A, b, c, d, a1, b1, c1和d1是对内存中对象的引用，它们由它们的id唯一标识。

An assignment operation takes a reference to the object in memory and assigns that reference to a new name. c=[1,2,3,4] is an assignment that creates a new list object containing those four integers, and assigns the reference to that object to c. c1=c is an assignment that takes the same reference to the same object and assigns that to c1. Since the list is mutable, anything that happens to that list will be visible regardless of whether you access it through c or c1, because they both reference the same object.

C1 =copy.copy(c)是一个“浅拷贝”，它创建一个新列表，并将对新列表的引用赋值给C1。C仍然指向原始的列表。所以，如果你在c1处修改列表，c所指向的列表不会改变。

复制的概念与整数和字符串等不可变对象无关。由于不能修改这些对象，因此不需要在内存的不同位置有相同值的两个副本。因此，整数和字符串，以及其他一些复制概念不适用的对象，只是简单地重新赋值。这就是为什么使用a和b的例子会得到相同的id。

c1=copy.deepcopy(c) is a "deep copy", but it functions the same as a shallow copy in this example. Deep copies differ from shallow copies in that shallow copies will make a new copy of the object itself, but any references inside that object will not themselves be copied. In your example, your list has only integers inside it (which are immutable), and as previously discussed there is no need to copy those. So the "deep" part of the deep copy does not apply. However, consider this more complex list:

E = [[1,2]，[4,5,6]，[7,8,9]]]

这是一个包含其他列表的列表(也可以将其描述为一个二维数组)。

If you run a "shallow copy" on e, copying it to e1, you will find that the id of the list changes, but each copy of the list contains references to the same three lists -- the lists with integers inside. That means that if you were to do e[0].append(3), then e would be [[1, 2, 3],[4, 5, 6],[7, 8, 9]]. But e1 would also be [[1, 2, 3],[4, 5, 6],[7, 8, 9]]. On the other hand, if you subsequently did e.append([10, 11, 12]), e would be [[1, 2, 3],[4, 5, 6],[7, 8, 9],[10, 11, 12]]. But e1 would still be [[1, 2, 3],[4, 5, 6],[7, 8, 9]]. That's because the outer lists are separate objects that initially each contain three references to three inner lists. If you modify the inner lists, you can see those changes no matter if you are viewing them through one copy or the other. But if you modify one of the outer lists as above, then e contains three references to the original three lists plus one more reference to a new list. And e1 still only contains the original three references.

“深度复制”不仅会复制外部列表，而且还会进入列表内部并复制内部列表，因此两个结果对象不包含任何相同的引用(就可变对象而言)。如果内部列表中有更多的列表(或其他对象，如字典)，它们也会被复制。这是“深层复制”的“深层”部分。

下面的代码显示了底层地址在复制、深度复制和赋值时是如何受到影响的。这类似于Sohaib Farooqi用列表展示的东西，但是用类。

from copy import deepcopy, copy

class A(object):
    """docstring for A"""
    def __init__(self):
        super().__init__()

class B(object):
    """docstring for B"""
    def __init__(self):
        super().__init__()
        self.myA = A()

a = B()
print("a is", a)
print("a.myA is", a.myA)
print("After copy")
b = copy(a)
print("b is", b)
print("b.myA is", b.myA)
b.myA = A()
print("-- after changing value")
print("a is", a)
print("a.myA is", a.myA)
print("b is", b)
print("b.myA is", b.myA)

print("Resetting")
print("*"*40)
a = B()
print("a is", a)
print("a.myA is", a.myA)
print("After deepcopy")
b = deepcopy(a)
print("b is", b)
print("b.myA is", b.myA)
b.myA = A()
print("-- after changing value")
print("a is", a)
print("a.myA is", a.myA)
print("b is", b)
print("b.myA is", b.myA)

print("Resetting")
print("*"*40)
a = B()
print("a is", a)
print("a.myA is", a.myA)
print("After assignment")
b = a
print("b is", b)
print("b.myA is", b.myA)
b.myA = A()
print("-- after changing value")
print("a is", a)
print("a.myA is", a.myA)
print("b is", b)
print("b.myA is", b.myA)

这段代码的输出如下:

