切片操作给我一个深或浅的副本？ [英] Does a slicing operation give me a deep or shallow copy?

问题描述

官方Python文档说，使用切片运算符并在Python中分配

The official Python docs say that using the slicing operator and assigning in Python makes a shallow copy of the sliced list.

但是当我编写代码的例子：

But when I write code for example:

o = [1, 2, 4, 5]
p = o[:]

当我写：

id(o)
id(p)

我获得不同的ID，并且添加一个列表不反映在其他列表中。

I get different id's and also appending one one list does not reflect in the other list. Isn't it creating a deep copy or is there somewhere I am going wrong?

推荐答案

您正在创建一个浅复制，因为嵌套值不会被复制，只是被引用。

You are creating a shallow copy, because nested values are not copied, merely referenced. A deep copy would create copies of the values referenced by the list too.

演示：

>>> lst = [{}]
>>> lst_copy = lst[:]
>>> lst_copy[0]['foo'] = 'bar'
>>> lst_copy.append(42)
>>> lst
[{'foo': 'bar'}]
>>> id(lst) == id(lst_copy)
False
>>> id(lst[0]) == id(lst_copy[0])
True

在这里，不复制嵌套的字典;它仅由两个列表引用。

Here the nested dictionary is not copied; it is merely referenced by both lists. The new element 42 is not shared.

请记住，Python中的所有内容都是一个对象，名称和列表元素仅仅是对那些对象的引用。列表的副本创建新的外部列表，但新列表仅接收对完全相同的对象的引用。

Remember that everything in Python is an object, and names and list elements are merely references to those objects. A copy of a list creates a new outer list, but the new list merely receives references to the exact same objects.

以递归方式创建列表中包含的每个对象的新副本：

A proper deep copy creates new copies of each and every object contained in the list, recursively:

>>> from copy import deepcopy
>>> lst_deepcopy = deepcopy(lst)
>>> id(lst_deepcopy[0]) == id(lst[0])
False

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