查找素数（缺少3！）

Question

I came up with this solution to calculate probability of prime numbers in user defined die's sides. But for some reason it doesn't like to include the number 3 only. Can anybody enlighten me why it hates number 3?

Output is look like: [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))

I would however expect: [2, 3, 5, 7, 11, 13, 17, 19, 23]

Answer 1

You don't reset count here when you find a prime:

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

Which means that you can't find two primes in a row. That lever affects 3 since 2 is the only even prime. Just set count back to 0 whether or not you found a prime to fix it. You can remove the else and don't need continue :

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

You also don't need the else or continue inside your inner for loop since it doesn't do anything different than normal looping behaviour:

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