如何在python中按值分配 [英] How can I assign by value in python

问题描述

我知道，由于Python的工作方式x = []; y = x; x.append(1); y将打印[1].但是，相反，

I understand that, due to the way Python works x = []; y = x; x.append(1); y will print [1]. However, the reverse, say,

z = [1,2]
temp = z
temp[1] = 3
z,temp

将打印([1,3],[1,3]).如果我理解正确，那么z和temp都指向同一列表，因此更改列表将更改另一个列表，因为列表是可变的.如何防止这种情况发生?即，我想进行一个for循环，将z复制到temp，以不同的方式进行更改，然后将其推送到队列中.为此，z必须始终包含基本数组，因此我需要更改temp时不要更改z.

will print ([1,3],[1,3]). If I understand correctly, both z and temp point to the same list, so changing one will change the other, seeing as lists are mutable. How can I prevent this from happening? Namely, I want to make a for loop that will copy z into temp, change it in different ways, and push it onto a queue. For that to work, z must always contain the base array, therefore I need that changing temp doesn't change z.

我尝试将z更改为元组，以便z=z,，然后调用z[0]而不是z.但这仍然不能解决我的问题.

I tried changing z into a tuple so that z=z,, then calling z[0] instead of z. Still this doesn't solve my problem.

推荐答案

复制列表很容易...只需对其切片:

Copying a list is easy ... Just slice it:

temp = z[:]

这将创建一个浅表副本-列表中元素的突变将显示在z中的元素中，但不会直接更改为temp.

This will create a shallow copy -- mutations to elements in the list will show up in the elements in z, but not changes to temp directly.

出于更一般的目的，python有一个copy模块可供您使用:

For more general purposes, python has a copy module that you can use:

temp = copy.copy(z)

或者，也许:

temp = copy.deepcopy(z)

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