# [Python-Dev] PEP 201 - Parallel iteration

**Ken Manheimer
**
klm@digicool.com

*Mon, 17 Jul 2000 20:40:33 -0400 (EDT)*

On Tue, 18 Jul 2000, Peter Schneider-Kamp wrote:
>* One important remark:
*>* What about unzip()? Shouldn't there be the reverse function?
*>* One example (which I really wrote not so long ago in an ANN context):
*
Evan Simpson (a colleague here at digicool, who'd be a good addition
to python-dev) noticed that unzip is unnecessary - zip is its own
inverse. Since this is a worthwhile observation, i'll indulge a bit
and present it in evan's terms - particularly noting a different
packaging that i think is pretty cool, as methods on sequences.
>* From evan@digicool.com Mon Jul 17 20:17:30 2000
*>* Date: Mon, 17 Jul 2000 14:36:18 -0400
*>* From: Evan Simpson <evan@digicool.com>
*>* To: Ken Manheimer <klm@digicool.com>
*>* Subject: Next Braid
*>*
*>* How's this:
*>*
*>* ([1, 2, 3], ['a', 'b', 'c']).braid() == ((1, 'a'), (2, 'b'), (3, 'c'))
*>* [[1, 2, 3], ['a', 'b', 'c']].braid() == [(1, 'a'), (2, 'b'), (3, 'c')]
*>*
*>* and
*>*
*>* [[1,2,3], ['a']].braid(None) == [(1, 'a'), (2, None), (3, None)]
*>*
*>* Also, notice that if 's' is a rectangular sequence of sequences, then for
*>* all 'a' and 'b':
*>*
*>* s.braid()[a][b] == s[b][a] by definition, so
*>*
*>* s.braid().braid()[a][b] == s.braid()[b][a] == s[a][b] and therefore,
*>*
*>* braid is its own inverse, and we don't need unbraid.
*>*
*>* Cheers,
*>*
*>* Evan @ digicool & 4-am
*
I can't do better than that, but i have one further elaboration i'd
like to see. If the subject sequence already contains some elements,
the result sequence is a copy of the originals with the new braided,
whoops zipped elements appended - this way, braids can be cumulative.
>>>* [(1, 3), (2, 4)].braid(('a', 'b'), ('x', 'y'))
*[(1, 3, 'a', 'x'), (2, 4, 'b', 'y')]
Anyway, we don't need an unzip/unbraid. !
Ken
klm@digicool.com