更改字典中的列表会更改所有列表 [英] Changing list in dictionary changes all lists

问题描述

我在这里阅读了这些帖子，确切地解释了我所遇到的问题，但没有一种解决方案有效:

I have read these posts here explaining exactly the problem I am having, but none of the solutions are working:

更改一个列表也会意外地更改另一个列表

更改一个dict值会更改所有值

我具有2D数组的特定形状，需要使用它来初始化字典中的条目.这是我的操作方式:

I have a specific shape of 2D array I need to initialize the entries in a dictionary with. Here is how I have done it:

empty = []
for x in range(2):
    empty.append([])
    for y in range(2):
        empty[x].append(False)

status = {k:[] for k in ["a", "b", "c"]}

status["a"] = list(empty)
status["b"] = list(empty)
status["c"] = list(empty)

print(status)
status["a"][0][0] = True
print(status)

(例如，简化列表的形状)

(Shape of list simplified for example)

此打印:

{'a': [[False, False], [False, False]], 'b': [[False, False], [False, False]], 'c': [[False, False], [False, False]]}
{'a': [[True, False], [False, False]], 'b': [[True, False], [False, False]], 'c': [[True, False], [False, False]]}

如您所见，设置列表值之一会更改所有列表.我不希望这样，我希望它们是具有不同值的单独列表(在一本词典中).

As you can see, setting one of the lists values changes all of the lists. I do not want this, I want them to be separate lists (in one dictionary) with different values.

最初，我认为我已经做了旧的 newlist = oldlist 错误，在该错误中我将newlist设置为与oldlist相同的对象，但是没有.如您在代码中所见，我正在使用 newlist = list(oldlist)制作单独的列表.我也尝试过 newlist = oldlist [:] ， newlist = oldlist.copy()等.

Initially, I thought I had done the old newlist = oldlist blunder where I set newlist to the same object as oldlist, but nope. As you can see in my code, I am making separate lists using newlist = list(oldlist). I have also tried newlist = oldlist[:], newlist = oldlist.copy(), etc.

我想念什么?

推荐答案

from copy import deepcopy
empty = []
for x in range(2):
    empty.append([])
    for y in range(2):
        empty[x].append(False)

status = {k:[] for k in ["a", "b", "c"]}

status["a"] = deepcopy(empty)
status["b"] = deepcopy(empty)
status["c"] = deepcopy(empty)

print(status)
status["a"][0][0] = True
print(status)

您与副本关系密切，但实际上需要深拷贝

you were close with the copy, but you actually neeed a deepcopy

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