基于第三个变量值设置2个变量的智能方法，无需使用if[duplicate]

Question

我有一个代码，我在whichweek中传递一个值，然后它返回dayf和dayt的值。我正在使用if语句，但它很长，一点都不“pythonic”。有什么优雅的方式来写这个吗？

我是这样做的：

if whichweek == 5:
    dayf = 0
    dayt = 32
elif whichweek == 1:
    dayf = 0
    dayt = 8
elif whichweek == 2:
    dayf = 7
    dayt = 15
elif whichweek == 3:
    dayf = 14
    dayt = 22
elif whichweek == 4:
    dayf = 21
    dayt = 29

谢谢帮忙。

Answer 1

我会使用字典：

days = {
    5: (0, 32),
    1: (0, 8),
    2: (7, 15),
    3: (14, 22),
    4: (21, 29)
}

dayf, dayt = days[whichweek]

Answer 2

是的，很少有人能让它更像python，我会用一本字典来整理所有的数据，比如：

import random

if __name__ == "__main__":
    random.seed(1)

    week_data = {
        1: (0, 32),
        2: (0, 8),
        3: (7, 15),
        4: (14, 22),
        5: (21, 29)
    }

    for i in range(10):
        whichweek = random.randint(1, 5)
        dayf, dayt = week_data[whichweek]
        print whichweek, dayf, dayt

如果你想处理错误，你可以使用这样的东西：

import random

if __name__ == "__main__":
    random.seed(1)

    week_data = {
        1: (0, 32),
        2: (0, 8),
        3: (7, 15),
        4: (14, 22),
        5: (21, 29)
    }

    for i in range(10):
        whichweek = random.randint(1, 10)
        res = week_data.get(whichweek, None)

        if res is None:
            print("Error: {0} is not in week_data".format(whichweek))
        else:
            dayf, dayt = res
            print whichweek, dayf, dayt

最后，如果要避免条件检查，请始终返回如下值：

import random

if __name__ == "__main__":
    random.seed(1)

    week_data = {
        1: (0, 32),
        2: (0, 8),
        3: (7, 15),
        4: (14, 22),
        5: (21, 29)
    }

    for i in range(10):
        whichweek = random.randint(1, 10)
        dayf, dayt = week_data.get(whichweek, (None, None))
        print whichweek, dayf, dayt

Answer 3

weeks = {1: (0, 8)}
try:
    dayf = weeks[1][0]
    dayt = weeks[1][1]
except (KeyError, IndexError):
    pass