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 "", 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 "", 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)
你学到了么?
延伸阅读
-
PEP 3132
-
Tuple unpacking in return and yield statements