Python:为什么在我实际上不更改列表时会更改它? [英] Python: why does my list change when I'm not actually changing it?

问题描述

新手有问题，请保持谦虚:

Newbie with a question, so please be gentle:

list = [1, 2, 3, 4, 5]
list2 = list

def fxn(list,list2):
    for number in list:
        print(number)
        print(list)
        list2.remove(number)
        print("after remove list is  ", list, " and list 2 is  ", list2)
    return list, list2

list, list2 = fxn(list, list2)
print("after fxn list is  ", list)
print("after fxn list2 is  ", list2)

结果是:

1
[1, 2, 3, 4, 5]
after remove list is   [2, 3, 4, 5]  and list 2 is   [2, 3, 4, 5]
3
[2, 3, 4, 5]
after remove list is   [2, 4, 5]  and list 2 is   [2, 4, 5]
5
[2, 4, 5]
after remove list is   [2, 4]  and list 2 is   [2, 4]
after fxn list is   [2, 4]
after fxn list2 is   [2, 4]

我不明白为什么仅在执行list2.remove()而不是list.remove()时列表会发生变化.我什至不知道要用什么搜索词来弄清楚.

I don't understand why list is changing when I am only doing list2.remove(), not list.remove(). I'm not even sure what search terms to use to figure it out.

推荐答案

这是因为list和list2在执行赋值list2=list之后都引用了相同的列表.

That's because both list and list2 are referring to the same list after you did the assignment list2=list.

尝试此操作以查看它们是指的是相同的对象还是不同的对象:

Try this to see if they are referring to the same objects or different:

id(list)
id(list2)

一个例子:

>>> list = [1, 2, 3, 4, 5]
>>> list2 = list
>>> id(list)
140496700844944
>>> id(list2)
140496700844944
>>> list.remove(3)
>>> list
[1, 2, 4, 5]
>>> list2
[1, 2, 4, 5]

如果您真的想创建list的副本，以使list2并不引用原始列表，而是引用列表的副本，请使用slice运算符:

If you really want to create a duplicate copy of list such that list2 doesn't refer to the original list but a copy of the list, use the slice operator:

list2 = list[:]

一个例子:

>>> list
[1, 2, 4, 5]
>>> list2
[1, 2, 4, 5]
>>> list = [1, 2, 3, 4, 5]
>>> list2 = list[:]
>>> id(list)
140496701034792
>>> id(list2)
140496701034864
>>> list.remove(3)
>>> list
[1, 2, 4, 5]
>>> list2
[1, 2, 3, 4, 5]

此外，请勿将list用作变量名，因为最初list是指类型列表，但是通过定义自己的list变量，您将隐藏引用的原始list类型列表.示例:

Also, don't use list as a variable name, because originally, list refers to the type list, but by defining your own list variable, you are hiding the original list that refers to the type list. Example:

>>> list
<type 'list'>
>>> type(list)
<type 'type'>
>>> list = [1, 2, 3, 4, 5]
>>> list
[1, 2, 3, 4, 5]
>>> type(list)
<type 'list'>

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