Building a defaultdict whose values are defaultdict

Question

Is there any particular reason that my application says the argument must be callable or none type? I'm pretty sure this is how you'd instantiate a defaultdict with a defaultdict as its values.

dict = defaultdict(defaultdict(set))

Answer 1

The argument provided to the defaultdict constructor must be a zero-argument callable (see the docs ), so defaultdict(defaultdict) does work, but defaultdict(defaultdict(set)) does not. You can 'cheat' a little though:

d = defaultdict(lambda: defaultdict(set))

That way you provide a zero-argument callable in the form of the lambda function which in turn, when called, returns the appropriate default value you want.

Answer 2

You need to provide a function for defaultdict constructor but defauldict(set) is defaultdict object instead. If you want to build a defaultdict whose values are defaultdicts you can use lambdas:

from collections import defaultdict

dd = lambda: defaultdict(dd)
x = dd()
x['foo']['bar']['foobar'] = 1