Safe hash comparison

comparehash.py

from itertools import izip
def compare_hash(hashone,hashtwo):
    if len(hashone) == len(hashtwo):  # Every hash should be the same length
        if sum((ord(o) ^ ord(t) for o,t in izip(hashone,hashtwo))):
            return False
        else:
            return True
    else:
        return False

Any time you find yourself doing "if statement: return True else: return False" just return the statement. For example line 4 could be:

return sum((ord(o) ^ ord(t) for o,t in izip(hashone,hashtwo))) == 0

Or even more simply the whole thing is an and of two booleans: so

def compare_hash(hashone, hashtwo):
  return len(hashone) == len(hashtwo) and sum((ord(o) ^ ord(t) for o,t in izip(hashone,hashtwo))) == 0

I agree, I just wrote this in a verbose manner so people will understand the goal behind the code.