Exposing `defaultdict` as a regular `dict`

问题

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) is exposed to the client code. At that point, I don't want anyone modifying the mapping.

And nobody does, intentionally. But sometimes, client code may by accident refer to an element that doesn't exist. At that point, a normal dictionary would have raised KeyError, but since the mapping is defaultdict, it simply creates a new element (an empty set) at that key. This is quite hard to catch, since everything happens silently. But I need to ensure this doesn't happen (the semantics actually doesn't break, but the mapping grows to a huge size).

What should I do? I can see these choices:

Find all the instances in current and future client code where a dictionary lookup is performed on the mapping, and convert it to mapping.get(k, {}) instead. This is just terrible.
"Freeze" defaultdict after the data structure is fully initialized, by converting it to dict. (I know it's not really frozen, but I trust client code to not actually write mapping[k] = v.) Inelegant, and a large performance hit.
Wrap defaultdict into a dict interface. What's an elegant way to do that? I'm afraid the performance hit may be huge though (this lookup is heavily used in tight loops).
Subclass defaultdict and add a method that "shuts down" all the defaultdict features, leaving it to behave as if it's a regular dict. It's a variant of 3 above, but I'm not sure if it's any faster. And I don't know if it's doable without relying on the implementation details.
Use regular dict in the data structure, rewriting all the code there to first check if the element is in the dictionary and adding it if it's not. Not good.

回答1:

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.

回答2:

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

my_dict = dict(my_default_dict)

The regular dict is of course effectively frozen.

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

回答3:

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

来源：https://stackoverflow.com/questions/13465681/exposing-defaultdict-as-a-regular-dict

标签

python

python-3.x

wrapper

defaultdict