是的,名单在Python可变的,这个操作是正常的。
MySecondList = MyList
简单地创建一个新的参考同一个列表对象,并list.append
修改同一个对象就地(其他操作,如+=
,list.extend
,list.pop
等也修改就地列表)
可以使用浅拷贝这里:
MySecondList = MyList[:]
演示:
>>> from sys import getrefcount
>>> lis = [1,2,3]
>>> foo = lis #creates a new reference to the same object [1,2,3]
>>> lis is foo
True
>>> getrefcount(lis) #number of references to the same object
3 #foo , lis and shell itself
#you can modify the list [1,2,3] from any of it's references
>>> foo.append(4)
>>> lis.append(5)
>>> foo,lis
([1, 2, 3, 4, 5], [1, 2, 3, 4, 5])
>>> lis = [1,2,3]
>>> foo = lis[:] #assigns a shallow copy of lis to foo
>>> foo is lis
False
>>> getrefcount(lis) #still 2(lis + shell_, as foo points to a different object
2
#different results here
>>> foo.append(4)
>>> lis.append(5)
>>> foo, lis
([1, 2, 3, 4], [1, 2, 3, 5])
对于列出的清单(或可变对象的列表)浅拷贝是不够的内部列表(或对象)是同一个对象只是新引用:
>>> lis = [[1,2,3],[4,5,6]]
>>> foo = lis[:]
>>> foo is lis #lis and foo are different
False
>>> [id(x) for x in lis] #but inner lists are still same
[3056076428L, 3056076716L]
>>> [id(x) for x in foo] #same IDs of inner lists, i.e foo[0] is lis[0] == True
[3056076428L, 3056076716L]
>>> foo[0][0] = 100 # modifying one will affect the other as well
>>> lis[0],foo[0]
([100, 2, 3], [100, 2, 3])
对于这样的情况下,使用copy.deepcopy
:
>>> from copy import deepcopy
>>> lis = [[1,2,3],[4,5,6]]
>>> foo = deepcopy(lis)
非常感谢你 –