参见英文答案 > a mutable type inside an immutable container                                    3个
我正在尝试修改元组中的列表,append方法可以正常工作,而=运算符工作但引发了异常,说无法修改元组.我知道一个元组是不可变的,但我不是想改变它.为什么会这样？

In [36]: t=([1,2],)

In [37]: t[0].append(123)

In [38]: t
Out[38]: ([1, 2, 123],)

In [39]: t[0]+=[4,5,]
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-39-b5b3001fbe03> in <module>()
----> 1 t[0]+=[4,5,]

TypeError: 'tuple' object does not support item assignment

In [40]: t
Out[40]: ([1, 2, 123, 4, 5],)

解决方法:

=是就地加法运算符.它做了两件事：

>它调用obj .__ iadd __(rhs)来给对象提供就地变异对象的机会.
>它重新引用obj .__ iadd __(rhs)调用返回的引用.

通过在存储在元组中的列表上使用=,第一步成功; t [0]列表就地改变,但第二步,将t [0]重新绑定到t [0] .__ iadd__的返回值失败,因为元组是不可变的.

需要后一步来支持可变和不可变对象上的相同运算符：

>>> reference = somestr = 'Hello'
>>> somestr += ' world!'
>>> somestr
'Hello world!'
>>> reference
'Hello'
>>> reference is somestr
False

这里添加了一个不可变的字符串,somestr被反弹到一个新对象,因为字符串是不可变的.

>>> reference = somelst = ['foo']
>>> somelst += ['bar']
>>> somelst
['foo', 'bar']
>>> reference
['foo', 'bar']
>>> reference is somestr
True

这里列表已就地更改,somestr被反弹到同一个对象,因为list .__ iadd __()可以就地改变列表对象.

从augmented arithmetic special method hooks documentation：

These methods are called to implement the augmented arithmetic assignments (+=, -=, *=, /=, //=, %=, **=, <<=, >>=, &=, ^=, |=). These methods should attempt to do the operation in-place (modifying self) and return the result (which could be, but does not have to be, self).

这里的解决方法是调用t [0] .extend()代替：

>>> t = ([1,2],)
>>> t[0].extend([3, 4, 5])
>>> t[0]
[1, 2, 3, 4, 5]