如何在python中为超过特定值的数字引发Valueerror？

Question

我试过创建一个函数，它接受一个非负整数n的阶乘，这部分工作得很好，但是如果输入值低于0或高于12，我也必须创建一个ValueError，这是行不通的。你知道吗

def factorial(n):
    countdown = n
    factorial_sum = 1

    while True:
        try:
            if n == 0:
                return 1
            if n < 0 or n > 12:
                raise ValueError
        except ValueError:
            return 'Error'

        if (countdown / n) > 0 and n <= 12:
            factorial_sum = factorial_sum * countdown
            countdown = countdown - 1

        if (n - countdown + 1) == n:
            return factorial_sum

        elif (n - 1) == 0:
            return factorial_sum

挑战来自codewar和state：

In mathematics, the factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n. For example: 5! = 5 * 4 * 3 * 2 * 1 = 120. By convention the value of 0! is 1.

Write a function to calculate factorial for a given input. If input is below 0 or above 12 throw an exception of type ArgumentOutOfRangeException (C#) or IllegalArgumentException (Java) or RangeException (PHP) or throw a RangeError (JavaScript) or ValueError (Python) or return -1 (C).

所有答案将不胜感激

Answer 1

替换

try:
    if n == 0:
        return 1
    if n < 0 or n > 12:
        raise ValueError
except ValueError:
    return 'Error'

由

if n == 0:
    return 1
if n < 0 or n > 12:
    raise ValueError