Python's implementation of Mutual Information

Question

I am having some issues implementing the Mutual Information Function that Python\'s machine learning libraries provide, in particular : sklearn.metrics.mutual_info_score(labels

Answer 1

I encountered the same issue today. After a few trials I found the real reason: you take log2 if you strictly followed NLP tutorial, but sklearn.metrics.mutual_info_score uses natural logarithm(base e, Euler's number). I didn't find this detail in sklearn documentation...

I verified this by:

import numpy as np
def computeMI(x, y):
    sum_mi = 0.0
    x_value_list = np.unique(x)
    y_value_list = np.unique(y)
    Px = np.array([ len(x[x==xval])/float(len(x)) for xval in x_value_list ]) #P(x)
    Py = np.array([ len(y[y==yval])/float(len(y)) for yval in y_value_list ]) #P(y)
    for i in xrange(len(x_value_list)):
        if Px[i] ==0.:
            continue
        sy = y[x == x_value_list[i]]
        if len(sy)== 0:
            continue
        pxy = np.array([len(sy[sy==yval])/float(len(y))  for yval in y_value_list]) #p(x,y)
        t = pxy[Py>0.]/Py[Py>0.] /Px[i] # log(P(x,y)/( P(x)*P(y))
        sum_mi += sum(pxy[t>0]*np.log2( t[t>0]) ) # sum ( P(x,y)* log(P(x,y)/( P(x)*P(y)) )
    return sum_mi

If you change this np.log2 to np.log, I think it would give you the same answer as sklearn. The only difference is that when this method returns 0, sklearn will return a number very near to 0. ( And of course, use sklearn if you don't care about log base, my piece of code is just for demo, it gives poor performance...)

FYI, 1)sklearn.metrics.mutual_info_score takes lists as well as np.array; 2) the sklearn.metrics.cluster.entropy uses also log, not log2

Edit: as for "same result", I'm not sure what you really mean. In general, the values in the vectors don't really matter, it is the "distribution" of values that matters. You care about P(X=x), P(Y=y) and P(X=x,Y=y), not the value x,y.

Answer 2

The code below should provided a result: 0.00011053558610110256

c=np.concatenate([np.ones(49), np.zeros(27652), np.ones(141), np.zeros(774106) ])
t=np.concatenate([np.ones(49), np.ones(27652), np.zeros(141), np.zeros(774106)])

computeMI(c,t)