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?
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)