python--深浅拷贝

例子:

    a = [21,56,['liangchen','zhang']]

    1、b = a

  •         b指向的列表地址就是a指向的列表地址
  •         b列表中元素的值改变,a列表中的元素的值也跟着改变
a = [21,56,['liangchen','zhang']]
b = a
b[0] = 22
print(id(a))
print(id(b))
print(a,b)
输出:
1342819623624
1342819623624
[22, 56, ['liangchen', 'zhang']] [22, 56, ['liangchen', 'zhang']] 

    2、浅拷贝:b = a.copy()

  •         b重新开辟了一个新的地址
  •         b列表中元素值的改变,a并不会变,因为两者并不是指向的同一地址,只是两者初始值元素相同而已
  •         如果b列表中还有一层列表,那么改变第二层的列表的元素,则a对应的元素也会改变,因为第二层列表的地址和a中的第二层的列表的地址相同
a = [21,56,['liangchen','zhang']]
b = a.copy()
b[0] = 22
print(id(a))
print(id(b))
print(a,b)
输出:
2066816787144
2066817291720
[21, 56, ['liangchen', 'zhang']] [22, 56, ['liangchen', 'zhang']]
a = [21,56,['liangchen','zhang']]
b = a.copy()
b[2][1] = 'chen'
print(id(a[2]))
print(id(b[2]))
print(a,b)
输出:
1897167741512
1897167741512
[21, 56, ['liangchen', 'chen']] [21, 56, ['liangchen', 'chen']]

    3、深拷贝(导入:import copy):b = copy.deepcopy()

  •             b重新开辟了一个新的地址
  •             b的第二层列表也重新开辟了新的地址
  •             b列表中元素值的改变,a并不会变
  •             b嵌套的第二层列表中的元素改变,a也不会变
import copy
a = [21,56,['liangchen','zhang']]
b = copy.deepcopy(a)
b[2][1] = 'chen'
print(id(a),id(b))
print(id(a[2]),id(b[2]))
print(a,b)
输出:
1706386976584 1706387605832
1706387465352 1706387605896
[21, 56, ['liangchen', 'zhang']] [21, 56, ['liangchen', 'chen']]
posted @ 2019-08-19 15:46  BruceTyler  阅读(177)  评论(0编辑  收藏  举报