Optimal way to compute pairwise mutual information

Question

For an m x n matrix, what's the optimal (fastest) way to compute the mutual information for all pairs of columns (n x n)?

By mutual information, I mean:

I(X, Y) = H(X) + H(Y) - H(X,Y)

where H(X) refers to the Shannon entropy of X.

Currently I'm using np.histogram2d and np.histogram to calculate the joint (X,Y) and individual (X or Y) counts. For a given matrix A (e.g. a 250000 X 1000 matrix of floats), I am doing a nested for loop,

    n = A.shape[1]
    for ix = arange(n)  
        for jx = arange(ix+1,n):
           matMI[ix,jx]= calc_MI(A[:,ix],A[:,jx])

Surely there must be better/faster ways to do this?

As an aside, I've also looked for mapping functions on columns (column-wise or row-wise operations) on arrays, but haven't found a good general answer yet.

Here is my full implementation, following the conventions in the Wiki page:

import numpy as np

def calc_MI(X,Y,bins):

   c_XY = np.histogram2d(X,Y,bins)[0]
   c_X = np.histogram(X,bins)[0]
   c_Y = np.histogram(Y,bins)[0]

   H_X = shan_entropy(c_X)
   H_Y = shan_entropy(c_Y)
   H_XY = shan_entropy(c_XY)

   MI = H_X + H_Y - H_XY
   return MI

def shan_entropy(c):
    c_normalized = c / float(np.sum(c))
    c_normalized = c_normalized[np.nonzero(c_normalized)]
    H = -sum(c_normalized* np.log2(c_normalized))  
    return H

A = np.array([[ 2.0,  140.0,  128.23, -150.5, -5.4  ],
              [ 2.4,  153.11, 130.34, -130.1, -9.5  ],
              [ 1.2,  156.9,  120.11, -110.45,-1.12 ]])

bins = 5 # ?
n = A.shape[1]
matMI = np.zeros((n, n))

for ix in np.arange(n):
    for jx in np.arange(ix+1,n):
        matMI[ix,jx] = calc_MI(A[:,ix], A[:,jx], bins)

Although my working version with nested for loops does it at reasonable speed, I'd like to know if there is a more optimal way to apply calc_MI on all the columns of A (to calculate their pairwise mutual information)?

I'd also like to know:

1. Whether there are efficient ways to map functions to operate on columns (or rows) of np.arrays (maybe like np.vectorize, which looks more like a decorator)?

2. Whether there are other optimal implementations for this specific calculation (mutual information)?

Answer 1

I can't suggest a faster calculation for the outer loop over the n*(n-1)/2 vectors, but your implementation of calc_MI(x, y, bins) can be simplified if you can use scipy version 0.13 or scikit-learn.

In scipy 0.13, the lambda_ argument was added to scipy.stats.chi2_contingency This argument controls the statistic that is computed by the function. If you use lambda_="log-likelihood" (or lambda_=0), the log-likelihood ratio is returned. This is also often called the G or G² statistic. Other than a factor of 2*n (where n is the total number of samples in the contingency table), this is the mutual information. So you could implement calc_MI as:

from scipy.stats import chi2_contingency

def calc_MI(x, y, bins):
    c_xy = np.histogram2d(x, y, bins)[0]
    g, p, dof, expected = chi2_contingency(c_xy, lambda_="log-likelihood")
    mi = 0.5 * g / c_xy.sum()
    return mi

The only difference between this and your implementation is that this implementation uses the natural logarithm instead of the base-2 logarithm (so it is expressing the information in "nats" instead of "bits"). If you really prefer bits, just divide mi by log(2).

If you have (or can install) sklearn (i.e. scikit-learn), you can use sklearn.metrics.mutual_info_score, and implement calc_MI as:

from sklearn.metrics import mutual_info_score

def calc_MI(x, y, bins):
    c_xy = np.histogram2d(x, y, bins)[0]
    mi = mutual_info_score(None, None, contingency=c_xy)
    return mi