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), (         


        

Answer 1

Sorting and itertools.groupby go a long way:

>>> L = [('a', 1), ('b', 2), ('c', 3), ('d', 4), ('a', 2), ('b', 3)]
>>> L.sort(key=lambda t:t[0])
>>> d = defaultdict(list, [(tup[0], [t[1] for t in tup[1]]) for tup in itertools.groupby(L, key=lambda t: t[0])])
>>> d
defaultdict(, {'a': [1, 2], 'c': [3], 'b': [2, 3], 'd': [4]})

To make this more of a one-liner:

L = [('a', 1), ('b', 2), ('c', 3), ('d', 4), ('a', 2), ('b', 3)]
d = defaultdict(list, [(tup[0], [t[1] for t in tup[1]]) for tup in itertools.groupby(sorted(L, key=operator.itemgetter(0)), key=lambda t: t[0])])

Hope this helps