如何缩短或清理此 python 代码？

Question

我有一个家庭作业，要编写一个程序，该程序将高速公路编号作为输入，并输出高速公路是主要还是辅助，走向东/西，北/南，如果是辅助，它服务的主要高速公路是什么。 这是我的代码，它给了我满分，但我是初学者，必须有更短的方法来编写它。 有人介意清理一下吗？

highway_number = int(input())

if highway_number >= 1 and highway_number <= 99:
    prim = 'is primary,'
    if (highway_number % 2) == 0:
        print('The', highway_number, prim, 'going east/west.')
    else:
        print('The', highway_number, prim, 'going north/south.')
elif highway_number >= 100 and highway_number <= 999:
    aux = 'is auxiliary,'
    if (highway_number % 2) == 0:
        print('The', highway_number, aux, 'serving the %d, going east/west.' % (highway_number%100))
    else:
        print('The', highway_number, aux, 'serving the %d, going north/south.' % (highway_number%100))
else:
    print(highway_number, 'is not a valid interstate highway number.')

这里有 290 个输出：

The 290 is auxiliary, serving the 90, going east/west.

Answer 1

用于

1) 链式比较

2) f弦

3) 利用 0/1 的布尔值的内联 if 语句

使代码更短

highway_number = int(input())

if 1 <= highway_number <= 99:
    direction = 'east/west' if highway_number % 2 else 'north/south'
    print(f'The {highway_number} is primary, going {direction}')
elif 100 <= highway_number <= 999:
    direction = 'north/south' if highway_number % 2 else 'east/west'
    print(f'The {highway_number} is auxiliary, serving the {highway_number%100}, going {direction}')
else:
    print(highway_number, 'is not a valid interstate highway number.')