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'rvalue', 'pvalue', 'stderr'))
""" 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.
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.
Examples -------- >>> import matplotlib.pyplot as plt >>> from scipy import stats >>> np.random.seed(12345678) >>> x = np.random.random(10) >>> y = np.random.random(10) >>> slope, intercept, r_value, p_value, std_err = stats.linregress(x, y)
To get coefficient of determination (r_squared)
>>> print("r-squared:", r_value**2) r-squared: 0.08040226853902833
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()
""" 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:
raise ValueError("Inputs must not be empty.")
# average sum of squares: r = 0.0 else: # test for numerical error propagation r = 1.0 r = -1.0
# 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:
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`.
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]
# 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] |