"str.contains(part)" or alternatives?
Bengt Richter
bokr at oz.net
Mon Sep 16 18:10:26 EDT 2002
On Wed, 11 Sep 2002 21:18:23 -0700, Greg Fortune <lists at gregfortune.com> wrote:
><snip>>
>>> I think the current proposal is to move to
>>>
>>> if ll in s:
>>>
>>> for this construct. Happy?
>>
>> I would like .contains more, but at least the proposal is _much_ better
>> than "if s.find(part) != -1". :-)
>>
>> "if part in s" disturbs the pattern "if item in sequence" somewhat but
>> that has already been weakened by "if key in dict" (that I like,
>> nonetheless).
>>
>> Stefan
>
>
>Why not simply use .count(part) > 0 ? I didn't catch the first part of the
>conversation, but if I'm looking for a contains, I just do.
>
>if(search_me.count(substr) > 0):
> print 'found'
>else:
> print 'not found'
>
>
Mostly it won't matter, but in cases like
search_me = 'a'*10000000
substr = 'a'
it will 'way suboptimal in comparison with find or index.
Regards,
Bengt Richter
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