为什么 Python 元组中列表的 += 会引发 TypeError 但无论如何都要修改列表? [英] Why does += of a list within a Python tuple raise TypeError but modify the list anyway?

问题描述

我刚刚遇到了一些很奇怪的事情.

<预><代码>>>>t = ([],)>>>t[0].append('你好')>>>吨(['你好']，)>>>t[0] += ['世界']回溯(最近一次调用最后一次):文件<stdin>"，第 1 行，在 <module> 中类型错误:元组"对象不支持项目分配>>>吨(['你好，世界']，)

为什么它会引发 TypeError 而改变 tuple 中的 list ?

解决方案

正如我在评论中开始提到的，+= 实际上修改了列表 就地 然后尝试将结果分配给元组中的第一个位置.来自数据模型文档:

<块引用>

调用这些方法来实现增广算术赋值(+=, -=, =,/=,//=, %=, *=, <<=, >>=, &=, ^=, |=).这些方法应该尝试就地执行操作(修改 self)并返回结果(可能是但不一定是 self).

+= 因此等价于:

t[0].extend(['world']);t[0] = t[0];

所以就地修改列表不是问题(1. step)，因为列表是可变的，但是将结果分配回元组是无效的(2. step)，这就是抛出错误的地方.

I just came across something that was quite strange.

>>> t = ([],)
>>> t[0].append('hello')
>>> t
(['hello'],)
>>> t[0] += ['world']
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'tuple' object does not support item assignment
>>> t
(['hello', 'world'],)

Why does it raise TypeError and yet change the list inside the tuple?

解决方案

As I started mentioning in comment, += actually modifies the list in-place and then tries to assign the result to the first position in the tuple. From the data model 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).

+= is therefore equivalent to:

t[0].extend(['world']);
t[0] = t[0];

So modifying the list in-place is not problem (1. step), since lists are mutable, but assigning the result back to the tuple is not valid (2. step), and that's where the error is thrown.

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