On Fri, Jul 24, 2009 at 3:01 AM, <davidwil...@safe-mail.net> wrote:
>> If ws_industry contains 25 items, it will probably be faster to search
>> it if it is a set rather than a list. Just use
>> ws_industry = set(('it', 'science'))
>
> ws_industry contains only unique valuesOK. It will still be faster to check membership if it is a set rather than a list. Checking membership in a list has to search the whole list in order. The time will vary depending on where, and if, the target value is found in the list: kent $ python -m timeit -s "m=map(str, range(25))" "'x' in m" 1000000 loops, best of 3: 0.902 usec per loop kent $ python -m timeit -s "m=map(str, range(25))" "'12' in m" 1000000 loops, best of 3: 0.649 usec per loop kent $ python -m timeit -s "m=map(str, range(25))" "'0' in m" 10000000 loops, best of 3: 0.0801 usec per loop Searching a set takes approximately the same amount of time regardless of the size of the set and whether the target is in the set or not: kent $ python -m timeit -s "m=set(map(str, range(25)))" "'x' in m" 10000000 loops, best of 3: 0.0854 usec per loop kent $ python -m timeit -s "m=set(map(str, range(25)))" "'12' in m" 10000000 loops, best of 3: 0.117 usec per loop kent $ python -m timeit -s "m=set(map(str, range(25)))" "'0' in m" 10000000 loops, best of 3: 0.0867 usec per loop Notice that the first two tests are significantly faster than searching the list; the last one is slightly slower but this is the best case for the list search (finding the first element). Kent _______________________________________________ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
