Python:为什么listA.append('a')影响listB? [英] python: Why listA.append('a') affects listB?

问题描述

这是我的代码:

In [8]: b=dict.fromkeys([1,2,3,4], [])

In [9]: b[1].append(1)

In [10]: b[2].append(2)

In [11]: b[1]
Out[11]: [1, 2]

In [12]: b[2]
Out[12]: [1, 2]

In [13]: b
Out[13]: {1: [1, 2], 2: [1, 2], 3: [1, 2], 4: [1, 2]}

我希望:{1:[1]，2:[2]，3:[]，4:[]}

Whereas I would expect: {1: [1], 2: [2], 3: [], 4: []}

我想这可能是b [X]只是一个引用"所致，它们都引用相同的列表.

I guess this is maybe caused by b[X] is just a "reference", they all refer to a same list.

然后我将[]替换为一个int对象.结果让我更加困惑:

Then I replace [] with an int object. The result makes me more confused:

In [15]: b=dict.fromkeys([1,2,3,4], 1)

In [16]: b[1] += 1

In [17]: b[2] += 1

In [18]: b
Out[18]: {1: 2, 2: 2, 3: 1, 4: 1}

在这种情况下，此int对象1不是引用.

This int object 1 isn't a referance in this case.

然后我将[]替换为['a']:

Then I replace [] with ['a']:

In [19]: b=dict.fromkeys([1,2,3,4], ['a'])

In [20]: b[1].append(1)

In [21]: b[2].append(2)

In [22]: b
Out[22]: {1: ['a', 1, 2], 2: ['a', 1, 2], 3: ['a', 1, 2], 4: ['a', 1, 2]}

现在['a']再次成为参考.

Now ['a'] is a reference again.

在第一种情况下，有人可以告诉我原因以及如何获得预期的结果"{1:[1]，2:[2]，3:[]，4:[]}".

Can someone tells me why, and how to get expected result "{1: [1], 2: [2], 3: [], 4: []}" in the first case.

任何有用的建议都会受到赞赏.

Any useful suggestions are appreciated.

推荐答案

由于dict中的所有值实际上都是对同一列表的引用，因此dict.fromkeys使用相同的列表对象并将其分配给每个键.由于list.append是就地操作，因此所有键都会受到影响.

Because all values in the dict are actually references to the same list, dict.fromkeys uses the same list object and assigns it to each key. As list.append is an in-place operation so all keys are affected.

>>> b = dict.fromkeys([1,2,3,4], [])
>>> [id(x) for x in b.values()]
[158948300, 158948300, 158948300, 158948300]

因此，对于可变值，请使用dict理解:

So, for mutable value use a dict comprehension:

>>> b = {k:[] for k in xrange(1, 5)}
>>> [id(x) for x in b.values()]
[158945580, 158948396, 158948108, 158946764]

或者按照@Bakuriu的建议，collections.defaultdict也可以正常工作:

Or as @Bakuriu suggested, collections.defaultdict will also work fine:

>>> from collections import defaultdict
>>> dic = defaultdict(list)
>>> dic[1].append(1)
>>> dic[2].append(2)
>>> dic
defaultdict(<type 'list'>, {1: [1], 2: [2]})
>>> dic[3]
[]

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