Jensen-Shannon Divergence

Question

I have another question that I was hoping someone could help me with.

I\'m using the Jensen-Shannon-Divergence to measure the similarity between two probability distribu

Answer 1

Note that the scipy entropy call below is the Kullback-Leibler divergence.

See: http://en.wikipedia.org/wiki/Jensen%E2%80%93Shannon_divergence

#!/usr/bin/env python
from scipy.stats import entropy
from numpy.linalg import norm
import numpy as np

def JSD(P, Q):
    _P = P / norm(P, ord=1)
    _Q = Q / norm(Q, ord=1)
    _M = 0.5 * (_P + _Q)
    return 0.5 * (entropy(_P, _M) + entropy(_Q, _M))

Also note that the test case in the Question looks erred?? The sum of the p distribution does not add to 1.0.

See: http://www.itl.nist.gov/div898/handbook/eda/section3/eda361.htm

Answer 2

A general version, for n probability distributions, in python

import numpy as np
from scipy.stats import entropy as H


def JSD(prob_distributions, weights, logbase=2):
    # left term: entropy of misture
    wprobs = weights * prob_distributions
    mixture = wprobs.sum(axis=0)
    entropy_of_mixture = H(mixture, base=logbase)

    # right term: sum of entropies
    entropies = np.array([H(P_i, base=logbase) for P_i in prob_distributions])
    wentropies = weights * entropies
    sum_of_entropies = wentropies.sum()

    divergence = entropy_of_mixture - sum_of_entropies
    return(divergence)

# From the original example with three distributions:
P_1 = np.array([1/2, 1/2, 0])
P_2 = np.array([0, 1/10, 9/10])
P_3 = np.array([1/3, 1/3, 1/3])

prob_distributions = np.array([P_1, P_2, P_3])
n = len(prob_distributions)
weights = np.empty(n)
weights.fill(1/n)

print(JSD(prob_distributions, weights))
#0.546621319446

Answer 3

Get some data for distributions with known divergence and compare your results against those known values.

BTW: the sum in KL_divergence may be rewritten using the zip built-in function like this:

sum(_p * log(_p / _q) for _p, _q in zip(p, q) if _p != 0)

This does away with lots of "noise" and is also much more "pythonic". The double comparison with 0.0 and 0 is not necessary.

Answer 4

Explicitly following the math in the Wikipedia article:

def jsdiv(P, Q):
    """Compute the Jensen-Shannon divergence between two probability distributions.

    Input
    -----
    P, Q : array-like
        Probability distributions of equal length that sum to 1
    """

    def _kldiv(A, B):
        return np.sum([v for v in A * np.log2(A/B) if not np.isnan(v)])

    P = np.array(P)
    Q = np.array(Q)

    M = 0.5 * (P + Q)

    return 0.5 * (_kldiv(P, M) +_kldiv(Q, M))

Answer 5

Since the Jensen-Shannon distance (distance.jensenshannon) has been included in Scipy 1.2, the Jensen-Shannon divergence can be obtained as the square of the Jensen-Shannon distance:

from scipy.spatial import distance

distance.jensenshannon([1.0/10, 9.0/10, 0], [0, 1.0/10, 9.0/10]) ** 2
# 0.5306056938642212