Python:列表和它们的副本 [英] Python: lists and copy of them

问题描述

我无法解释以下行为:

l1 = [1, 2, 3, 4]
l1[:][0] = 888
print(l1) # [1, 2, 3, 4]
l1[:] = [9, 8, 7, 6]
print(l1) # [9, 8, 7, 6]

似乎l1[:][0]是指副本，而l1[:]是指对象本身.

It seems to be that l1[:][0] refers to a copy, whereas l1[:] refers to the object itself.

推荐答案

l1[:][0] = 888首先获取l1(l1[:])中所有元素的一部分，(根据列表语义)返回一个新的列表对象包含l1中的所有对象-它是l1的浅表副本.

l1[:][0] = 888 first takes a slice of all the elements in l1 (l1[:]), which (as per list semantics) returns a new list object containing all the objects in l1 -- it's a shallow copy of l1.

然后用整数888([0] = 888)替换该复制列表的第一个元素.

It then replaces the first element of that copied list with the integer 888 ([0] = 888).

然后，复制的列表将被丢弃，因为它没有执行任何操作.

Then, the copied list is discarded because nothing is done with it.

第二个示例l1[:] = [9, 8, 7, 6] 将l1中的所有元素替换为列表[9, 8, 7, 6]中的元素.这是切片任务.

Your second example l1[:] = [9, 8, 7, 6] replaces all the elements in l1 with those in the list [9, 8, 7, 6]. It's a slice assignment.

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