Simple prime number generator in Python
There are some problems:
- Why do you print out count when it didn't divide by x? It doesn't mean it's prime, it means only that this particular x doesn't divide it
continue
moves to the next loop iteration - but you really want to stop it usingbreak
Here's your code with a few fixes, it prints out only primes:
import mathdef main(): count = 3 while True: isprime = True for x in range(2, int(math.sqrt(count) + 1)): if count % x == 0: isprime = False break if isprime: print count count += 1
For much more efficient prime generation, see the Sieve of Eratosthenes, as others have suggested. Here's a nice, optimized implementation with many comments:
# Sieve of Eratosthenes# Code by David Eppstein, UC Irvine, 28 Feb 2002# http://code.activestate.com/recipes/117119/def gen_primes(): """ Generate an infinite sequence of prime numbers. """ # Maps composites to primes witnessing their compositeness. # This is memory efficient, as the sieve is not "run forward" # indefinitely, but only as long as required by the current # number being tested. # D = {} # The running integer that's checked for primeness q = 2 while True: if q not in D: # q is a new prime. # Yield it and mark its first multiple that isn't # already marked in previous iterations # yield q D[q * q] = [q] else: # q is composite. D[q] is the list of primes that # divide it. Since we've reached q, we no longer # need it in the map, but we'll mark the next # multiples of its witnesses to prepare for larger # numbers # for p in D[q]: D.setdefault(p + q, []).append(p) del D[q] q += 1
Note that it returns a generator.
def is_prime(num): """Returns True if the number is prime else False.""" if num == 0 or num == 1: return False for x in range(2, num): if num % x == 0: return False else: return True>> filter(is_prime, range(1, 20)) [2, 3, 5, 7, 11, 13, 17, 19]
We will get all the prime numbers upto 20 in a list.I could have used Sieve of Eratosthenes but you saidyou want something very simple. ;)