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Python3.8对「可迭代解包」的改进

 4 years ago
source link: https://www.tuicool.com/articles/iieuiqU
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标题想了很久，最终还是直接表述了。O(∩_∩)O

Python 3的可迭代解包

在 PEP 3132 - Extended Iterable Unpacking 里面描述了一种对可迭代对象的解包用法，Python 3可用： 

In : a, *b, c = range(5)

In : a, c
Out: (0, 4)

In : b
Out: [1, 2, 3]

In : *a, = range(5)

In : a
Out: [0, 1, 2, 3, 4]

In : for a, *b in [(1, 2, 3), (4, 5, 6, 7)]:
...:     print(b)
...:
[2, 3]
[5, 6, 7]

挺方便的，ES6也有对应的 destructuring assignment 语法实现解包数组，不过人家实现的更彻底，还可以解包对象。

不过在Python 3.2时引入了一个BUG。看一个例子：

In : def a():
...:     rest = (4, 5, 6)
...:     t = 1, 2, 3, *rest
...:     return t
...:

In : for i in a():
...:     print(i)
...:
1
2
3
4
5
6

其实这是一个非常简化的写法，要不然需要把1，2，3放在一个元组中，再+rest:

In : (1, 2, 3) + (3, 4, 5, 6)
Out: (1, 2, 3, 4, 5, 6)

但是上面这个例子稍微改一下：

In : def b():
...:     rest = (4, 5, 6)
...:     return 1, 2, 3, *rest
File "<ipython-input-38-b5a7115853e2>", line 3
  return 1, 2, 3, *rest
                    ^
SyntaxError: invalid syntax

不用变量t, 直接返回就会抛SyntaxError，另外就是yield(当然要注意return和yield本身的意见不同啊):

In : def c():
...:     rest = (4, 5, 6)
...:     yield 1, 2, 3, *rest
File "<ipython-input-39-cf6991fcff64>", line 3
  yield 1, 2, 3, *rest
                   ^
SyntaxError: invalid syntax

Python 3.8

在Python 3.8，修复了这个问题：

>>> def b():
...     rest = (4, 5, 6)
...     return 1, 2, 3, *rest
...
>>> for i in b():
...     print(i)
...
1
2
3
4
5
6
>>> def c():
...     rest = (4, 5, 6)
...     yield 1, 2, 3, *rest
...
>>> for i in c():
...     print(i)
...
(1, 2, 3, 4, 5, 6)

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