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Groups > comp.lang.python > #35783
| Date | 2012-12-29 16:40 -0500 |
|---|---|
| From | Mitya Sirenef <msirenef@lightbird.net> |
| Subject | Re: dict comprehension question. |
| References | <724d4fea-606a-4503-b538-87442f6bcebc@ci3g2000vbb.googlegroups.com> <50DF4C00.2020505@lightbird.net> <50DF4E15.1030700@lightbird.net> <CAPM-O+w8aYZ2zJcmc+ZyHOHmA7yt8OYPYAqhGWSB5tQgbnqRdA@mail.gmail.com> |
| Newsgroups | comp.lang.python |
| Message-ID | <mailman.1447.1356817208.29569.python-list@python.org> (permalink) |
On 12/29/2012 03:15 PM, Joel Goldstick wrote:
>
>
>
> On Sat, Dec 29, 2012 at 3:09 PM, Mitya Sirenef <msirenef@lightbird.net
> <mailto:msirenef@lightbird.net>> wrote:
>
> On 12/29/2012 03:01 PM, Mitya Sirenef wrote:
>
> On 12/29/2012 02:48 PM, Quint Rankid wrote:
>
> >> Newbie question. I've googled a little and haven't found the
> answer.
> >>
> >> Given a list like:
> >> w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
> >> I would like to be able to do the following as a dict
> comprehension.
> >> a = {}
> >> for x in w:
> >> a[x] = a.get(x,0) + 1
> >> results in a having the value:
> >> {1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
> >>
> >> I've tried a few things
> >> eg
> >> a1 = {x:self.get(x,0)+1 for x in w}
> >> results in error messages.
> >>
> >> And
> >> a2 = {x:a2.get(x,0)+1 for x in w}
> >> also results in error messages.
> >>
> >> Trying to set a variable to a dict before doing the comprehension
> >> a3 = {}
> >> a3 = {x:a3.get(x,0)+1 for x in w}
> >> gets this result, which isn't what I wanted.
> >> {1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1}
> >>
> >> I'm not sure that it's possible to do this, and if not, perhaps the
> >> most obvious question is what instance does the get method bind to?
> >>
> >> TIA
> >
> > Will this do?:
> >
> > >>> w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
> > >>> {x: w.count(x) for x in w}
> > {1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
> >
> >
> > - mitya
> >
>
> I should probably add that this might be inefficient for large
> lists as
> it repeats count for each item. If you need it for large lists,
> profile
> against the 'for loop' version and decide if performance is good
> enough
> for you, for small lists it's a nice and compact solution.
>
> In a more general case, you can't refer to the list/dict/etc
> comprehension as it's being constructed, that's just not a design goal
> of comprehensions.
>
>
> Would this help:
>
> >>> w = [1,2,3,1,2,4,4,5,6,1]
> >>> s = set(w)
> >>> s
> set([1, 2, 3, 4, 5, 6])
> >>> {x:w.count(x) for x in s}
> {1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
> >>>
Indeed, this is much better -- I didn't think of it..
--
Lark's Tongue Guide to Python: http://lightbird.net/larks/
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dict comprehension question. Quint Rankid <qbr567@gmail.com> - 2012-12-29 11:48 -0800
Re: dict comprehension question. Roy Smith <roy@panix.com> - 2012-12-29 14:58 -0500
Re: dict comprehension question. Mitya Sirenef <msirenef@lightbird.net> - 2012-12-29 15:01 -0500
Re: dict comprehension question. Mitya Sirenef <msirenef@lightbird.net> - 2012-12-29 15:09 -0500
Re: dict comprehension question. Joel Goldstick <joel.goldstick@gmail.com> - 2012-12-29 15:15 -0500
Re: dict comprehension question. Peter Otten <__peter__@web.de> - 2012-12-29 21:32 +0100
Re: dict comprehension question. MRAB <python@mrabarnett.plus.com> - 2012-12-29 20:39 +0000
Re: dict comprehension question. Mitya Sirenef <msirenef@lightbird.net> - 2012-12-29 16:40 -0500
Re: dict comprehension question. Terry Reedy <tjreedy@udel.edu> - 2012-12-29 18:22 -0500
Re: dict comprehension question. Terry Reedy <tjreedy@udel.edu> - 2012-12-29 18:56 -0500
Re: dict comprehension question. Steven D'Aprano <steve+comp.lang.python@pearwood.info> - 2013-01-01 03:10 +0000
Re: dict comprehension question. 88888 Dihedral <dihedral88888@googlemail.com> - 2013-01-01 00:40 -0800
Re: dict comprehension question. Tim Chase <python.list@tim.thechases.com> - 2012-12-29 18:26 -0600
Re: dict comprehension question. Joel Goldstick <joel.goldstick@gmail.com> - 2012-12-29 23:10 -0500
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