Computing the correlation coefficient between two multi-dimensional arrays

Question

I have two arrays that have the shapes N X T and M X T. I'd like to compute the correlation coefficient across T between every possible pair of rows n and m (from N and M, respectively).

What's the fastest, most pythonic way to do this? (Looping over N and M would seem to me to be neither fast nor pythonic.) I'm expecting the answer to involve numpy and/or scipy. Right now my arrays are numpy arrays, but I'm open to converting them to a different type.

I'm expecting my output to be an array with the shape N X M.

N.B. When I say "correlation coefficient," I mean the Pearson product-moment correlation coefficient.

Here are some things to note:

• The numpy function correlate requires input arrays to be one-dimensional.
• The numpy function corrcoef accepts two-dimensional arrays, but they must have the same shape.
• The scipy.stats function pearsonr requires input arrays to be one-dimensional.

Answer 1

Correlation (default 'valid' case) between two 2D arrays:

You can simply use matrix-multiplication np.dot like so -

out = np.dot(arr_one,arr_two.T)

Correlation with the default "valid" case between each pairwise row combinations (row1,row2) of the two input arrays would correspond to multiplication result at each (row1,row2) position.

---

Row-wise Correlation Coefficient calculation for two 2D arrays:

def corr2_coeff(A, B):
    # Rowwise mean of input arrays & subtract from input arrays themeselves
    A_mA = A - A.mean(1)[:, None]
    B_mB = B - B.mean(1)[:, None]

    # Sum of squares across rows
    ssA = (A_mA**2).sum(1)
    ssB = (B_mB**2).sum(1)

    # Finally get corr coeff
    return np.dot(A_mA, B_mB.T) / np.sqrt(np.dot(ssA[:, None],ssB[None]))

This is based upon this solution to How to apply corr2 functions in Multidimentional arrays in MATLAB

Benchmarking

This section compares runtime performance with the proposed approach against generate_correlation_map & loopy pearsonr based approach listed in the other answer.(taken from the function test_generate_correlation_map() without the value correctness verification code at the end of it). Please note the timings for the proposed approach also include a check at the start to check for equal number of columns in the two input arrays, as also done in that other answer. The runtimes are listed next.

Case #1:

In [106]: A = np.random.rand(1000, 100)

In [107]: B = np.random.rand(1000, 100)

In [108]: %timeit corr2_coeff(A, B)
100 loops, best of 3: 15 ms per loop

In [109]: %timeit generate_correlation_map(A, B)
100 loops, best of 3: 19.6 ms per loop

Case #2:

In [110]: A = np.random.rand(5000, 100)

In [111]: B = np.random.rand(5000, 100)

In [112]: %timeit corr2_coeff(A, B)
1 loops, best of 3: 368 ms per loop

In [113]: %timeit generate_correlation_map(A, B)
1 loops, best of 3: 493 ms per loop

Case #3:

In [114]: A = np.random.rand(10000, 10)

In [115]: B = np.random.rand(10000, 10)

In [116]: %timeit corr2_coeff(A, B)
1 loops, best of 3: 1.29 s per loop

In [117]: %timeit generate_correlation_map(A, B)
1 loops, best of 3: 1.83 s per loop

The other loopy pearsonr based approach seemed too slow, but here are the runtimes for one small datasize -

In [118]: A = np.random.rand(1000, 100)

In [119]: B = np.random.rand(1000, 100)

In [120]: %timeit corr2_coeff(A, B)
100 loops, best of 3: 15.3 ms per loop

In [121]: %timeit generate_correlation_map(A, B)
100 loops, best of 3: 19.7 ms per loop

In [122]: %timeit pearsonr_based(A, B)
1 loops, best of 3: 33 s per loop

Answer 2

@Divakar provides a great option for computing the unscaled correlation, which is what I originally asked for.

In order to calculate the correlation coefficient, a bit more is required:

import numpy as np

def generate_correlation_map(x, y):
    """Correlate each n with each m.

    Parameters
    ----------
    x : np.array
      Shape N X T.

    y : np.array
      Shape M X T.

    Returns
    -------
    np.array
      N X M array in which each element is a correlation coefficient.

    """
    mu_x = x.mean(1)
    mu_y = y.mean(1)
    n = x.shape[1]
    if n != y.shape[1]:
        raise ValueError('x and y must ' +
                         'have the same number of timepoints.')
    s_x = x.std(1, ddof=n - 1)
    s_y = y.std(1, ddof=n - 1)
    cov = np.dot(x,
                 y.T) - n * np.dot(mu_x[:, np.newaxis],
                                  mu_y[np.newaxis, :])
    return cov / np.dot(s_x[:, np.newaxis], s_y[np.newaxis, :])

Here's a test of this function, which passes:

from scipy.stats import pearsonr

def test_generate_correlation_map():
    x = np.random.rand(10, 10)
    y = np.random.rand(20, 10)
    desired = np.empty((10, 20))
    for n in range(x.shape[0]):
        for m in range(y.shape[0]):
            desired[n, m] = pearsonr(x[n, :], y[m, :])[0]
    actual = generate_correlation_map(x, y)
    np.testing.assert_array_almost_equal(actual, desired)

Answer 3

For those interested in computing the Pearson correlation coefficient between a 1D and 2D array, I wrote the following function, where x is a 1D array and y a 2D array.

def pearsonr_2D(x, y):
    """computes pearson correlation coefficient
       where x is a 1D and y a 2D array"""

    upper = np.sum((x - np.mean(x)) * (y - np.mean(y, axis=1)[:,None]), axis=1)
    lower = np.sqrt(np.sum(np.power(x - np.mean(x), 2)) * np.sum(np.power(y - np.mean(y, axis=1)[:,None], 2), axis=1))
    
    rho = upper / lower
    
    return rho

Example run:

>>> x
Out[1]: array([1, 2, 3])

>>> y
Out[2]: array([[ 1,  2,  3],
               [ 6,  7, 12],
               [ 9,  3,  1]])

>>> pearsonr_2D(x, y)
Out[3]: array([ 1.        ,  0.93325653, -0.96076892])