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
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