a is <__main__.B object at 0x7f1d8ff59760>
a.myA is <__main__.A object at 0x7f1d8fe8f970>
After copy
b is <__main__.B object at 0x7f1d8fe43280>
b.myA is <__main__.A object at 0x7f1d8fe8f970>
-- after changing value
a is <__main__.B object at 0x7f1d8ff59760>
a.myA is <__main__.A object at 0x7f1d8fe8f970>
b is <__main__.B object at 0x7f1d8fe43280>
b.myA is <__main__.A object at 0x7f1d8fe85820>
Resetting
****************************************
a is <__main__.B object at 0x7f1d8fe85370>
a.myA is <__main__.A object at 0x7f1d8fe43310>
After deepcopy
b is <__main__.B object at 0x7f1d8fde3040>
b.myA is <__main__.A object at 0x7f1d8fde30d0>
-- after changing value
a is <__main__.B object at 0x7f1d8fe85370>
a.myA is <__main__.A object at 0x7f1d8fe43310>
b is <__main__.B object at 0x7f1d8fde3040>
b.myA is <__main__.A object at 0x7f1d8fe43280>
Resetting
****************************************
a is <__main__.B object at 0x7f1d8fe432b0>
a.myA is <__main__.A object at 0x7f1d8fe85820>
After assignment
b is <__main__.B object at 0x7f1d8fe432b0>
b.myA is <__main__.A object at 0x7f1d8fe85820>
-- after changing value
a is <__main__.B object at 0x7f1d8fe432b0>
a.myA is <__main__.A object at 0x7f1d8fe85370>
b is <__main__.B object at 0x7f1d8fe432b0>
b.myA is <__main__.A object at 0x7f1d8fe85370>

深度复制与嵌套结构有关。如果你有列表的列表，那么deepcopy也复制嵌套的列表，所以它是递归复制。通过复制，您有一个新的外部列表，但内部列表是引用。赋值不复制。 为前

import copy
spam = [[0, 1, 2, 3], 4, 5]
cheese = copy.copy(spam)
cheese.append(3)
cheese[0].append(3)
print(spam)
print(cheese)

输出

[[0,1,2,3,3]， 4,5] [[0,1,2,3,3]， 4,5,3] 复制方法将外部列表的内容复制到新列表，但两个列表的内部列表仍然相同，因此如果你在任何列表的内部列表中做出更改，它将影响两个列表。

但是如果你使用深度复制，它也会为内部列表创建新的实例。

import copy
spam = [[0, 1, 2, 3], 4, 5]
cheese = copy.deepcopy(spam)
cheese.append(3)
cheese[0].append(3)
print(spam)
print(cheese)

输出

[0, 1, 2, 3] [[0, 1, 2, 3, 3], 4, 5, 3]

对于不可变对象，不需要复制，因为数据永远不会改变，所以Python使用相同的数据;id总是相同的。对于可变对象，因为它们可能会改变，[浅]复制创建一个新对象。

深度复制与嵌套结构有关。如果你有列表的列表，那么deepcopy也复制嵌套的列表，所以它是递归复制。通过复制，您有一个新的外部列表，但内部列表是引用。

赋值不复制。它只是将引用设置为旧数据。因此，您需要复制来创建一个具有相同内容的新列表。

下面的代码演示了赋值、使用复制方法的浅复制、使用(slice)[:]的浅复制和深度复制之间的区别。下面的示例使用嵌套列表，使差异更加明显。

from copy import deepcopy

########"List assignment (does not create a copy) ############
l1 = [1,2,3, [4,5,6], [7,8,9]]
l1_assigned = l1

print(l1)
print(l1_assigned)

print(id(l1), id(l1_assigned))
print(id(l1[3]), id(l1_assigned[3]))
print(id(l1[3][0]), id(l1_assigned[3][0]))

l1[3][0] = 100
l1.pop(4)
l1.remove(1)


print(l1)
print(l1_assigned)
print("###################################")

########"List copy using copy method (shallow copy)############

l2 = [1,2,3, [4,5,6], [7,8,9]]
l2_copy = l2.copy()

print(l2)
print(l2_copy)

print(id(l2), id(l2_copy))
print(id(l2[3]), id(l2_copy[3]))
print(id(l2[3][0]), id(l2_copy[3][0]))
l2[3][0] = 100
l2.pop(4)
l2.remove(1)


print(l2)
print(l2_copy)

print("###################################")

########"List copy using slice (shallow copy)############

l3 = [1,2,3, [4,5,6], [7,8,9]]
l3_slice = l3[:]

print(l3)
print(l3_slice)

print(id(l3), id(l3_slice))
print(id(l3[3]), id(l3_slice[3]))
print(id(l3[3][0]), id(l3_slice[3][0]))

l3[3][0] = 100
l3.pop(4)
l3.remove(1)


print(l3)
print(l3_slice)

print("###################################")

########"List copy using deepcopy ############

l4 = [1,2,3, [4,5,6], [7,8,9]]
l4_deep = deepcopy(l4)

print(l4)
print(l4_deep)

print(id(l4), id(l4_deep))
print(id(l4[3]), id(l4_deep[3]))
print(id(l4[3][0]), id(l4_deep[3][0]))

l4[3][0] = 100
l4.pop(4)
l4.remove(1)

print(l4)
print(l4_deep)
print("##########################")
print(l4[2], id(l4[2]))
print(l4_deep[3], id(l4_deep[3]))

print(l4[2][0], id(l4[2][0]))
print(l4_deep[3][0], id(l4_deep[3][0]))

让我们在一个图形示例中看看下面的代码是如何执行的:

import copy

class Foo(object):
    def __init__(self):
        pass


a = [Foo(), Foo()]
shallow = copy.copy(a)
deep = copy.deepcopy(a)