有什么办法可以使这小段代码更短？ （蟒蛇）

Question

first  = [0, 0, 0, 0]
second = [0, 0, 0, 0]
third  = [0, 0, 0, 0]
fourth = [0, 0, 0, 0]

while 0 in first and 0 in second and 0 in third and 0 in fourth:

顶部的列表以每个值保持为0开头。随着程序的进行，我计划将列表中的值从0更改为其他数字。实际上，我想知道是否有办法重写'while'语句以检查0是否在任何列表中，而没有'while not in __ and not in __'等长链。 干杯

Answer 1

while all(0 in i for i in [first,second,third,fourth]):
   ...

，如果要检查列表中是否包含0，请执行以下操作：

while any(0 in i for i in [first,second,third,fourth]):
   ...

Answer 2

您可以串联/合并所有列表并检查真实性：

while not all(first+second+third+fourth):

这将检查任何False-y值，如果为0，则返回True。

Answer 3

while 0 in first + second + third + fourth:
    # do stuff

Answer 4

您也可以map

ls = [first, second, third, fourth]
func = lambda x: 0 in x

while all(map(func, ls)):
    #dostuff

---

如果要使用all，请使用0 in first and 0 in second and ...。如果需要any或0 in first等，请使用0 in second

Answer 5

while not any(map(all, [first, second, ...])):