Python dictionary that defaults to key?

I'd override the __missing__ method of dict:

>>> class MyDefaultDict(dict):...     def __missing__(self, key):...         self[key] = key...         return key...>>> d = MyDefaultDict()>>> d['joe']'joe'>>> d{'joe': 'joe'}

Edit: Oops, I just realized that code in my file originally came from another stackoverflow answer! https://stackoverflow.com/a/2912455/456876, go upvote that one.

This is what I use - it's a defaultdict variant that passes the key as an argument to the default-value factory function that's passed as an argument to init, instead of no arguments:

class keybased_defaultdict(defaultdict):    def __missing__(self, key):        if self.default_factory is None:            raise KeyError(key)        else:            value = self[key] = self.default_factory(key)            return value

This is the use you want:

>>> d = keybased_defaultdict(lambda x: x)>>> d[1]1>>> d['a']'a'

Other possibilities:

>>> d = keybased_defaultdict(lambda x: len(x))>>> d['a']1>>> d['abc']3

If you don't want to subclass dict, you can try using

d.get('a', 'a')d.get('b', 'b')d.get('c', 'c')

Which I think is clearer and less magical for this purpose

If you are a DRY fanatic and only have single char keys, you can do this :)

d.get(*'a'*2)d.get(*'b'*2)d.get(*'c'*2)