ``y - medslope*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``. ``medintercept`` 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. 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 -------- >>> 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("Incompatible lengths ! (%s<>%s)" % (len(y), len(x))) deltax = x[:, np.newaxis] - x deltay = y[:, np.newaxis] - y slopes, intercepts = [], [] for j in range(len(x)): id_nonzero = deltax[j, :] != 0 slopes_j = deltay[j, id_nonzero] / deltax[j, id_nonzero] medslope_j = np.median(slopes_j) slopes.append(medslope_j) if method == 'separate': z = y*x[j] - y[j]*x medintercept_j = np.median(z[id_nonzero] / deltax[j, id_nonzero]) intercepts.append(medintercept_j) medslope = np.median(np.asarray(slopes)) if method == "separate": medinter = np.median(np.asarray(intercepts)) else: medinter = np.median(y - medslope*x) return medslope, medinter