Optimal way to compute pairwise mutual information using numpy

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