Distance correlation with p-value

distcorr.py

from scipy.spatial.distance import pdist, squareform
import numpy as np
import copy


def distcorr(Xval, Yval, pval=True, nruns=500):
    """ Compute the distance correlation function, returning the p-value.
    Based on Satra/distcorr.py (gist aa3d19a12b74e9ab7941)

    >>> a = [1,2,3,4,5]
    >>> b = np.array([1,2,9,4,4])
    >>> distcorr(a, b)
    (0.76267624241686671, 0.266)
    """
    X = np.atleast_1d(Xval)
    Y = np.atleast_1d(Yval)
    if np.prod(X.shape) == len(X):
        X = X[:, None]
    if np.prod(Y.shape) == len(Y):
        Y = Y[:, None]
    X = np.atleast_2d(X)
    Y = np.atleast_2d(Y)
    n = X.shape[0]
    if Y.shape[0] != X.shape[0]:
        raise ValueError('Number of samples must match')
    a = squareform(pdist(X))
    b = squareform(pdist(Y))
    A = a - a.mean(axis=0)[None, :] - a.mean(axis=1)[:, None] + a.mean()
    B = b - b.mean(axis=0)[None, :] - b.mean(axis=1)[:, None] + b.mean()

    dcov2_xy = (A * B).sum() / float(n * n)
    dcov2_xx = (A * A).sum() / float(n * n)
    dcov2_yy = (B * B).sum() / float(n * n)
    dcor = np.sqrt(dcov2_xy) / np.sqrt(np.sqrt(dcov2_xx) * np.sqrt(dcov2_yy))

    if pval:
        greater = 0
        for i in range(nruns):
            Y_r = copy.copy(Yval)
            np.random.shuffle(Y_r)
            if distcorr(Xval, Y_r, pval=False) > dcor:
                greater += 1
        return (dcor, greater / float(nruns))
    else:
        return dcor

Hi @B1azingB1ade, thanks for your comments! I have changed the code to use np.random.shuffle. According to the docs, it only shuffles along the first axis: https://docs.scipy.org/doc/numpy-1.14.0/reference/generated/numpy.random.shuffle.html… they seem to be cooking a way to do this: numpy/numpy#5173.

@kirk86 yes, it seems like "correlation distance" is different from "distance correlation".

@rmill040, correct! Thanks for spotting this bug, I just fixed it.

Hi,
I think Xval should be shuffled with Yval. or your distcorr([1,2,3],[2,4,6]) will not give the correct p-value, 1 in this case.

I tried running distcorr([1,2,3],[2,4,6], nruns=100000) both shuffling Xval and not shuffling it. Both cases returns a distance correlation of 1 with p-value of 0.33. This means that in both cases, 33% of the random shuffles also produce a distance correlation of 1. If we change if distcorr(Xval, Y_r, pval=False) >= dcor to > dcor, then no random shuffling makes that line be true, and the p-value is zero. I'm unsure whether we should use > instead of >=, but I've made the change anyways, because in the example you described it makes sense that the the p-value is zero.