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name Punditsdkoslkdosdkoskdo

Nested defaultdict of defaultdict

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.

For an arbitrary number of levels:

def rec_dd():
    return defaultdict(rec_dd)

>>> x = rec_dd()
>>> x['a']['b']['c']['d']
defaultdict(<function rec_dd at 0x7f0dcef81500>, {})
>>> print json.dumps(x)
{"a": {"b": {"c": {"d": {}}}}}

Of course you could also do this with a lambda, but I find lambdas to be less readable. In any case it would look like this:

rec_dd = lambda: defaultdict(rec_dd)
  • 167
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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))
  • 166
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      • 1
    • Ooops, yes, the lambda form is correct--because the defaultdict(something) returns a dictionary-like object, but defaultdict expects a callable! Thank you!
      • 2
    • One draw back of the lambda approach is that the objects it generates can't be pickled... but you can get around this by casting to a regular dict(result) before the pickle

Similar to BrenBarn's solution, but doesn't contain the name of the variable tree twice, so it works even after changes to the variable dictionary:

tree = (lambda f: f(f))(lambda a: (lambda: defaultdict(a(a))))

Then you can create each new x with x = tree().


For the def version, we can use function closure scope to protect the data structure from the flaw where existing instances stop working if the tree name is rebound. It looks like this:

from collections import defaultdict

def tree():
    def the_tree():
        return defaultdict(the_tree)
    return the_tree()
  • 22
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    • i'll have to think about this one (it's a little more complex). but i think your point is that if do x = tree(), but then someone comes by later and does tree=None, this one would still work, and that would wouldn't?

I would also propose more OOP-styled implementation, which supports infinite nesting as well as properly formatted repr.

class NestedDefaultDict(defaultdict):
    def __init__(self, *args, **kwargs):
        super(NestedDefaultDict, self).__init__(NestedDefaultDict, *args, **kwargs)

    def __repr__(self):
        return repr(dict(self))

Usage:

my_dict = NestedDefaultDict()
my_dict['a']['b'] = 1
my_dict['a']['c']['d'] = 2
my_dict['b']

print(my_dict)  # {'a': {'b': 1, 'c': {'d': 2}}, 'b': {}}
  • 11
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      • 2
    • Neat ! I added the passthrough of *args and **kwargs which allows it to function like the defaultdict, namely to create a dict with keyword arguments. This is useful for passing NestedDefaultDict into json.load

here is a recursive function to convert a recursive default dict to a normal dict

def defdict_to_dict(defdict, finaldict):
    # pass in an empty dict for finaldict
    for k, v in defdict.items():
        if isinstance(v, defaultdict):
            # new level created and that is the new value
            finaldict[k] = defdict_to_dict(v, {})
        else:
            finaldict[k] = v
    return finaldict

defdict_to_dict(my_rec_default_dict, {})
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I based this of Andrew's answer here. If you are looking to load data from a json or an existing dict into the nester defaultdict see this example:

def nested_defaultdict(existing=None, **kwargs):
    if existing is None:
        existing = {}
    if not isinstance(existing, dict):
        return existing
    existing = {key: nested_defaultdict(val) for key, val in existing.items()}
    return defaultdict(nested_defaultdict, existing, **kwargs)

https://gist.github.com/nucklehead/2d29628bb49115f3c30e78c071207775

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