Simple prime number 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).

Answer 1

This seems homework-y, so I'll give a hint rather than a detailed explanation. Correct me if I've assumed wrong.

You're doing fine as far as bailing out when you see an even divisor.

But you're printing 'count' as soon as you see even one number that doesn't divide into it. 2, for instance, does not divide evenly into 9. But that doesn't make 9 a prime. You might want to keep going until you're sure no number in the range matches.

(as others have replied, a Sieve is a much more efficient way to go... just trying to help you understand why this specific code isn't doing what you want)

Answer 2

Similar to user107745, but using 'all' instead of double negation (a little bit more readable, but I think same performance):

import math
[x for x in xrange(2,10000) if all(x%t for t in xrange(2,int(math.sqrt(x))+1))]

Basically it iterates over the x in range of (2, 100) and picking only those that do not have mod == 0 for all t in range(2,x)

Another way is probably just populating the prime numbers as we go:

primes = set()
def isPrime(x):
  if x in primes:
    return x
  for i in primes:
    if not x % i:
      return None
  else:
    primes.add(x)
    return x

filter(isPrime, range(2,10000))

Answer 3

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 said you want something very simple. ;)

Answer 4

Using generator:

def primes(num):
    if 2 <= num:
        yield 2
    for i in range(3, num + 1, 2):
        if all(i % x != 0 for x in range(3, int(math.sqrt(i) + 1))):
            yield i

Usage:

for i in primes(10):
    print(i)

2, 3, 5, 7

Answer 5

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 6

If you wanted to find all the primes in a range you could do this:

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
num = 0
itr = 0
tot = ''
while itr <= 100:
    itr = itr + 1
    num = num + 1
    if is_prime(num) == True:
        print(num)
        tot = tot + ' ' + str(num)
print(tot)

Just add while its <= and your number for the range.
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 97 101