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')
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.
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