查找质数(缺少3!)

Question

我想出了这个解决方案来计算用户定义的骰子侧面质数的概率。但出于某种原因，它不喜欢只包含数字3。有人能告诉我为什么它讨厌3号吗？

输出如下：[2, 5, 7, 11, 13, 17, 19, 23] --> Missing 3！

def cal_probability (event, sample_space):
    return len(event)/ len(sample_space)

def cal_prime_numbers (s_space):
    count = 0
    prime_nums = []
    for num in s_space:
        for i in s_space:
            if num % i == 0:
                count += 1
            else:
                continue
        if count == 2:
            prime_nums.append (num)
        else:
            count = 0
            continue
    return prime_nums

if __name__ == '__main__':
    sides = input ('Enter the number of sides: ')
    sample_space = list (range (1, int(sides)+1))
    print (sample_space)
    event = cal_prime_numbers (sample_space)
    print (event)
    p = cal_probability (event, sample_space)
    print ('The probabilty of prime numbers to happen in {0} sides die is: {1}%'.format (sides, p*100))

然而，我期望：[2, 3, 5, 7, 11, 13, 17, 19, 23]

Answer 1

当你找到一个素数时，你不能在这里重置计数：

if count == 2:
    prime_nums.append (num)
else:
    count = 0
    continue

这意味着你不能在一行中找到两个素数。这个杠杆会影响3，因为2是唯一的质数。无论你是否找到一个素数来修复它，只要将count设置回0即可。您可以删除else，并且不需要continue

if count == 2:
    prime_nums.append (num)
count = 0

您也不需要在内部for循环中使用else或continue，因为它与正常的循环行为没有任何不同：

def cal_prime_numbers (s_space):
    count = 0
    prime_nums = []
    for num in s_space:        
        for i in s_space:    
            if num % i == 0:
                count += 1
        if count == 2:
            prime_nums.append (num)
        count = 0
    return prime_nums