python defaultdict: 0 vs. int and [] vs list

Question

Is there any difference between passing int and lambda: 0 as arguments? Or between list and lambda: []?

It looks like they do the same thing:

from collections import defaultdict
dint1 = defaultdict(lambda: 0)
dint2 = defaultdict(int)
dlist1 = defaultdict(lambda: [])
dlist2 = defaultdict(list)

for ch in 'abracadabra':
    dint1[ch] += 1
    dint2[ch] += 1
    dlist1[ch].append(1)
    dlist2[ch].append(1)

print dint1.items()
print dint2.items()
print dlist1.items()
print dlist2.items()
## -- Output: --
[('a', 5), ('r', 2), ('b', 2), ('c', 1), ('d', 1)]
[('a', 5), ('r', 2), ('b', 2), ('c', 1), ('d', 1)]
[('a', [1, 1, 1, 1, 1]), ('r', [1, 1]), ('b', [1, 1]), ('c', [1]), ('d', [1])]
[('a', [1, 1, 1, 1, 1]), ('r', [1, 1]), ('b', [1, 1]), ('c', [1]), ('d', [1])]

but are there any cases where they'll have different behavior, or is it merely a notational difference?

Answer 1

All that defaultdict requires is a callable object that will return what should be used as a default value when called with no parameters.

If you were to call the int constructor, it would return 0 and if you were to call lambda: 0, it would return 0. Same with the lists. The only difference here is that the constructor will always use it's logic to create the object. A lambda, you could add additional logic if you chose to do so.

e.g.,

# alternating between `0` and `[]`
from itertools import count
factory = lambda c=count(): 0 if next(c) % 2 else []
superdict = defaultdict(factory)