content: from collections import namedtuple
  
  import numpy as np
  
  from . import distributions
  
  
  __all__ = ['_find_repeats', 'linregress', 'theilslopes']
  
  
  def linregress(x, y=None):
      """
      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.
          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.
  
      Returns
      -------
      slope : float
          slope of the regression line
      intercept : float
          intercept of the regression line
      rvalue : float
          correlation coefficient
      pvalue : float
          two-sided p-value for a hypothesis test whose null hypothesis is
          that the slope is zero.
      stderr : float
          Standard error of the estimated gradient.
  
      See also
      --------
      optimize.curve_fit : Use non-linear least squares to fit a function to data.
      optimize.leastsq : Minimize the sum of squares of a set of equations.
  
      Examples
      --------
      >>> from scipy import stats
      >>> np.random.seed(12345678)
      >>> x = np.random.random(10)
      >>> y = np.random.random(10)
      >>> slope, intercept, r_value, p_value, std_err = stats.linregress(x,y)
  
      # To get coefficient of determination (r_squared)
  
      >>> print("r-squared:", r_value**2)
      ('r-squared:', 0.080402268539028335)
  
      """
      TINY = 1.0e-20
      if y is None:  # x is a (2, N) or (N, 2) shaped array_like
          x = np.asarray(x)
          if x.shape[0] == 2:
              x, y = x
          elif x.shape[1] == 2:
              x, y = x.T
          else:
              msg = ("If only `x` is given as input, it has to be of shape "
                     "(2, N) or (N, 2), provided shape was %s" % str(x.shape))
              raise ValueError(msg)
      else:
          x = np.asarray(x)
          y = np.asarray(y)
  
      if x.size == 0 or y.size == 0:
          raise ValueError("Inputs must not be empty.")
  
      n = len(x)
      xmean = np.mean(x, None)
      ymean = np.mean(y, None)
  
      # average sum of squares:
      ssxm, ssxym, ssyxm, ssym = np.cov(x, y, bias=1).flat
      r_num = ssxym
      r_den = np.sqrt(ssxm * ssym)
      if r_den == 0.0:
          r = 0.0
      else:
          r = r_num / r_den
          # test for numerical error propagation
          if r > 1.0:
              r = 1.0
          elif r < -1.0:
              r = -1.0
  
      df = n - 2
      t = r * np.sqrt(df / ((1.0 - r + TINY)*(1.0 + r + TINY)))
      prob = 2 * distributions.t.sf(np.abs(t), df)
      slope = r_num / ssxm
      intercept = ymean - slope*xmean
      sterrest = np.sqrt((1 - r**2) * ssym / ssxm / df)
  
      LinregressResult = namedtuple('LinregressResult', ('slope', 'intercept',
                                                         'rvalue', 'pvalue',
                                                         'stderr'))
      return LinregressResult(slope, intercept, r, prob, sterrest)
  
  
  def theilslopes(y, x=None, alpha=0.95):
      r"""
      Computes the Theil-Sen estimator for a set of points (x, y).
  
      `theilslopes` implements a method for robust linear regression.  It
      computes the slope as the median of all slopes between paired values.
  
      Parameters
      ----------
      y : array_like
          Dependent variable.
      x : array_like or None, optional
          Independent variable. If None, use ``arange(len(y))`` instead.
      alpha : float, optional
          Confidence degree between 0 and 1. Default is 95% confidence.
          Note that `alpha` is symmetric around 0.5, i.e. both 0.1 and 0.9 are
          interpreted as "find the 90% confidence interval".
  
      Returns
      -------
      medslope : float
          Theil slope.
      medintercept : float
          Intercept of the Theil line, as ``median(y) - medslope*median(x)``.
      lo_slope : float
          Lower bound of the confidence interval on `medslope`.
      up_slope : float
          Upper bound of the confidence interval on `medslope`.
  
