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Is there any difference between passing
lambda: 0 as arguments? Or between
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', ), ('d', )] [('a', [1, 1, 1, 1, 1]), ('r', [1, 1]), ('b', [1, 1]), ('c', ), ('d', )]
but are there any cases where they’ll have different behavior, or is it merely a notational difference?
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.
# alternating between `0` and `` from itertools import count factory = lambda c=count(): 0 if next(c) % 2 else  superdict = defaultdict(factory)