siegelslopes#
- scipy.stats.siegelslopes(y, x=None, method='hierarchical', *, axis=None, nan_policy='propagate', keepdims=False)[source]#
Computes the Siegel estimator for a set of points (x, y).
siegelslopesimplements 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:
- yarray_like
Dependent variable.
- xarray_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.- axisint or None, default: None
If an int, the axis of the input along which to compute the statistic. The statistic of each axis-slice (e.g. row) of the input will appear in a corresponding element of the output. If
None, the input will be raveled before computing the statistic.- nan_policy{‘propagate’, ‘omit’, ‘raise’}
Defines how to handle input NaNs.
propagate: if a NaN is present in the axis slice (e.g. row) along which the statistic is computed, the corresponding entry of the output will be NaN.omit: NaNs will be omitted when performing the calculation. If insufficient data remains in the axis slice along which the statistic is computed, the corresponding entry of the output will be NaN.raise: if a NaN is present, aValueErrorwill be raised.
- keepdimsbool, default: False
If this is set to True, the axes which are reduced are left in the result as dimensions with size one. With this option, the result will broadcast correctly against the input array.
- Returns:
- result
SiegelslopesResultinstance The return value is an object with the following attributes:
- slopefloat
Estimate of the slope of the regression line.
- interceptfloat
Estimate of the intercept of the regression line.
- result
See also
theilslopesa similar technique without repeated medians
Notes
With
n = len(y), computem_jas the median of the slopes from the point(x[j], y[j])to all other n-1 points.slopeis then the median of all slopesm_j. Two ways are given to estimate the intercept in [1] which can be chosen via the parametermethod. The hierarchical approach uses the estimated slopeslopeand computesinterceptas the median ofy - 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 mediani_j.interceptis the median of thei_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
namedtupleof length 2, with fieldsslopeandintercept, so one can continue to write:slope, intercept = siegelslopes(y, x)
Beginning in SciPy 1.9,
np.matrixinputs (not recommended for new code) are converted tonp.ndarraybefore the calculation is performed. In this case, the output will be a scalar ornp.ndarrayof appropriate shape rather than a 2Dnp.matrix. Similarly, while masked elements of masked arrays are ignored, the output will be a scalar ornp.ndarrayrather than a masked array withmask=False.References
[1] (1,2)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()