Oscar Benjamin wrote:
> On 5 September 2012 10:48, Peter Otten <__pete...@web.de> wrote:
>> def count_common(a, b):
[sorry for seriously broken implementation]
> This function will return 1 if the first character differs. It does not
> count the number of common characters but rather the more relevant
> quantity which is the number of comparisons required to decide if the
> strings are equal.
I remember stumbling over the non-zero frequency for len(word) but somehow
plodded on with beautifying the broken results :(
def count_common(a, b):
"""
>>> count_common("alpha", "beta")
0
>>> count_common("", "")
0
>>> count_common("alpha", "argument")
1
>>> count_common("alpha", "alpha")
5
>>> count_common("alpha", "alphx")
4
"""
i = 0
for x, y in zip(a, b):
if x != y:
break
i += 1
return i
$ python3 reverse_compare2.py compare /usr/share/dict/words
499500 combinations
average common chars (head) 0.0535
average common chars (tail) 0.322
comparing every pair in a sample of 1000 8-char words
taken from '/usr/share/dict/words'
head
0: 477371 ************************************************************
1: 18310 **
2: 3191
3: 523
4: 79
5: 15
6: 10
7: 1
tail
0: 385069 ************************************************************
1: 78742 ************
2: 26734 ****
3: 7462 *
4: 1297
5: 168
6: 22
7: 6
--
http://mail.python.org/mailman/listinfo/python-list
