Fast hamming distance computation between binary numpy arrays

问题

I have two numpy arrays of the same length that contain binary values

import numpy as np
a=np.array([1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0])
b=np.array([1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1])

I want to compute the hamming distance between them as fast as possible since I have millions of such distance computations to make.

A simple but slow option is this (taken from wikipedia):

%timeit sum(ch1 != ch2 for ch1, ch2 in zip(a, b))
10000 loops, best of 3: 79 us per loop

I have come up with faster options, inspired by some answers here on stack overflow.

%timeit np.sum(np.bitwise_xor(a,b))
100000 loops, best of 3: 6.94 us per loop

%timeit len(np.bitwise_xor(a,b).nonzero()[0])
100000 loops, best of 3: 2.43 us per loop

I'm wondering if there are even faster ways to compute this, possibly using cython?

回答1:

There is a ready numpy function which beats len((a != b).nonzero()[0]) ;)

np.count_nonzero(a!=b)

回答2:

Compared to 1.07µs for np.count_nonzero(a!=b) on my platform, gmpy2.hamdist gets its down to about 143ns after conversion of each array to an mpz (multiple-precison integer):

import numpy as np
from gmpy2 import mpz, hamdist, pack

a = np.array([1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0])
b = np.array([1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1])

Based on a tip from @casevh, conversion from a 1D array of ones and zeros to a gmpy2 mpz object can be done reasonably efficiently with gmpy2.pack(list(reversed(list(array))),1).

# gmpy2.pack reverses bit order but that does not affect
# hamdist since both its arguments are reversed
ampz = pack(list(a),1) # takes about 4.29µs
bmpz = pack(list(b),1)

hamdist(ampz,bmpz)
Out[8]: 7

%timeit hamdist(ampz,bmpz)
10000000 loops, best of 3: 143 ns per loop

for relative comparison, on my platform:

%timeit np.count_nonzero(a!=b)
1000000 loops, best of 3: 1.07 µs per loop

%timeit len((a != b).nonzero()[0])
1000000 loops, best of 3: 1.55 µs per loop

%timeit len(np.bitwise_xor(a,b).nonzero()[0])
1000000 loops, best of 3: 1.7 µs per loop

%timeit np.sum(np.bitwise_xor(a,b))
100000 loops, best of 3: 5.8 µs per loop

回答3:

Using pythran can bring extra benefit here:

$ cat hamm.py
#pythran export hamm(int[], int[])
from numpy import nonzero
def hamm(a,b):
    return len(nonzero(a != b)[0])

As a reference (without pythran):

$ python -m timeit -s 'import numpy as np; a = np.random.randint(0,2, 100); b = np.random.randint(0,2, 100); from hamm import hamm' 'hamm(a,b)'
100000 loops, best of 3: 4.66 usec per loop

While after pythran compilation:

$ python -m pythran.run hamm.py
$ python -m timeit -s 'import numpy as np; a = np.random.randint(0,2, 100); b = np.random.randint(0,2, 100); from hamm import hamm' 'hamm(a,b)'
1000000 loops, best of 3: 0.745 usec per loop

That's roughly a 6x speedup over the numpy implementation, as pythran skips the creation of an intermediate array when evaluating the element wise comparison.

I also measured:

def hamm(a,b):
    return count_nonzero(a != b)

And I get 3.11 usec per loop for the Python version and 0.427 usec per loop with the Pythran one.

Disclaimer: I'm one of the Pythran dev.

来源：https://stackoverflow.com/questions/32730202/fast-hamming-distance-computation-between-binary-numpy-arrays

标签

python

arrays

numpy

cython

hamming-distance