要回答你的问题，这里有一种方法，它将确定一个数字是否是素数比你写的更有效，尽管对于大数，它仍然不是最好的。此外，该解决方案解决了用户输入0、1或2的情况。

def is_prime(n):
    """
    Integer -> Boolean
    returns True if n is a prime number
    """
    if n == 2 or n == 3:
        return True
    if n < 2 or n % 2 == 0:
        return False
    if n < 9:
        return True
    if n % 3 == 0:
        return False

    r = int(sqrt(n))
    f = 5
    while f <= r:
        if n % f == 0:
            return False
        if n % (f + 2) == 0:
            return False
        f += 6
    return True

我发现Itprorh66的答案很难用r和f变量来阅读。Jaosnharper关于只循环到平方根的评论是一个好主意。下面是一些代码，可以让这个过程更快：

import math

def is_prime(number: int) -> bool:
    """
    Summary:
        Check if an integer is a prime number.

    Returns:
        bool: If the input (number) is a prime number or not.
    """

    # One, zero, and negative numbers do not follow the definition of a prime
    # number.
    if number < 2:
        return False

    # Two is an exception with even numbers when considering whether they are
    # prime or not, so it is handled here.
    if number == 2:
        return True

    # Even numbers are always composite (except for two).
    if number % 2 == 0:
        return False

    # Plus one is used to include the final number while in the for-loop.
    square_root_number = math.isqrt(number) + 1

    # There is an increase of two instead of one to iterate over odd divisors
    # only.
    # The for-loop starts at three because numbers lower than three
    # have already been handled.
    for divisor in range(3, square_root_number, 2):
        if number % divisor == 0:
            return False

    return True