Nested defaultdict of defaultdict

Question

Is there a way to make a defaultdict also be the default for the defaultdict? (i.e. infinite-level recursive defaultdict?)

I want to be able to do:

x = defaultdict(...stuff...)
x[0][1][0]
{}

So, I can do x = defaultdict(defaultdict), but that's only a second level:

x[0]
{}
x[0][0]
KeyError: 0

There are recipes that can do this. But can it be done simply just using the normal defaultdict arguments?

Note this is asking how to do an infinite-level recursive defaultdict, so it's distinct to Python: defaultdict of defaultdict?, which was how to do a two-level defaultdict.

I'll probably just end up using the bunch pattern, but when I realized I didn't know how to do this, it got me interested.

Answer 1

The other answers here tell you how to create a defaultdict which contains "infinitely many" defaultdict, but they fail to address what I think may have been your initial need which was to simply have a two-depth defaultdict.

You may have been looking for:

defaultdict(lambda: defaultdict(dict))

The reasons why you might prefer this construct are:

• It is more explicit than the recursive solution, and therefore likely more understandable to the reader.
• This enables the "leaf" of the defaultdict to be something other than a dictionary, e.g.,: defaultdict(lambda: defaultdict(list)) or defaultdict(lambda: defaultdict(set))