列表索引更改多个元素 [英] List index changes multiple elements

问题描述

我找不到任何与我的问题相符的内容，所以希望这在某处并没有被提及过，我发现它太愚蠢了。

I couldn't find anything matching my problem, so hopefully this wasn't already mentioned somewhere and I'm just too stupid to find it.

thelist = []
a = [0]
for i in range(5):
    thelist.append(a)
print(thelist)

此时，程序返回[[0]，[0]，[0]，[0 ]，[0]]

At this point, the program returns [[0], [0], [0], [0], [0]]

thelist[0].append(1)
print(thelist)

在追加这个之后我会期望它返回相同的但是修改了第一个元素，如下所示：

After appending this I would expect it to return the same but with the first element modified, like this:

[[0, 1], [0], [0], [0], [0]]

实际发生的是，每个元素都以相同的方式被修改，我得到以下内容。

What actually happens, is that every element was modified in the same way and I get the following.

[[0, 1], [0, 1], [0, 1], [0, 1], [0, 1]]

我发现，如果我用a的值替换第2行中的a，一切正常。但是，当我多次附加变量时，为什么这不起作用？

I have found out, that if I replace the a in line 2 with the value of a, everything works fine. But why does this not work when I append a variable multiple times??

推荐答案

创建列表时，它会保存对该列表的引用每个职位都有相同的对象（a）。当您追加到第一个元素时，它实际上附加到所有thelist插槽中。您需要做的是在构造期间对对象进行深层复制：

When creating thelist, it's saving a reference to the same object (a) in each position. When you append to the first element, it's actually appending to a in all the "thelist" slots. What you need to do is a deep copy on the object during construction:

thelist = []
a = [0]
for i in range(5):
    thelist.append(list(a))
print(thelist)

然后以下附加内容将按预期工作。

Then the following append will work as desired.

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