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Tue Aug 30 20:44:44 CEST 2005 · http://mail.python.org/pipermail/python-dev/2005-August/055853.html

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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[Python-Dev] partition() (was: Remove str.find in 3.0?)