Simple Prime Generator in Python

Question

Could someone please tell me what I'm doing wrong with this code? It is just printing 'count' anyway. I just want a very simple prime generator (nothing fancy).

import math

def main():
    count = 3
    one = 1
    while one == 1:
        for x in range(2, int(math.sqrt(count) + 1)):
            if count % x == 0: 
                continue
            if count % x != 0:
                print count

        count += 1

Answer 1

import time

maxnum=input("You want the prime number of 1 through....")

n=2
prime=[]
start=time.time()

while n<=maxnum:

    d=2.0
    pr=True
    cntr=0

    while d<n**.5:

        if n%d==0:
            pr=False
        else:
            break
        d=d+1

    if cntr==0:

        prime.append(n)
        #print n

    n=n+1

print "Total time:",time.time()-start

Answer 2

Here's a simple (Python 2.6.2) solution... which is in-line with the OP's original request (now six-months old); and should be a perfectly acceptable solution in any "programming 101" course... Hence this post.

import math

def isPrime(n):
    for i in range(2, int(math.sqrt(n)+1)):
        if n % i == 0: 
            return False;
    return n>1;

print 2
for n in range(3, 50):
    if isPrime(n):
        print n

This simple "brute force" method is "fast enough" for numbers upto about about 16,000 on modern PC's (took about 8 seconds on my 2GHz box).

Obviously, this could be done much more efficiently, by not recalculating the primeness of every even number, or every multiple of 3, 5, 7, etc for every single number... See the Sieve of Eratosthenes (see eliben's implementation above), or even the Sieve of Atkin if you're feeling particularly brave and/or crazy.

Caveat Emptor: I'm a python noob. Please don't take anything I say as gospel.

Answer 3

How about this if you want to compute the prime directly:

def oprime(n):
counter = 0
b = 1
if n == 1:
    print 2
while counter < n-1:
    b = b + 2
    for a in range(2,b):
        if b % a == 0:
            break
    else:
        counter = counter + 1
        if counter == n-1:
            print b

Answer 4

re is powerful:

import re


def isprime(n):
    return re.compile(r'^1?$|^(11+)\1+$').match('1' * n) is None

print [x for x in range(100) if isprime(x)]

###########Output#############
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]