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scipy.stats.

# linregress

scipy.stats.linregress(x, y=None, alternative='two-sided')

  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 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 where `y=None` and x is a 2xN array, `linregress(x)` is equivalent to `linregress(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.

  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)

  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