矩阵的每列与向量之间的numpy协方差

Question

Based on this post , I can get covariance between two vectors using np.cov((x,y), rowvar=0) . I have a matrix MxN and a vector Mx1. I want to find the covariance between each column of the matrix and the given vector. I know that I can use for loop to write. I was wondering if I can somehow use np.cov() to get the result directly.

Answer 1

As Warren Weckesser said, the numpy.cov(X, Y) is a poor fit for the job because it will simply join the arrays in one M by (N+1) array and find the huge (N+1) by (N+1) covariance matrix. But we'll always have the definition of covariance and it's easy to use:

A = np.sqrt(np.arange(12).reshape(3, 4))   # some 3 by 4 array 
b = np.array([[2], [4], [5]])              # some 3 by 1 vector
cov = np.dot(b.T - b.mean(), A - A.mean(axis=0)) / (b.shape[0]-1)

This returns the covariances of each column of A with b.

array([[ 2.21895142,  1.53934466,  1.3379221 ,  1.20866607]])

The formula I used is for sample covariance (which is what numpy.cov computes, too), hence the division by (b.shape[0] - 1). If you divide by b.shape[0] you get the unadjusted population covariance .

For comparison, the same computation using np.cov :

import numpy as np
A = np.sqrt(np.arange(12).reshape(3, 4))
b = np.array([[2], [4], [5]])
np.cov(A, b, rowvar=False)[-1, :-1]

Same output, but it takes about twice this long (and for large matrices, the difference will be much larger). The slicing at the end is because np.cov computes a 5 by 5 matrix, in which only the first 4 entries of the last row are what you wanted. The rest is covariance of A with itself, or of b with itself.

Correlation coefficient

The correlation coefficientis obtained by dividing by square roots of variances. Watch out for that -1 adjustment mentioned earlier: numpy.var does not make it by default, to make it happen you need ddof=1 parameter.

corr = cov / np.sqrt(np.var(b, ddof=1) * np.var(A, axis=0, ddof=1)) 

Check that the output is the same as the less efficient version

np.corrcoef(A, b, rowvar=False)[-1, :-1]