On Tue, 16 Aug 2011 12:12 pm Steven D'Aprano wrote:
> On Tue, 16 Aug 2011 09:26 am Johannes wrote:
>
>> hi list,
>> what is the best way to check if a given list (lets call it l1) is
>> totally contained in a second list (l2)?
>
> This is not the most efficient algorithm, but for short lists it should be
> plenty fast enough:
Nope, sorry, that was buggy. Here's another version which should be accurate
but may be slower.
def search(source, target, start=0, end=None):
"""Naive search for target in source."""
m = len(source)
n = len(target)
if end is None:
end = m
if n == 0 or m < n:
return None
for i in range(start, end-n+1):
if source[i:i+n] == target:
return i
return None
--
Steven
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