如何从 deepcopy 中排除特定引用？

Question

我有一个对象，它有自己的内容（即某物的列表）和对另一个对象的引用，它与之链接。如何排除对其他对象的引用被深度复制？

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]]

Answer 1

您的 class 可以实现 __deepcopy__ 方法来控制它的复制方式。来自 copy module documentation:

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.

只是 return 您的 class 的一个新实例，您不想被深度复制的参考只是按原样进行。使用deepcopy()函数复制其他对象：

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)

演示：

>>> 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

如何从 deepcopy 中排除特定引用？

How to exclude specific references from deepcopy?

python

deep-copy