you can use the "is" for identity test. l1 = [v for v in array if not v is 0] l2 = [v for v in array if v is 0]
On Jul 6, 2017, at 10:31 PM, Sayth Renshaw <flebber.c...@gmail.com<mailto:flebber.c...@gmail.com>> wrote: I was trying to solve a problem and cannot determine how to filter 0's but not false. Given a list like this ["a",0,0,"b",None,"c","d",0,1,False,0,1,0,3,[],0,1,9,0,0,{},0,0,9] I want to be able to return this list ["a","b",None,"c","d",1,False,1,3,[],1,9,{},9,0,0,0,0,0,0,0,0,0,0] However if I filter like this def move_zeros(array): l1 = [v for v in array if v != 0] l2 = [v for v in array if v == 0] return l1 + l2 I get this ['a', 'b', None, 'c', 'd', 1, 1, 3, [], 1, 9, {}, 9, 0, 0, 0, False, 0, 0, 0, 0, 0, 0, 0] I have tried or conditions of v == False etc but then the 0's being false also aren't moved. How can you check this at once? Cheers Sayth -- https://mail.python.org/mailman/listinfo/python-list -- https://mail.python.org/mailman/listinfo/python-list
