Paul Rudin wrote:
> Peter Otten <__pete...@web.de> writes:
>
>
>> OP: you may be looking for
>>
>>>>> a = "a bb ccc"
>>>>> a[::-1].find(" ")
>> 3
>
>
> But you should be aware of the effeciency implications of doing
> this. a[::-1] constructs a new list.
A new string, yes.
> It's probably faster to do e.g.:
> len(a) - a.rfind(..) - 1
Yes, especially if the string is "long":
$ python -m timeit -s'a = "word1 word2 word3"' 'a[::-1].rfind(" ")'
1000000 loops, best of 3: 0.834 usec per loop
$ python -m timeit -s'a = "word1 word2 word3"*100' 'a[::-1].rfind(" ")'
100000 loops, best of 3: 5.04 usec per loop
$ python -m timeit -s'a = "word1 word2 word3"' 'len(a)-a.rfind(" ")-1'
1000000 loops, best of 3: 0.587 usec per loop
$ python -m timeit -s'a = "word1 word2 word3"*100' 'len(a)-a.rfind(" ")-1'
1000000 loops, best of 3: 0.592 usec per loop
But be aware of the following difference:
>>> a[::-1].rfind("not there")
-1
>>> len(a) - a.rfind("not there") -1
8
Peter
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