In article <[email protected]>, Chris Angelico <[email protected]> wrote:
> On Wed, Jun 12, 2013 at 1:28 AM, Serhiy Storchaka <[email protected]> wrote: > > 11.06.13 01:50, Chris Angelico напиÑав(ла): > > > >> On Tue, Jun 11, 2013 at 6:34 AM, Roy Smith <[email protected]> wrote: > >>> > >>> new_songs = [s for s in songs if s.is_new()] > >>> old_songs = [s for s in songs if not s.is_new()] > >> > >> > >> Hmm. Would this serve? > >> > >> old_songs = songs[:] > >> new_songs = [songs.remove(s) or s for s in songs if s.is_new()] > > > > > > O(len(songs)**2) complexity. If I didn't want to make two passes over songs, I probably don't want something that's O(len(songs)^2) :-) > 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. Assume is_new() is cheap. It's essentially: return (datetime.utcnow() - self.create_time) < [[a pre-defined timedelta]]
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