Fastest bitwise xor between two multibyte binary data variables

Question

What is the fastest way to implementat the following logic:

def xor(data, key):
    l = len(key)

    buff = \"\"
    for i in range(0, len(data)):
        b         


        

Answer 1

Here's a version that only uses Python built-in and standard modules which seems very fast -- although I haven't compared it to your numpy version. It uses a couple of optimized conversion functions from the Python Cryptography Toolkit as indicated.

# Part of the Python Cryptography Toolkit
# found here:
# http://www.google.com/codesearch/p?hl=en#Y_gnTlD6ECg/trunk/src/gdata/Crypto/Util/number.py&q=lang:python%20%22def%20long_to_bytes%22&sa=N&cd=1&ct=rc

# Improved conversion functions contributed by Barry Warsaw, after
# careful benchmarking

import struct

def long_to_bytes(n, blocksize=0):
    """long_to_bytes(n:long, blocksize:int) : string
    Convert a long integer to a byte string.

    If optional blocksize is given and greater than zero, pad the front of the
    byte string with binary zeros so that the length is a multiple of
    blocksize.
    """
    # after much testing, this algorithm was deemed to be the fastest
    s = ''
    n = long(n)
    pack = struct.pack
    while n > 0:
        s = pack('>I', n & 0xffffffffL) + s
        n = n >> 32
    # strip off leading zeros
    for i in range(len(s)):
        if s[i] != '\000':
            break
    else:
        # only happens when n == 0
        s = '\000'
        i = 0
    s = s[i:]
    # add back some pad bytes.  this could be done more efficiently w.r.t. the
    # de-padding being done above, but sigh...
    if blocksize > 0 and len(s) % blocksize:
        s = (blocksize - len(s) % blocksize) * '\000' + s
    return s

def bytes_to_long(s):
    """bytes_to_long(string) : long
    Convert a byte string to a long integer.

    This is (essentially) the inverse of long_to_bytes().
    """
    acc = 0L
    unpack = struct.unpack
    length = len(s)
    if length % 4:
        extra = (4 - length % 4)
        s = '\000' * extra + s
        length = length + extra
    for i in range(0, length, 4):
        acc = (acc << 32) + unpack('>I', s[i:i+4])[0]
    return acc


# original code in SO question
def xor_orig(data, key):
    l = len(key)

    buff = ""
    for i in range(0, len(data)):
        buff += chr(ord(data[i]) ^ ord(key[i % l]))
    return buff

# faster pure python version
def xor_new(data, key):
    import math

    # key multiplication in order to match the data length
    key = (key*int( math.ceil(float(len(data))/float(len(key)))))[:len(data)]

    # convert key and data to long integers
    key_as_long = bytes_to_long(key)
    data_as_long = bytes_to_long(data)

    # xor the numbers together and convert the result back to a byte string
    return long_to_bytes(data_as_long ^ key_as_long)

if __name__=='__main__':
    import random
    import sha

    TEST_DATA_LEN = 100000

    data = ''.join(chr(random.randint(0, 255)) for i in xrange(TEST_DATA_LEN))
    key = sha.new(data).digest()

    assert xor_new(data, key) == xor_orig(data, key)
    print 'done'