Nested defaultdict of defaultdict

There is a nifty trick for doing that:

tree = lambda: defaultdict(tree)

Then you can create your x with x = tree().

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()

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))

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': {}}

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, {})

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

@nucklehead’s response can be extended to handle arrays in JSON as well:

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

Here is a function for an arbitrary base defaultdict for an arbitrary depth of nesting.

(cross posting from Can't pickle defaultdict)

def wrap_defaultdict(instance, times=1):
    """Wrap an instance an arbitrary number of `times` to create nested defaultdict.
    
    Parameters
    ----------
    instance - list, dict, int, collections.Counter
    times - the number of nested keys above `instance`; if `times=3` dd[one][two][three] = instance
    
    Notes
    -----
    using `x.copy` allows pickling (loading to ipyparallel cluster or pkldump)
        - thanks https://stackoverflow.com/questions/16439301/cant-pickle-defaultdict
    """
    from collections import defaultdict

    def _dd(x):
        return defaultdict(x.copy)

    dd = defaultdict(instance)
    for i in range(times-1):
        dd = _dd(dd)

    return dd

Based on Chris W answer, however, to address the type annotation concern, you could make it a factory function that defines the detailed types. For example this is the final solution to my problem when I was researching this question:

def frequency_map_factory() -> dict[str, dict[str, int]]:
    """
    Provides a recorder of: per X:str, frequency of Y:str occurrences.
    """
    return defaultdict(lambda: defaultdict(int))

Here’s a solution similar to @Stanislav’s answer that works with multiprocessing and also allows for termination of the nesting:

from collections import defaultdict
from functools import partial

class NestedDD(defaultdict):
    def __init__(self, n, *args, **kwargs):
        self.n = n
        factory = partial(build_nested_dd, n=n - 1) if n > 1 else int
        super().__init__(factory, *args, **kwargs)

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

def build_nested_dd(n):
    return NestedDD(n)