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# scipy.stats.mstats.sen_seasonal_slopes

scipy.stats.mstats.sen_seasonal_slopes(x)

  Computes seasonal Theil-Sen and Kendall slope estimators.

  The seasonal generalization of Sen’s slope computes the slopes between all pairs of values within a “season” (column) of a 2D array. It returns an array containing the median of these “within-season” slopes for each season (the Theil-Sen slope estimator of each season), and it returns the median of the within-season slopes across all seasons (the seasonal Kendall slope estimator).

  Parameters:
    x 2D array_like

      Each column of x contains measurements of the dependent variable within a season. The independent variable (usually time) of each season is assumed to be `np.arange(x.shape[0])`.

  Returns:
    result `SenSeasonalSlopesResult` instance

      The return value is an object with the following attributes:

      intra_slopendarray

        For each season, the Theil-Sen slope estimator: the median of within-season slopes.

      inter_slopefloat

        The seasonal Kendall slope estimateor: the median of within-season slopes across all seasons.

  See also

  `theilslopes`

    the analogous function for non-seasonal data

  `scipy.stats.theilslopes`

    non-seasonal slopes for non-masked arrays

  Notes

  The slopes \(d_{ijk}\) within season \(i\) are:

  \[d_{ijk} = \frac{x_{ij} - x_{ik}} {j - k}\]

  for pairs of distinct integer indices \(j, k\) of \(x\).

  Element \(i\) of the returned intra_slope array is the median of the \(d_{ijk}\) over all \(j < k\); this is the Theil-Sen slope estimator of season \(i\). The returned inter_slope value, better known as the seasonal Kendall slope estimator, is the median of the \(d_{ijk}\) over all \(i, j, k\).

  References

  [1]

  Hirsch, Robert M., James R. Slack, and Richard A. Smith. “Techniques of trend analysis for monthly water quality data.” Water Resources Research 18.1 (1982): 107-121.

  Examples

  Suppose we have 100 observations of a dependent variable for each of four seasons:

      >>> import numpy as np
      >>> rng = np.random.default_rng()
      >>> x = rng.random(size=(100, 4))

  We compute the seasonal slopes as:

      >>> from scipy import stats
      >>> intra_slope, inter_slope = stats.mstats.sen_seasonal_slopes(x)

  If we define a function to compute all slopes between observations within a season:

      >>> def dijk(yi):
      ...     n = len(yi)
      ...     x = np.arange(n)
      ...     dy = yi - yi[:, np.newaxis]
      ...     dx = x - x[:, np.newaxis]
      ...     # we only want unique pairs of distinct indices
      ...     mask = np.triu(np.ones((n, n), dtype=bool), k=1)
      ...     return dy[mask]/dx[mask]

  then element `i` of `intra_slope` is the median of `dijk[x[:, i]]`:

      >>> i = 2
      >>> np.allclose(np.median(dijk(x[:, i])), intra_slope[i])
      True

  and `inter_slope` is the median of the values returned by `dijk` for all seasons:

      >>> all_slopes = np.concatenate([dijk(x[:, i]) for i in range(x.shape[1])])
      >>> np.allclose(np.median(all_slopes), inter_slope)
      True

  Because the data are randomly generated, we would expect the median slopes to be nearly zero both within and across all seasons, and indeed they are:

      >>> intra_slope.data
      array([ 0.00124504, -0.00277761, -0.00221245, -0.00036338])
      >>> inter_slope
      -0.0010511779872922058