On 16/09/05, Kent Johnson <[EMAIL PROTECTED]> wrote: > Alan Gauld wrote: > > Bearing in mind that shorter is not necessarily better... > > > > [condition(i) and list1.append(i) or list2.append(i) for i in > > original] > > Hmm, no, this will always evaluate list2.append(i) because the value of > list1.append(i) is None. You could use > > [ (condition(i) and list1 or list2).append(i) for i in original ]
This will not work either, because initially, list1 will be [] which is false. So everything will end up in list2. >>> list1, list2 = [], [] >>> [(i%2==1 and list1 or list2).append(i) for i in range(10)] [None, None, None, None, None, None, None, None, None, None] >>> list1, list2 ([], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]) You can get around this with a dodge.. >>> list1, list2 = [], [] >>> [(i%2 == 1 and [list1] or [list2])[0].append(i) for i in range(10)] [None, None, None, None, None, None, None, None, None, None] >>> list1, list2 ([1, 3, 5, 7, 9], [0, 2, 4, 6, 8]) But I'm sure that using side-effects in a list comprehension is prohibited by at least one major religion, so really, I would just stick to a for loop :-) -- John. _______________________________________________ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
