Fastest way to compute entropy of each numpy array row?

Question

I have a array in size MxN and I like to compute the entropy value of each row. What would be the fastest way to do so ?

Answer 1

As @Warren pointed out, it's unclear from your question whether you are starting out from an array of probabilities, or from the raw samples themselves. In my answer I've assumed the latter, in which case the main bottleneck will be computing the bin counts over each row.

Assuming that each vector of samples is relatively long, the fastest way to do this will probably be to use np.bincount:

import numpy as np

def entropy(x):
    """
    x is assumed to be an (nsignals, nsamples) array containing integers between
    0 and n_unique_vals
    """
    x = np.atleast_2d(x)
    nrows, ncols = x.shape
    nbins = x.max() + 1

    # count the number of occurrences for each unique integer between 0 and x.max()
    # in each row of x
    counts = np.vstack((np.bincount(row, minlength=nbins) for row in x))

    # divide by number of columns to get the probability of each unique value
    p = counts / float(ncols)

    # compute Shannon entropy in bits
    return -np.sum(p * np.log2(p), axis=1)

Although Warren's method of computing the entropies from the probability values using entr is slightly faster than using the explicit formula, in practice this is likely to represent a tiny fraction of the total runtime compared to the time taken to compute the bin counts.

Test correctness for a single row:

vals = np.arange(3)
prob = np.array([0.1, 0.7, 0.2])
row = np.random.choice(vals, p=prob, size=1000000)

print("theoretical H(x): %.6f, empirical H(x): %.6f" %
      (-np.sum(prob * np.log2(prob)), entropy(row)[0]))
# theoretical H(x): 1.156780, empirical H(x): 1.157532

Test speed:

In [1]: %%timeit x = np.random.choice(vals, p=prob, size=(1000, 10000))
   ....: entropy(x)
   ....: 
10 loops, best of 3: 34.6 ms per loop

If your data don't consist of integer indices between 0 and the number of unique values, you can convert them into this format using np.unique:

y = np.random.choice([2.5, 3.14, 42], p=prob, size=(1000, 10000))
unq, x = np.unique(y, return_inverse=True)
x.shape = y.shape

Answer 2

scipy.special.entr computes -x*log(x) for each element in an array. After calling that, you can sum the rows.

Here's an example. First, create an array p of positive values whose rows sum to 1:

In [23]: np.random.seed(123)

In [24]: x = np.random.rand(3, 10)

In [25]: p = x/x.sum(axis=1, keepdims=True)

In [26]: p
Out[26]: 
array([[ 0.12798052,  0.05257987,  0.04168536,  0.1013075 ,  0.13220688,
         0.07774843,  0.18022149,  0.1258417 ,  0.08837421,  0.07205402],
       [ 0.08313743,  0.17661773,  0.1062474 ,  0.01445742,  0.09642919,
         0.17878489,  0.04420998,  0.0425045 ,  0.12877228,  0.1288392 ],
       [ 0.11793032,  0.15790292,  0.13467074,  0.11358463,  0.13429674,
         0.06003561,  0.06725376,  0.0424324 ,  0.05459921,  0.11729367]])

In [27]: p.shape
Out[27]: (3, 10)

In [28]: p.sum(axis=1)
Out[28]: array([ 1.,  1.,  1.])

Now compute the entropy of each row. entr uses the natural logarithm, so to get the base-2 log, divide the result by log(2).

In [29]: from scipy.special import entr

In [30]: entr(p).sum(axis=1)
Out[30]: array([ 2.22208731,  2.14586635,  2.22486581])

In [31]: entr(p).sum(axis=1)/np.log(2)
Out[31]: array([ 3.20579434,  3.09583074,  3.20980287])

If you don't want the dependency on scipy, you can use the explicit formula:

In [32]: (-p*np.log2(p)).sum(axis=1)
Out[32]: array([ 3.20579434,  3.09583074,  3.20980287])