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

问题描述

传递 int 和 lambda：0 之间是否有什么区别？或列表和 lambda：[] ？

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

看起来他们做同样的事情：

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])]

$ b

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

推荐答案

如果您要调用 int 构造函数，则会返回 0 ，如果您调用 lambda：0 ，它将返回 0 。与列表相同。这里唯一的区别是，构造函数总是使用它的逻辑来创建对象。如果您选择这样做，您可以添加其他逻辑。

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.

例如，

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

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