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

I think most of this is a lot of smoke and mirrors to avoid a simple for loop:

di={}
for k,v in [('a', 1), ('b', 2), ('c', 3), ('d', 4), ('a', 2),('b', 3)]:
    di.setdefault(k,[]).append(v)
# di={'a': [1, 2], 'c': [3], 'b': [2, 3], 'd': [4]}

If your goal is one line and you want abusive syntax that I cannot at all endorse or support you can use a side effect comprehension:

>>> li=[('a', 1), ('b', 2), ('c', 3), ('d', 4), ('a', 2),('b', 3)]
>>> di={};{di.setdefault(k[0],[]).append(k[1]) for k in li}
set([None])
>>> di
{'a': [1, 2], 'c': [3], 'b': [2, 3], 'd': [4]}

If you really want to go overboard into the unreadable:

>>> {k1:[e for _,e in v1] for k1,v1 in {k:filter(lambda x: x[0]==k,li) for k,v in li}.items()}
{'a': [1, 2], 'c': [3], 'b': [2, 3], 'd': [4]}

You don't want to do that. Use the for loop Luke!