Python:在复制的列表中交换列表元素而不会影响原始列表 [英] Python: swap list elements in a copied list without affecting the original list

问题描述

我有一个列表a和一个列表b，该列表(应该是)列表a的副本.

I have a list a and a list b which is (should be) a copy of list a.

a = [[['a'], ['b'], ['c']], [['A'], ['B'], ['C']]]
b = a[:][:]
b[0][1], b[0][2] = b[0][2], b[0][1]

如果我现在查看a和b，我将得到以下信息:

If I now look at a and b I get the following:

a = [[['a'], ['c'], ['b']], [['A'], ['B'], ['C']]]
b = [[['a'], ['c'], ['b']], [['A'], ['B'], ['C']]]

为什么列表b中的交换也会影响原始列表a?

Why does the swap in list b also affects the original list a?

谢谢.

推荐答案

b = a[:][:]只是b = (a[:])[:]或原始列表副本的副本.原始列表中的列表仍然被引用，并且当您更改它们时，它会显示在两个列表中.

b = a[:][:] is just b = (a[:])[:] or a copy of a copy of the original list. The lists inside the original list are still referenced and when you change them it shows in both lists.

你可以做

b = [l[:] for l in a] # a new list, consisting of copies each sublist

或

from copy import deepcopy
b = deepcopy(a)

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