是的,这是正常的,因为列表在 python 中是可变的并且这个操作:
MySecondList = MyList
只需创建对同一列表对象的新引用并list.append
就地修改同一对象。(其他操作如+=
,list.extend
等list.pop
也会就地修改列表)
您可以在此处使用浅拷贝:
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])
对于列表列表(或可变对象列表),浅拷贝是不够的,因为内部列表(或对象)只是对同一对象的新引用:
>>> 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])
对于这种情况,请使用copy.deepcopy
:
>>> from copy import deepcopy
>>> lis = [[1,2,3],[4,5,6]]
>>> foo = deepcopy(lis)