One-step initialization of defaultdict that appends to list?

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)

What you're apparently missing is that defaultdict is a straightforward (not especially "magical") subclass of dict. All the first argument does is provide a factory function for missing keys. When you initialize a defaultdict, you're initializing a dict.

If you want to produce

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

you should be initializing it the way you would initialize any other dict whose values are lists:

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

If your initial data has to be in the form of tuples whose 2nd element is always an integer, then just go with the for loop. You call it one extra step; I call it the clear and obvious way to do it.