scipy.stats.

linregress#

scipy.stats.linregress(x, y, alternative='two-sided', *, axis=0, nan_policy='propagate', keepdims=False)[source]#

Calculate a linear least-squares regression for two sets of measurements.

Parameters:
x, yarray_like

Two sets of measurements. Both arrays should have the same length N.

alternative{‘two-sided’, ‘less’, ‘greater’}, optional

Defines the alternative hypothesis. Default is ‘two-sided’. The following options are available:

  • ‘two-sided’: the slope of the regression line is nonzero

  • ‘less’: the slope of the regression line is less than zero

  • ‘greater’: the slope of the regression line is greater than zero

Added in version 1.7.0.

axisint or None, default: 0

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, a ValueError will 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:
resultLinregressResult instance

The return value is an object with the following attributes:

slopefloat

Slope of the regression line.

interceptfloat

Intercept of the regression line.

rvaluefloat

The Pearson correlation coefficient. The square of rvalue is equal to the coefficient of determination.

pvaluefloat

The 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. See alternative above for alternative hypotheses.

stderrfloat

Standard error of the estimated slope (gradient), under the assumption of residual normality.

intercept_stderrfloat

Standard error of the estimated intercept, under the assumption of residual normality.

See also

scipy.optimize.curve_fit

Use non-linear least squares to fit a function to data.

scipy.optimize.leastsq

Minimize the sum of squares of a set of equations.

Notes

For compatibility with older versions of SciPy, the return value acts like a namedtuple of length 5, with fields slope, intercept, rvalue, pvalue and stderr, so one can continue to write:

slope, intercept, r, p, se = linregress(x, y)

With that style, however, the standard error of the intercept is not available. To have access to all the computed values, including the standard error of the intercept, use the return value as an object with attributes, e.g.:

result = linregress(x, y)
print(result.intercept, result.intercept_stderr)

Beginning in SciPy 1.9, np.matrix inputs (not recommended for new code) are converted to np.ndarray before the calculation is performed. In this case, the output will be a scalar or np.ndarray of appropriate shape rather than a 2D np.matrix. Similarly, while masked elements of masked arrays are ignored, the output will be a scalar or np.ndarray rather than a masked array with mask=False.

Examples

>>> import numpy as np
>>> import matplotlib.pyplot as plt
>>> from scipy import stats
>>> rng = np.random.default_rng()

Generate some data:

>>> x = rng.random(10)
>>> y = 1.6*x + rng.random(10)

Perform the linear regression:

>>> res = stats.linregress(x, y)

Coefficient of determination (R-squared):

>>> print(f"R-squared: {res.rvalue**2:.6f}")
R-squared: 0.717533

Plot the data along with the fitted line:

>>> plt.plot(x, y, 'o', label='original data')
>>> plt.plot(x, res.intercept + res.slope*x, 'r', label='fitted line')
>>> plt.legend()
>>> plt.show()
../../_images/scipy-stats-linregress-1_00_00.png

Calculate 95% confidence interval on slope and intercept:

>>> # Two-sided inverse Students t-distribution
>>> # p - probability, df - degrees of freedom
>>> from scipy.stats import t
>>> tinv = lambda p, df: abs(t.ppf(p/2, df))

>>> ts = tinv(0.05, len(x)-2)
>>> print(f"slope (95%): {res.slope:.6f} +/- {ts*res.stderr:.6f}")
slope (95%): 1.453392 +/- 0.743465
>>> print(f"intercept (95%): {res.intercept:.6f}"
...       f" +/- {ts*res.intercept_stderr:.6f}")
intercept (95%): 0.616950 +/- 0.544475
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