Amir Amir - 6 months ago 6
Python Question

Efficient way of computing Kullback–Leibler divergence in Python

I have to compute the Kullback-Leibler Divergence (KLD) between thousands of discrete probability vectors. Currently I am using the following code but it's way too slow for my purposes. I was wondering if there is any faster way to compute KL Divergence?

import numpy as np
import scipy.stats as sc

#n is the number of data points
kld = np.zeros(n, n)
for i in range(0, n):
for j in range(0, n):
if(i != j):
kld[i, j] = sc.entropy(distributions[i, :], distributions[j, :])


Here's the answer (by @Divakar):

kld = sc.entropy(distrib.T[:,:,None], distrib.T[:,None,:])


Scipy's stats.entropy in its default sense invites inputs as 1D arrays giving us a scalar, which is being done in the listed question. Internally this function also allows broadcasting, which we can abuse in here for a vectorized solution.

From the docs -

scipy.stats.entropy(pk, qk=None, base=None)

If only probabilities pk are given, the entropy is calculated as S = -sum(pk * log(pk), axis=0).

If qk is not None, then compute the Kullback-Leibler divergence S = sum(pk * log(pk / qk), axis=0).

In our case, we are doing these entropy calculations for each row against all rows, performing sum reductions to have a scalar at each iteration with those two nested loops. Thus, the output array would be of shape (M,M), where M is the number of rows in input array.

Now, the catch here is that stats.entropy() would sum along axis=0, so we will feed it two versions of distributions, both of whom would have the rowth-dimension brought to axis=0 for reduction along it and the other two axes interleaved - (M,1) & (1,M) to give us a (M,M) shaped output array using broadcasting.

Thus, a vectorized and much more efficient way to solve our case would be -

from scipy import stats
kld = stats.entropy(distributions.T[:,:,None], distributions.T[:,None,:])

Runtime tests and verify -

In [15]: def entropy_loopy(distrib):
    ...:     n = distrib.shape[0] #n is the number of data points
    ...:     kld = np.zeros((n, n))
    ...:     for i in range(0, n):
    ...:         for j in range(0, n):
    ...:             if(i != j):
    ...:                 kld[i, j] = stats.entropy(distrib[i, :], distrib[j, :])
    ...:     return kld

In [16]: distrib = np.random.randint(0,9,(100,100)) # Setup input

In [17]: out = stats.entropy(distrib.T[:,:,None], distrib.T[:,None,:])

In [18]: np.allclose(entropy_loopy(distrib),out) # Verify
Out[18]: True

In [19]: %timeit entropy_loopy(distrib)
1 loops, best of 3: 800 ms per loop

In [20]: %timeit stats.entropy(distrib.T[:,:,None], distrib.T[:,None,:])
10 loops, best of 3: 104 ms per loop