linregress#
- scipy.stats.linregress(x, y=None, alternative='two-sided')[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. 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 wherey=None
and x is a 2xN array,linregress(x)
is equivalent tolinregress(x[0], x[1])
.Deprecated since version 1.14.0: Inference of the two sets of measurements from a single argument x is deprecated will result in an error in SciPy 1.16.0; the sets must be specified separately as x and y.
- 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.
- Returns:
- result
LinregressResult
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.
- result
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 fieldsslope
,intercept
,rvalue
,pvalue
andstderr
, 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)
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()
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