One-step initialization of defaultdict that appends to list?

Question

It would be convenient if a defaultdict could be initialized along the following lines

d = defaultdict(list, (('a', 1), ('b', 2), ('c', 3), ('d', 4), ('a', 2),
   ('b', 3)))

to produce

defaultdict(<type 'list'>, {'a': [1, 2], 'c': [3], 'b': [2, 3], 'd': [4]})

Instead, I get

defaultdict(<type 'list'>, {'a': 2, 'c': 3, 'b': 3, 'd': 4})

To get what I need, I end up having to do this:

d = defaultdict(list)
for x, y in (('a', 1), ('b', 2), ('c', 3), ('d', 4), ('a', 2), ('b', 3)):
    d[x].append(y)

This is IMO one step more than should be necessary, am I missing something here?

Answer 1

the behavior you describe would not be consistent with the defaultdicts other behaviors. Seems like what you want is FooDict such that

>>> f = FooDict()
>>> f['a'] = 1
>>> f['a'] = 2
>>> f['a']
[1, 2]

We can do that, but not with defaultdict; lets call it AppendDict

import collections

class AppendDict(collections.MutableMapping):
    def __init__(self, container=list, append=None, pairs=()):
        self.container = collections.defaultdict(container)
        self.append = append or list.append
        for key, value in pairs:
            self[key] = value

    def __setitem__(self, key, value):
        self.append(self.container[key], value)

    def __getitem__(self, key): return self.container[key]
    def __delitem__(self, key): del self.container[key]
    def __iter__(self): return iter(self.container)
    def __len__(self): return len(self.container)