Python equivalent of sum() using xor()

Question

I like the Python sum function :

>>> z = [1] * 11
>>> zsum = sum(z)
>>> zsum == 11
True

I want the same functionality with using xor (^) not add (+). I want to use map. But I can not work out how to do this. Any hints?

I am not satisfied with this :

def xor(l):
    r = 0
    for v in l: r ^= v
    return v

I want a 1 liner using map. Hints?

Answer 1

zxor = reduce(lambda a, b: a ^ b, z, 0)

import operator
zxor = reduce(operator.xor, z, 0)

Answer 2

Note that starting Python 3.8, and the introduction of assignment expressions (PEP 572) (:= operator), we can use and update a variable within a list comprehension and thus reduce a list to the xor of its elements:

zxor = 0
[zxor := zxor ^ x for x in [1, 0, 1, 0, 1, 0]]
# zxor = 1