defaultdict : first argument must be callable or None

For a defaultdict the default value is usually not really a value, it a factory: a method that generates a new value. You can solve this issue by using a lambda expression that generates a list:

lst = lambda:list(range(0,5))d = defaultdict(lst)

This is also a good idea here, since otherwise all default values would reference the same list. For instance here:

d[1].append(14)

will not have impact on d[2] (given both d[1] and d[2] did not exist).

You can however achieve this with:

val = list(range(0,5))lst = lambda:vald = defaultdict(lst)

But this can have unwanted side effects: if you here perform d[1].append(14) then d[2] will be [1,2,3,4,5,14] and d[1] is d[2] will be True:

$ python3Python 3.5.2 (default, Nov 17 2016, 17:05:23) [GCC 5.4.0 20160609] on linuxType "help", "copyright", "credits" or "license" for more information.>>> from collections import defaultdict>>> val = list(range(0,5))>>> lst = lambda:val>>> d = defaultdict(lst)>>> d[1][0, 1, 2, 3, 4]>>> d[1].append(14)>>> d[2][0, 1, 2, 3, 4, 14]>>> d[1] is d[2]True

whereas:

$ python3Python 3.5.2 (default, Nov 17 2016, 17:05:23) [GCC 5.4.0 20160609] on linuxType "help", "copyright", "credits" or "license" for more information.>>> from collections import defaultdict>>> lst = lambda:list(range(0,5))>>> d = defaultdict(lst)>>> d[1][0, 1, 2, 3, 4]>>> d[1].append(14)>>> d[2][0, 1, 2, 3, 4]>>> d[1] is d[2]False

You should make the parameter a callable, say, using lambda:

from collections import defaultdictd = defaultdict(lambda: list(range(0,5)))print(d[0])# [0, 1, 2, 3, 4]

Default dictionary accepts callable as first argument(which is default factory for not defined values), so what you want to do is the next step:

from collections import defaultdictdefault_factory = (lambda: list(range(0,5))d = defaultdict(default_factory)

See more about callable in What is a "callable" in Python? SO question.