构建一个值为defaultdict的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