On 18 Feb, 19:33, Jacob Kaplan-Moss <jacob.kaplanm...@gmail.com>
wrote:
> On Wed, Feb 18, 2009 at 12:24 PM, Deniz Dogan <deniz.a.m.do...@gmail.com>
> wrote:
> > I'm sorry about that, I meant something like this:
>
> > { 2008-01-02 : [Bike 1, Bike 2],
> > 2008-02-09 : [Bike 7, Bike 4] }
>
> Well, you can't do this with the aggregation API because you can't do
> that is SQL. Remember: there's no such thing as a "list" in a
> relational database.
Oops, didn't really think about that!
> However, this is pretty easy to do just in Python using
> ``itertools.groupby``
> (seehttp://docs.python.org/library/itertools.html#itertools.groupby)::
>
> >>> bikes = Bike.objects.all()
> >>> for d, bl in itertools.groupby(bikes, lambda b: b.production_date):
> ... print d, list(bl)
>
> 2008-01-01 [<Bike: 1>, <Bike: 2>]
> 2008-01-02 [<Bike: 3>]
> 2008-01-03 [<Bike: 4>]
> 2008-01-04 [<Bike: 5>, <Bike: 6>]
> 2008-01-05 [<Bike: 7>]
>
> The ORM's not a crutch; it doesn't bother with tasks you can easily do
> using the tools Python gives you. Mm, and in case you're keeping
> track, this is quite easy on the database; just a single query.
>
> Hope that's closer to what you had in mind,
>
> Jacob
Thanks a lot! I had just completed a little hack of my own for this,
but I'll make sure to check itertools out, because it seems pretty
cool.
Deniz
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