基本上，b = a指向的b是任何a指向的地方，没有别的。

您要问的是可变类型。数字，字符串，元组，frozensets，布尔值None，是不可变的。列表，字典，集合，字节数组是可变的。

如果我将类型设为可变，例如list：

>>> a = [1, 2]  # create an object in memory that points to 1 and 2, and point a at it
>>> b = a       # point b to wherever a points
>>> a[0] = 2    # change the object that a points to by pointing its first item at 2
>>> a
[2, 2]
>>> b
[2, 2]

他们都将仍然指向同一项目。

我也会评论您的原始代码：

>>>a=5     # '5' is interned, so it already exists, point a at it in memory
>>>b=a     # point b to wherever a points
>>>a=6     # '6' already exists in memory, point a at it
>>>print b # b still points at 5 because you never moved it
5

通过执行操作，您始终可以看到内存中指向的位置id(something)。

>>> id(5)
77519368
>>> a = 5
>>> id(a)
77519368     # the same as what id(5) showed us, 5 is interned
>>> b = a
>>> id(b)
77519368     # same again
>>> id(6)
77519356
>>> a = 6
>>> id(a)
77519356     # same as what id(6) showed us, 6 is interned
>>> id(b)
77519368     # still pointing at 5.    
>>> b
5

您可以使用copy，当你想要做一个结构的副本。但是，它
仍然不会复制被 拘禁 的东西
。这包括整数小于256，True，False，None，短字符串喜欢a。基本上，除非您确定不会被实习生搞砸 ， 否则几乎
不 应该 使用它 。

再看一个示例，该示例即使使用可变类型也可以显示，将一个变量指向新变量仍然不会更改旧变量：

>>> a = [1, 2]
>>> b = a
>>> a = a[:1]    # copy the list a points to, starting with item 2, and point a at it
>>> b            # b still points to the original list
[1, 2]
>>> a
[1]
>>> id(b)
79367984
>>> id(a)
80533904

切片列表（无论何时使用:）都会复制一份。