On Wed, Jun 12, 2013 at 1:28 AM, Serhiy Storchaka <storch...@gmail.com> wrote: > 11.06.13 01:50, Chris Angelico написав(ла): > >> On Tue, Jun 11, 2013 at 6:34 AM, Roy Smith <r...@panix.com> 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.
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 -- http://mail.python.org/mailman/listinfo/python-list