      Notes
      -----
      The implementation of `theilslopes` follows [1]_. The intercept is
      not defined in [1]_, and here it is defined as ``median(y) -
      medslope*median(x)``, which is given in [3]_. Other definitions of
      the intercept exist in the literature. A confidence interval for
      the intercept is not given as this question is not addressed in
      [1]_.
  
      References
      ----------
      .. [1] P.K. Sen, "Estimates of the regression coefficient based on Kendall's tau",
             J. Am. Stat. Assoc., Vol. 63, pp. 1379-1389, 1968.
      .. [2] H. Theil, "A rank-invariant method of linear and polynomial
             regression analysis I, II and III",  Nederl. Akad. Wetensch., Proc.
             53:, pp. 386-392, pp. 521-525, pp. 1397-1412, 1950.
      .. [3] W.L. Conover, "Practical nonparametric statistics", 2nd ed.,
             John Wiley and Sons, New York, pp. 493.
  
      Examples
      --------
      >>> 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, intercept and 90% confidence interval.  For comparison,
      also compute the least-squares fit with `linregress`:
  
      >>> res = stats.theilslopes(y, x, 0.90)
      >>> lsq_res = stats.linregress(x, y)
  
      Plot the results. The Theil-Sen regression line is shown in red, with the
      dashed red lines illustrating the confidence interval of the slope (note
      that the dashed red lines are not the confidence interval of the regression
      as the confidence interval of the intercept is not included). 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, res[1] + res[2] * x, 'r--')
      >>> ax.plot(x, res[1] + res[3] * x, 'r--')
      >>> ax.plot(x, lsq_res[1] + lsq_res[0] * x, 'g-')
      >>> plt.show()
  
      """
      # We copy both x and y so we can use _find_repeats.
      y = np.array(y).flatten()
      if x is None:
          x = np.arange(len(y), dtype=float)
      else:
          x = np.array(x, dtype=float).flatten()
          if len(x) != len(y):
              raise ValueError("Incompatible lengths ! (%s<>%s)" % (len(y), len(x)))
  
      # Compute sorted slopes only when deltax > 0
      deltax = x[:, np.newaxis] - x
      deltay = y[:, np.newaxis] - y
      slopes = deltay[deltax > 0] / deltax[deltax > 0]
      slopes.sort()
      medslope = np.median(slopes)
      medinter = np.median(y) - medslope * np.median(x)
      # Now compute confidence intervals
      if alpha > 0.5:
          alpha = 1. - alpha
  
      z = distributions.norm.ppf(alpha / 2.)
      # This implements (2.6) from Sen (1968)
      _, nxreps = _find_repeats(x)
      _, nyreps = _find_repeats(y)
      nt = len(slopes)       # N in Sen (1968)
      ny = len(y)            # n in Sen (1968)
      # Equation 2.6 in Sen (1968):
      sigsq = 1/18. * (ny * (ny-1) * (2*ny+5) -
                       np.sum(k * (k-1) * (2*k + 5) for k in nxreps) -
                       np.sum(k * (k-1) * (2*k + 5) for k in nyreps))
      # Find the confidence interval indices in `slopes`
      sigma = np.sqrt(sigsq)
      Ru = min(int(np.round((nt - z*sigma)/2.)), len(slopes)-1)
      Rl = max(int(np.round((nt + z*sigma)/2.)) - 1, 0)
      delta = slopes[[Rl, Ru]]
      return medslope, medinter, delta[0], delta[1]
  
  
  def _find_repeats(arr):
      # This function assumes it may clobber its input.
      if len(arr) == 0:
          return np.array(0, np.float64), np.array(0, np.intp)
  
      # XXX This cast was previously needed for the Fortran implementation,
      # should we ditch it?
      arr = np.asarray(arr, np.float64).ravel()
      arr.sort()
  
      # Taken from NumPy 1.9's np.unique.
      change = np.concatenate(([True], arr[1:] != arr[:-1]))
      unique = arr[change]
      change_idx = np.concatenate(np.nonzero(change) + ([arr.size],))
      freq = np.diff(change_idx)
      atleast2 = freq > 1
      return unique[atleast2], freq[atleast2]
  