How to “perfectly” override a dict?

Question

How can I make as \"perfect\" a subclass of dict as possible? The end goal is to have a simple dict in which the keys are lowercase.

It would seem

Answer 1

You can write an object that behaves like a dict quite easily with ABCs (Abstract Base Classes) from the collections.abc module. It even tells you if you missed a method, so below is the minimal version that shuts the ABC up.

from collections.abc import MutableMapping


class TransformedDict(MutableMapping):
    """A dictionary that applies an arbitrary key-altering
       function before accessing the keys"""

    def __init__(self, *args, **kwargs):
        self.store = dict()
        self.update(dict(*args, **kwargs))  # use the free update to set keys

    def __getitem__(self, key):
        return self.store[self._keytransform(key)]

    def __setitem__(self, key, value):
        self.store[self._keytransform(key)] = value

    def __delitem__(self, key):
        del self.store[self._keytransform(key)]

    def __iter__(self):
        return iter(self.store)
    
    def __len__(self):
        return len(self.store)

    def _keytransform(self, key):
        return key

You get a few free methods from the ABC:

class MyTransformedDict(TransformedDict):

    def _keytransform(self, key):
        return key.lower()


s = MyTransformedDict([('Test', 'test')])

assert s.get('TEST') is s['test']   # free get
assert 'TeSt' in s                  # free __contains__
                                    # free setdefault, __eq__, and so on

import pickle
# works too since we just use a normal dict
assert pickle.loads(pickle.dumps(s)) == s

I wouldn't subclass dict (or other builtins) directly. It often makes no sense, because what you actually want to do is implement the interface of a dict. And that is exactly what ABCs are for.