import numpy as np

from . import distributions

from .._lib._bunch import _make_tuple_bunch

__all__ = ['_find_repeats', 'linregress', 'theilslopes', 'siegelslopes']

# This is not a namedtuple for backwards compatibility. See PR #12983
LinregressResult = _make_tuple_bunch('LinregressResult',
                                     ['slope', 'intercept', 'rvalue',
                                      'pvalue', 'stderr'],
                                     extra_field_names=['intercept_stderr'])


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. In
        the case where ``y=None`` and `x` is a 2x2 array, ``linregress(x)`` is
        equivalent to ``linregress(x[0], x[1])``.

    Returns
    -------
    result : ``LinregressResult`` instance
        The return value is an object with the following attributes:

        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, using Wald Test with t-distribution of
            the test statistic.
        stderr : float
            Standard error of the estimated slope (gradient), under the
            assumption of residual normality.
        intercept_stderr : float
            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
    -----
    Missing values are considered pair-wise: if a value is missing in `x`,
    the corresponding value in `y` is masked.

    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 matplotlib.pyplot as plt
    >>> from scipy import stats

    Generate some data:

    >>> np.random.seed(12345678)
    >>> x = np.random.random(10)
    >>> y = 1.6*x + np.random.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: 0.735498

    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%): 1.944864 +/- 0.950885
    >>> print(f"intercept (95%): {res.intercept:.6f}"
    ...       f" +/- {ts*res.intercept_stderr:.6f}")
    intercept (95%): 0.268578 +/- 0.488822

    """
    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:
            raise ValueError("If only `x` is given as input, it has to "
                             "be of shape (2, N) or (N, 2); provided shape "
                             f"was {x.shape}.")
    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 sums of square differences from the mean
    #   ssxm = mean( (x-mean(x))^2 )
    #   ssxym = mean( (x-mean(x)) * (y-mean(y)) )
    ssxm, ssxym, _, ssym = np.cov(x, y, bias=1).flat

    # R-value
    #   r = ssxym / sqrt( ssxm * ssym )
    if ssxm == 0.0 or ssym == 0.0:
        # If the denominator was going to be 0
        r = 0.0
    else:
        r = ssxym / np.sqrt(ssxm * ssym)
        # Test for numerical error propagation (make sure -1 < r < 1)
        if r > 1.0:
            r = 1.0
        elif r < -1.0:
            r = -1.0

    slope = ssxym / ssxm
    intercept = ymean - slope*xmean
    if n == 2:
        # handle case when only two points are passed in
        if y[0] == y[1]:
            prob = 1.0
        else:
            prob = 0.0
        slope_stderr = 0.0
        intercept_stderr = 0.0
    else:
        df = n - 2  # Number of degrees of freedom
        # n-2 degrees of freedom because 2 has been used up
        # to estimate the mean and standard deviation
        t = r * np.sqrt(df / ((1.0 - r + TINY)*(1.0 + r + TINY)))
        prob = 2 * distributions.t.sf(np.abs(t), df)
        slope_stderr = np.sqrt((1 - r**2) * ssym / ssxm / df)

        # Also calculate the standard error of the intercept
        # The following relationship is used:
        #   ssxm = mean( (x-mean(x))^2 )
        #        = ssx - sx*sx
        #        = mean( x^2 ) - mean(x)^2
        intercept_stderr = slope_stderr * np.sqrt(ssxm + xmean**2)

    return LinregressResult(slope=slope, intercept=intercept, rvalue=r,
                            pvalue=prob, stderr=slope_stderr,
                            intercept_stderr=intercept_stderr)


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`.

    See also
    --------
    siegelslopes : a similar technique using repeated medians

    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) -
                     sum(k * (k-1) * (2*k + 5) for k in nxreps) -
                     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]


def siegelslopes(y, x=None, method="hierarchical"):
    r"""
    Computes the Siegel estimator for a set of points (x, y).

    `siegelslopes` implements a method for robust linear regression
    using repeated medians (see [1]_) to fit a line to the points (x, y).
    The method is robust to outliers with an asymptotic breakdown point
    of 50%.

    Parameters
    ----------
    y : array_like
        Dependent variable.
    x : array_like or None, optional
        Independent variable. If None, use ``arange(len(y))`` instead.
    method : {'hierarchical', 'separate'}
        If 'hierarchical', estimate the intercept using the estimated
        slope ``medslope`` (default option).
        If 'separate', estimate the intercept independent of the estimated
        slope. See Notes for details.

    Returns
    -------
    medslope : float
        Estimate of the slope of the regression line.
    medintercept : float
        Estimate of the intercept of the regression line.

    See also
    --------
    theilslopes : a similar technique without repeated medians

    Notes
    -----
    With ``n = len(y)``, compute ``m_j`` as the median of
    the slopes from the point ``(x[j], y[j])`` to all other `n-1` points.
    ``medslope`` is then the median of all slopes ``m_j``.
    Two ways are given to estimate the intercept in [1]_ which can be chosen
    via the parameter ``method``.
    The hierarchical approach uses the estimated slope ``medslope``
    and computes ``medintercept`` as the median of ``y - medslope*x``.
    The other approach estimates the intercept separately as follows: for
    each point ``(x[j], y[j])``, compute the intercepts of all the `n-1`
    lines through the remaining points and take the median ``i_j``.
    ``medintercept`` is the median of the ``i_j``.

    The implementation computes `n` times the median of a vector of size `n`
    which can be slow for large vectors. There are more efficient algorithms
    (see [2]_) which are not implemented here.

    References
    ----------
    .. [1] A. Siegel, "Robust Regression Using Repeated Medians",
           Biometrika, Vol. 69, pp. 242-244, 1982.

    .. [2] A. Stein and M. Werman, "Finding the repeated median regression
           line", Proceedings of the Third Annual ACM-SIAM Symposium on
           Discrete Algorithms, pp. 409-413, 1992.

    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 and intercept.  For comparison, also compute the
    least-squares fit with `linregress`:

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

    Plot the results. The Siegel regression line is shown in red. 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, lsq_res[1] + lsq_res[0] * x, 'g-')
    >>> plt.show()

    """
    if method not in ['hierarchical', 'separate']:
        raise ValueError("method can only be 'hierarchical' or 'separate'")
    y = np.asarray(y).ravel()
    if x is None:
        x = np.arange(len(y), dtype=float)
    else:
        x = np.asarray(x, dtype=float).ravel()
        if len(x) != len(y):
            raise ValueError("Incompatible lengths ! (%s<>%s)" %
                             (len(y), len(x)))

    deltax = x[:, np.newaxis] - x
    deltay = y[:, np.newaxis] - y
    slopes, intercepts = [], []

    for j in range(len(x)):
        id_nonzero = deltax[j, :] != 0
        slopes_j = deltay[j, id_nonzero] / deltax[j, id_nonzero]
        medslope_j = np.median(slopes_j)
        slopes.append(medslope_j)
        if method == 'separate':
            z = y*x[j] - y[j]*x
            medintercept_j = np.median(z[id_nonzero] / deltax[j, id_nonzero])
            intercepts.append(medintercept_j)

    medslope = np.median(np.asarray(slopes))
    if method == "separate":
        medinter = np.median(np.asarray(intercepts))
    else:
        medinter = np.median(y - medslope*x)

    return medslope, medinter
