Numpy sum of operator results without allocating an unnecessary array

On my machine this is faster:

(a == b).sum()

If you don't want to use any extra storage, than I would suggest using numba.I'm not too familiar with it, but this seems to work well.I ran into some trouble getting Cython to take a boolean NumPy array.

from numba import autojitdef pysumeq(a, b):    tot = 0    for i in xrange(a.shape[0]):        for j in xrange(a.shape[1]):            if a[i,j] == b[i,j]:                tot += 1    return tot# make numba versionnbsumeq = autojit(pysumeq)A = (rand(10,10)<.5)B = (rand(10,10)<.5)# do a simple dry run to get it to compile# for this specific use casenbsumeq(A, B)

If you don't have numba, I would suggest using the answer by @user2357112

Edit: Just got a Cython version working, here's the .pyx file. I'd go with this.

from numpy cimport ndarray as arcimport numpy as npcimport cython@cython.boundscheck(False)@cython.wraparound(False)def cysumeq(ar[np.uint8_t,ndim=2,cast=True] a, ar[np.uint8_t,ndim=2,cast=True] b):    cdef int i, j, h=a.shape[0], w=a.shape[1], tot=0    for i in xrange(h):        for j in xrange(w):            if a[i,j] == b[i,j]:                tot += 1    return tot

To start with you can skip then A*B step:

>>> aarray([ True, False,  True, False,  True], dtype=bool)>>> barray([False,  True,  True, False,  True], dtype=bool)>>> np.sum(~(a^b))3

If you do not mind destroying array a or b, I am not sure you will get faster then this:

>>> a^=b   #In place xor operator>>> np.sum(~a)3

If the problem is allocation and deallocation, maintain a single output array and tell numpy to put the results there every time:

out = np.empty_like(a) # Allocate this outside a loop and use it every iterationnum_eq = np.equal(a, b, out).sum()

This'll only work if the inputs are always the same dimensions, though. You may be able to make one big array and slice out a part that's the size you need for each call if the inputs have varying sizes, but I'm not sure how much that slows you down.