Split a list into two parts based on a filter?

[Peter Otten]

Chris Angelico wrote:

> On Wed, Jun 12, 2013 at 1:28 AM, Serhiy Storchaka <storchaka at gmail.com>
> wrote:
>> 11.06.13 01:50, Chris Angelico написав(ла):
>>
>>> On Tue, Jun 11, 2013 at 6:34 AM, Roy Smith <roy at 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()]

I think you meant old_songs.remove(s).

>> 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.

But is it correct? In the general case, no:

>>> numbers = [1, 1.0, 2.0, 2]
>>> ints = numbers[:]
>>> floats = [ints.remove(n) or n for n in numbers if isinstance(n, float)]
>>> floats
[1.0, 2.0]
>>> ints
[1.0, 2] # hmm