如何从Deepcopy中排除特定引用？ [英] How to exclude specific references from deepcopy?

问题描述

我有一个对象，该对象具有自己的内容（即某物的列表），并且具有与另一个对象相关联的引用。如何从深度复制中排除对另一个对象的引用？

I have object which has it's own content (i.e. list of something) and a reference to another object, with which it is linked. How can I exclude the reference to the other object from being deep-copied?

from copy import deepcopy
class Foo:
    def __init__(self, content, linked_to):
        self.content = content
        self.linked_to = linked_to

a1 = Foo([[1,2],[3,4]], None)
a2 = Foo([[5,6],[7,8]], a1)

a3 = deepcopy(a2) # <- I don't want there, that a3.linked_to will be copied
# I want that a3.linked_to will still point to a1

a3.linked_to.content.append([9,10])
print a1.content # [[1,2],[3,4]], but I want [[1,2],[3,4], [9,10]] 

推荐答案

您的类可以实现 __deepcopy __ 方法来控制其复制方式。来自 copy 模块文档：

Your class can implement a __deepcopy__ method to control how it is copied. From the copy module documentation:

为了让一个类定义自己的副本实现，它可以定义特殊方法 __ copy __（ ）和 __ deepcopy __（）。前者被称为实现浅拷贝操作；没有传递其他参数。后者被称为实现深度复制操作。它传递了一个参数，即备忘字典。如果 __ deepcopy __（）实现需要对组件进行深拷贝，则应调用 deepcopy（）函数

In order for a class to define its own copy implementation, it can define special methods __copy__() and __deepcopy__(). The former is called to implement the shallow copy operation; no additional arguments are passed. The latter is called to implement the deep copy operation; it is passed one argument, the memo dictionary. If the __deepcopy__() implementation needs to make a deep copy of a component, it should call the deepcopy() function with the component as first argument and the memo dictionary as second argument.

简单地返回类的新实例，并带有引用不想被复制原样。使用 deepcopy（）函数复制其他对象：

Simply return a new instance of your class, with the reference you don't want to be deep-copied just taken across as-is. Use the deepcopy() function to copy other objects:

from copy import deepcopy

class Foo:
    def __init__(self, content, linked_to):
        self.content = content
        self.linked_to = linked_to

    def __deepcopy__(self, memo):
        # create a copy with self.linked_to *not copied*, just referenced.
        return Foo(deepcopy(self.content, memo), self.linked_to)

Demo ：

>>> a1 = Foo([[1, 2], [3, 4]], None)
>>> a2 = Foo([[5, 6], [7, 8]], a1)
>>> a3 = deepcopy(a2)
>>> a3.linked_to.content.append([9, 10])  # still linked to a1
>>> a1.content
[[1, 2], [3, 4], [9, 10]]
>>> a1 is a3.linked_to
True
>>> a2.content is a3.content  # content is no longer shared
False

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