Numpy and scipy ways to calculate KL Divergence.

kulback_leibler_divergence.py

"""
Specifically, the Kullback–Leibler divergence from Q to P, denoted DKL(P‖Q), is
a measure of the information gained when one revises one's beliefs from the
prior probability distribution Q to the posterior probability distribution P. In
other words, it is the amount of information lost when Q is used to approximate
P.
"""
import numpy as np
from scipy.stats import entropy

def kl(p, q):
    """Kullback-Leibler divergence D(P || Q) for discrete distributions

    Parameters
    ----------
    p, q : array-like, dtype=float, shape=n
    Discrete probability distributions.
    """
    p = np.asarray(p, dtype=np.float)
    q = np.asarray(q, dtype=np.float)

    return np.sum(np.where(p != 0, p * np.log(p / q), 0))


def kl(p, q):
    """Kullback-Leibler divergence D(P || Q) for discrete distributions

    Parameters
    ----------
    p, q : array-like, dtype=float, shape=n
    Discrete probability distributions.
    """
    p = np.asarray(p, dtype=np.float)
    q = np.asarray(q, dtype=np.float)

    return np.sum(np.where(p != 0, p * np.log(p / q), 0))


p = [0.1, 0.9]
q = [0.1, 0.9]
assert entropy(p, q) == kl(p, q)

@rodrigobdz please note that those are equivalent except for the sign and the formulation of the KL-divergence with np.log(q/p) hence has a leading negation which is not the case here, meaning the script is correct this way (cf. wikipedia)

Hi
You mentioned about p, q discrete probabilities which you created manually. but in real life meaching learning, what value we can use, e.g. If I am using RandomForest classifier it gives me predict_proba() a probability values can I use them if Yes then would it be P or Q and if P then from where can I get Q vise versa?

Unless I am mistaken. The p != 0 should be q != 0. Because you can multiply by 0 but you cannot divide by 0. And in your fliped KL implementation you are dividing by q not p.

Note that scipy.stats.entropy(pk, qk=None, base=None, axis=0) does compute KL if qk is not None.