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Date: Wed, 12 Jun 2013 03:28:34 +1000
Subject: Re: Split a list into two parts based on a filter?
From: Chris Angelico <rosuav@gmail.com>
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Xref: csiph.com comp.lang.python:47686

On Wed, Jun 12, 2013 at 1:28 AM, Serhiy Storchaka <storchaka@gmail.com> wro=
te:
> 11.06.13 01:50, Chris Angelico =D0=BD=D0=B0=D0=BF=D0=B8=D1=81=D0=B0=D0=B2=
(=D0=BB=D0=B0):
>
>> On Tue, Jun 11, 2013 at 6:34 AM, Roy Smith <roy@panix.com> wrote:
>>>
>>> new_songs =3D [s for s in songs if s.is_new()]
>>> old_songs =3D [s for s in songs if not s.is_new()]
>>
>>
>> Hmm. Would this serve?
>>
>> old_songs =3D songs[:]
>> new_songs =3D [songs.remove(s) or s for s in songs if s.is_new()]
>
>
> O(len(songs)**2) complexity.

Which isn't significant if len(songs) is low. We weren't told the
relative costs - is the is_new call ridiculously expensive? Everything
affects algorithmic choice.

ChrisA