fast XORing bytes in python 3 [duplicate]

Question

I need to xor 2 bytes objects. I use this code:

def bxor(b1, b2): # use xor for bytes
    result = b""
    for b1, b2 in zip(b1, b2):
        result += bytes([b1 ^ b2])
    return result

It works fine when the bytes objects are small, but if I xor big objects (a few MB) it takes very long time (a few hours). How can I make it faster?

Answer 1

Adding this in another answer, 'cause it is one:

If you want something faster than the "manual" methods given, there's always Numpy:

import numpy

def bxor_numpy(b1, b2):
    n_b1 = numpy.fromstring(b1, dtype='uint8')
    n_b2 = numpy.fromstring(b2, dtype='uint8')

    return (n_b1 ^ n_b2).tostring()

and it's fast:

first_random = urandom(100000)
second_random = urandom(100000)

min(Timer(partial(bxor_inplace, first_random, second_random)).repeat(10, 100))
#>>> 1.5381054869794752
min(Timer(partial(bxor_append, first_random, second_random)).repeat(10, 100))
#>>> 1.5624085619929247
min(Timer(partial(bxor_numpy, first_random, second_random)).repeat(10, 100))
#>>> 0.009930026979418471

So it's 150x faster than the best alternatives posted here.