import copy

a = "deepak"
b = 1, 2, 3, 4
c = [1, 2, 3, 4]
d = {1: 10, 2: 20, 3: 30}

a1 = copy.copy(a)
b1 = copy.copy(b)
c1 = copy.copy(c)
d1 = copy.copy(d)


print("immutable - id(a)==id(a1)", id(a) == id(a1))
print("immutable - id(b)==id(b1)", id(b) == id(b1))
print("mutable - id(c)==id(c1)", id(c) == id(c1))
print("mutable - id(d)==id(d1)", id(d) == id(d1))

我得到了以下结果:

immutable - id(a)==id(a1) True
immutable - id(b)==id(b1) True
mutable - id(c)==id(c1) False
mutable - id(d)==id(d1) False

如果执行deepcopy:

a1 = copy.deepcopy(a)
b1 = copy.deepcopy(b)
c1 = copy.deepcopy(c)
d1 = copy.deepcopy(d)

结果是一样的:

immutable - id(a)==id(a1) True
immutable - id(b)==id(b1) True
mutable - id(c)==id(c1) False
mutable - id(d)==id(d1) False

如果我做赋值操作:

a1 = a
b1 = b
c1 = c
d1 = d

结果如下:

immutable - id(a)==id(a1) True
immutable - id(b)==id(b1) True
mutable - id(c)==id(c1) True
mutable - id(d)==id(d1) True

谁能解释一下这些副本之间的区别是什么?它与可变和不可变对象有关吗?如果有,你能解释一下吗?


当前回答

对于不可变对象,创建一个副本没有多大意义,因为它们不会改变。对于可变对象赋值,copy和deepcopy的行为不同。让我们用例子来讨论每一个。

赋值操作只是将源的引用赋值给目标,例如:

>>> i = [1,2,3]
>>> j=i
>>> hex(id(i)), hex(id(j))
>>> ('0x10296f908', '0x10296f908') #Both addresses are identical

现在i和j技术上指向同一个列表。i和j都有相同的内存地址。任何更新 其中一个会被另一个反射。例句:

>>> i.append(4)
>>> j
>>> [1,2,3,4] #Destination is updated

>>> j.append(5)
>>> i
>>> [1,2,3,4,5] #Source is updated

另一方面,copy和deepcopy会创建一个新的变量副本。所以现在对原始变量的更改将不会被反映 复制变量,反之亦然。但是复制(浅复制),不创建嵌套对象的副本,而只是 复制嵌套对象的引用。Deepcopy递归复制所有嵌套对象。

一些演示copy和deepcopy行为的例子:

使用copy的平面列表示例:

>>> import copy
>>> i = [1,2,3]
>>> j = copy.copy(i)
>>> hex(id(i)), hex(id(j))
>>> ('0x102b9b7c8', '0x102971cc8') #Both addresses are different

>>> i.append(4)
>>> j
>>> [1,2,3] #Updation of original list didn't affected copied variable

使用copy的嵌套列表示例:

>>> import copy
>>> i = [1,2,3,[4,5]]
>>> j = copy.copy(i)

>>> hex(id(i)), hex(id(j))
>>> ('0x102b9b7c8', '0x102971cc8') #Both addresses are still different

>>> hex(id(i[3])), hex(id(j[3]))
>>> ('0x10296f908', '0x10296f908') #Nested lists have same address

>>> i[3].append(6)
>>> j
>>> [1,2,3,[4,5,6]] #Updation of original nested list updated the copy as well

使用deepcopy的平面列表示例:

>>> import copy
>>> i = [1,2,3]
>>> j = copy.deepcopy(i)
>>> hex(id(i)), hex(id(j))
>>> ('0x102b9b7c8', '0x102971cc8') #Both addresses are different

>>> i.append(4)
>>> j
>>> [1,2,3] #Updation of original list didn't affected copied variable

使用deepcopy的嵌套列表示例:

>>> import copy
>>> i = [1,2,3,[4,5]]
>>> j = copy.deepcopy(i)

>>> hex(id(i)), hex(id(j))
>>> ('0x102b9b7c8', '0x102971cc8') #Both addresses are still different

>>> hex(id(i[3])), hex(id(j[3]))
>>> ('0x10296f908', '0x102b9b7c8') #Nested lists have different addresses

>>> i[3].append(6)
>>> j
>>> [1,2,3,[4,5]] #Updation of original nested list didn't affected the copied variable    

其他回答

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]))

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>
>>lst=[1,2,3,4,5]

>>a=lst

>>b=lst[:]

>>> b
[1, 2, 3, 4, 5]

>>> a
[1, 2, 3, 4, 5]

>>> lst is b
False

>>> lst is a
True

>>> id(lst)
46263192

>>> id(a)
46263192 ------>  See here id of a and id of lst is same so its called deep copy and even boolean answer is true

>>> id(b)
46263512 ------>  See here id of b and id of lst is not same so its called shallow copy and even boolean answer is false although output looks same.

不确定上面是否提到过,但理解.copy()创建对原始对象的引用是非常重要的。如果你改变了复制的对象-你改变了原始对象。 .deepcopy()创建新对象并真正复制原始对象到新对象。改变新的深度复制对象不会影响原始对象。

是的,.deepcopy()递归复制原始对象,而.copy()创建一个引用对象到原始对象的一级数据。

因此.copy()和.deepcopy()之间的复制/引用差异是显著的。