Exposing `defaultdict` as a regular `dict`

Question

I am using defaultdict(set) to populate an internal mapping in a very large data structure. After it\'s populated, the whole structure (including the mapping) i

Answer 1

Once you have finished populating your defaultdict, you can simply create a regular dict from it:

my_dict = dict(my_default_dict)

One can optionally use the typing.Final type annotation.

If the default dict is a recursive default dict, see this answer which uses a recursive solution.

Answer 2

You could make a class that holds a reference to your dict and prevent setitem()

from collections import Mapping

class MyDict(Mapping):
    def __init__(self, d):
        self.d = d;

    def __getitem__(self, k):
        return self.d[k]

    def __iter__(self):
        return self.__iter__()

    def __setitem__(self, k, v):
        if k not in self.d.keys():
            raise KeyError
        else:
            self.d[k] = v

Answer 3

defaultdict docs say for default_factory:

If the default_factory attribute is None, this raises a KeyError exception with the key as argument.

What if you just set your defaultdict's default_factory to None? E.g.,

>>> d = defaultdict(int)
>>> d['a'] += 1
>>> d
defaultdict(<type 'int'>, {'a': 1})
>>> d.default_factory = None
>>> d['b'] += 2
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
KeyError: 'b'
>>> 

Not sure if this is the best approach, but seems to work.