Phillip J. Eby wrote:
> At 02:25 PM 8/30/2005 -0400, Raymond Hettinger wrote:
>
>> That case should be handled with consecutive partitions:
>>
>> # keep everything after the second 'X'
>> head, found, s = s.partition('X')
>> head, found, s = s.partition('x')
I was thinking of cases where head is everything before the second 'X'.
A posible use case might be getting items in comma delimited string.
> Or:
>
> s=s.partition('X')[2].partition('X')[2]
>
> which actually suggests a shorter, clearer way to do it:
>
> s = s.after('X').after('X')
>
> And the corresponding 'before' method, of course, such that if sep in s:
>
> s.before(sep), sep, s.after(sep) == s.partition(sep)
>
> Technically, these should probably be before_first and after_first, with
> the corresponding before_last and after_last corresponding to rpartition.
Do you really think these are easer than:
head, found, tail = s.partition('X',2)
I don't feel there is a need to avoid numbers entirely. In this case I
think it's the better way to find the n'th seperator and since it's an
optional value I feel it doesn't add a lot of complication. Anyway...
It's just a suggestion.
Cheers,
Ron
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