how to check among big prime numbers for condition

Question

Out of prime numbers less than 201920190 how to find those that dont have digit '7' in their str representation? I need to count the number(count) of such prime numbers that have no '7' in them.

Std sieve implementation is too slow, it just stuck computing..

def SieveOfEratosthenes(num):
    prime = [True for i in range(num+1)]
# boolean array
    p = 2
    while (p * p <= num):
 
        # If prime[p] is not
        # changed, then it is a prime
        if (prime[p] == True):
 
            # Updating all multiples of p
            for i in range(p * p, num+1, p):
                prime[i] = False
        p += 1
 
    # Print all prime numbers
    for p in range(2, num+1):
        if prime[p]:
                  #here added additional check to only print those that have no digit 7
                  if '7' not in str(p):
                     print(p)
 
 
# Driver code
if __name__ == '__main__':
    num = 201920190
    print("Following are the prime numbers smaller"),
    print("than or equal to", num)
    SieveOfEratosthenes(num)

How to do it better?

Answer 1

Perhaps cheating, but if your priority is simply to get a number you could just download a list of all the primes up to 201920190, e.g. from https://primes.utm.edu/lists/small/millions/ (there are about 12 million of them), and then count through that list.

It's not nearly as satisfying as getting your own sieve code to work, but it's pretty efficient!