Python对象混乱：a = b，修改b并进行更改！

Question

我以为我知道Python，直到今晚。做这样的事情的正确方法是什么？这是我的代码：

a = ["one", "two", "three"]
b = a  # here I want a complete copy that when b is changed, has absolutely no effect on a
b.append["four"]
print a  # a now has "four" in it

基本上，我想知道，而不是该b = a步骤，我将如何正确地制作列表或字典的副本，以便b更改时a不会随之更改？

Answer 1

您正在体验的是引用的概念。Python中的所有对象都有一个引用，当您将一个名称分配给两个名称a和时b，这将导致两者a并b指向 _同一_对象。

>>> a = range(3)
>>> b = a                     # same object
>>> b.append(3)
>>> a, b                      # same contents
([0, 1, 2, 3], [0, 1, 2, 3])

随着名单，你可以创建一个新的列表b是一个 副本， 另一个a使用b = a[:]。

>>> a = range(3)
>>> b = a[:]                  # make b a new copy of a
>>> b.append(3)
>>> a, b                      # a is left unchanged
([0, 1, 2], [0, 1, 2, 3])

有关任何对象的更通用解决方案，请使用copy模块。浅表副本将复制要复制的对象中存储的引用，而深表副本将递归地创建所有对象的新副本。

>>> a = [range(2), range(3)]
>>> b = copy.copy(a)          # shallow copy of a, equivalent to a[:]
>>> b[0] = range(4)
>>> a, b                      # setting an element of b leaves a unchanged
([[0, 1], [0, 1, 2]], [[0, 1, 2, 3], [0, 1, 2]])
>>> b[1].append(3)
>>> a, b                      # modifying an element of b modifies the element in a
([[0, 1], [0, 1, 2, 3]], [[0, 1, 2, 3], [0, 1, 2, 3]])

>>> a = [range(2), range(3)]
>>> b = copy.deepcopy(a)      # deep recursive copy of a
>>> b[1].append(3)
>>> a, b                      # modifying anything in b leaves a unchanged
([[0, 1], [0, 1, 2]], [[0, 1], [0, 1, 2, 3]])