Jensen-Shannon Divergence

问题:

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 distributions. The similarity scores appear to be correct in the sense that they fall between 1 and 0 given that one uses the base 2 logarithm, with 0 meaning that the distributions are equal.

However, I'm not sure whether there is in fact an error somewhere and was wondering whether someone might be able to say 'yes it's correct' or 'no, you did something wrong'.

Here is the code:

from numpy import zeros, array
from math import sqrt, log


class JSD(object):
    def __init__(self):
        self.log2 = log(2)


    def KL_divergence(self, p, q):
        """ Compute KL divergence of two vectors, K(p || q)."""
        return sum(p[x] * log((p[x]) / (q[x])) for x in range(len(p)) if p[x] != 0.0 or p[x] != 0)

    def Jensen_Shannon_divergence(self, p, q):
        """ Returns the Jensen-Shannon divergence. """
        self.JSD = 0.0
        weight = 0.5
        average = zeros(len(p)) #Average
        for x in range(len(p)):
            average[x] = weight * p[x] + (1 - weight) * q[x]
            self.JSD = (weight * self.KL_divergence(array(p), average)) + ((1 - weight) * self.KL_divergence(array(q), average))
        return 1-(self.JSD/sqrt(2 * self.log2))

if __name__ == '__main__':
    J = JSD()
    p = [1.0/10, 9.0/10, 0]
    q = [0, 1.0/10, 9.0/10]
    print J.Jensen_Shannon_divergence(p, q)

The problem is that I feel that the scores are not high enough when comparing two text documents, for instance. However, this is purely a subjective feeling.

Any help is, as always, appreciated.

回答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

回答2:

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.

回答3:

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

回答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))

回答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