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).

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

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 using break

Here's your code with a few fixes, it prints out only primes:

import math

def 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.

Answer 2

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 3

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]

Answer 4

def primes(n): # simple sieve of multiples 
   odds = range(3, n+1, 2)
   sieve = set(sum([list(range(q*q, n+1, q+q)) for q in odds], []))
   return [2] + [p for p in odds if p not in sieve]

>>> primes(50)
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]

To test if a number is prime:

>>> 541 in primes(541)
True
>>> 543 in primes(543)
False

Answer 5

print [x for x in range(2,100) if not [t for t in range(2,x) if not x%t]]

Answer 6

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 7

SymPy is a Python library for symbolic mathematics. It provides several functions to generate prime numbers.

isprime(n)              # Test if n is a prime number (True) or not (False).

primerange(a, b)        # Generate a list of all prime numbers in the range [a, b).
randprime(a, b)         # Return a random prime number in the range [a, b).
primepi(n)              # Return the number of prime numbers less than or equal to n.

prime(nth)              # Return the nth prime, with the primes indexed as prime(1) = 2. The nth prime is approximately n*log(n) and can never be larger than 2**n.
prevprime(n, ith=1)     # Return the largest prime smaller than n
nextprime(n)            # Return the ith prime greater than n

sieve.primerange(a, b)  # Generate all prime numbers in the range [a, b), implemented as a dynamically growing sieve of Eratosthenes. 

Here are some examples.

>>> import sympy
>>> 
>>> sympy.isprime(5)
True
>>> list(sympy.primerange(0, 100))
[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]
>>> sympy.randprime(0, 100)
83
>>> sympy.randprime(0, 100)
41
>>> sympy.prime(3)
5
>>> sympy.prevprime(50)
47
>>> sympy.nextprime(50)
53
>>> list(sympy.sieve.primerange(0, 100))
[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]

Answer 8

Here is a numpy version of Sieve of Eratosthenes having both okay complexity (lower than sorting an array of length n) and vectorization.

import numpy as np 
def generate_primes(n):
    is_prime = np.ones(n+1,dtype=bool)
    is_prime[0:2] = False
    for i in range(int(n**0.5)+1):
        if is_prime[i]:
            is_prime[i*2::i]=False
    return np.where(is_prime)[0]

Timings:

import time    
for i in range(2,10):
    timer =time.time()
    generate_primes(10**i)
    print('n = 10^',i,' time =', round(time.time()-timer,6))

>> n = 10^ 2  time = 5.6e-05
>> n = 10^ 3  time = 6.4e-05
>> n = 10^ 4  time = 0.000114
>> n = 10^ 5  time = 0.000593
>> n = 10^ 6  time = 0.00467
>> n = 10^ 7  time = 0.177758
>> n = 10^ 8  time = 1.701312
>> n = 10^ 9  time = 19.322478

Answer 9

python 3 (generate prime number)

import math

i = 2
while True:
    for x in range(2, int(math.sqrt(i) + 1)):
        if i%x==0:
            break
    else:
        print(i)
    i += 1

Answer 10

To my opinion it is always best to take the functional approach,

So I create a function first to find out if the number is prime or not then use it in loop or other place as necessary.

def isprime(n):
      for x in range(2,n):
        if n%x == 0:
            return False
    return True

Then run a simple list comprehension or generator expression to get your list of prime,

[x for x in range(1,100) if isprime(x)]

Answer 11

Another simple example, with a simple optimization of only considering odd numbers. Everything done with lazy streams (python generators).

Usage: primes = list(create_prime_iterator(1, 30))

import math
import itertools

def create_prime_iterator(rfrom, rto):
    """Create iterator of prime numbers in range [rfrom, rto]"""
    prefix = [2] if rfrom < 3 and rto > 1 else [] # include 2 if it is in range separately as it is a "weird" case of even prime
    odd_rfrom = 3 if rfrom < 3 else make_odd(rfrom) # make rfrom an odd number so that  we can skip all even nubers when searching for primes, also skip 1 as a non prime odd number.
    odd_numbers = (num for num in xrange(odd_rfrom, rto + 1, 2))
    prime_generator = (num for num in odd_numbers if not has_odd_divisor(num))
    return itertools.chain(prefix, prime_generator)

def has_odd_divisor(num):
    """Test whether number is evenly divisable by odd divisor."""
    maxDivisor = int(math.sqrt(num))
    for divisor in xrange(3, maxDivisor + 1, 2):
        if num % divisor == 0:
            return True
    return False

def make_odd(number):
    """Make number odd by adding one to it if it was even, otherwise return it unchanged"""
    return number | 1

Answer 12

This is my implementation. Im sure there is a more efficient way, but seems to work. Basic flag use.

def genPrime():
    num = 1
    prime = False
    while True:
        # Loop through all numbers up to num
        for i in range(2, num+1):
            # Check if num has remainder after the modulo of any previous numbers
            if num % i == 0:
                prime = False
                # Num is only prime if no remainder and i is num
                if i == num:
                    prime = True
                break

        if prime:
            yield num
            num += 1
        else:
            num += 1

prime = genPrime()
for _ in range(100):
    print(next(prime))

Answer 13

Just studied the topic, look for the examples in the thread and try to make my version:

from collections import defaultdict
# from pprint import pprint

import re


def gen_primes(limit=None):
    """Sieve of Eratosthenes"""
    not_prime = defaultdict(list)
    num = 2
    while limit is None or num <= limit:
        if num in not_prime:
            for prime in not_prime[num]:
                not_prime[prime + num].append(prime)
            del not_prime[num]
        else:  # Prime number
            yield num
            not_prime[num * num] = [num]
        # It's amazing to debug it this way:
        # pprint([num, dict(not_prime)], width=1)
        # input()
        num += 1


def is_prime(num):
    """Check if number is prime based on Sieve of Eratosthenes"""
    return num > 1 and list(gen_primes(limit=num)).pop() == num


def oneliner_is_prime(num):
    """Simple check if number is prime"""
    return num > 1 and not any([num % x == 0 for x in range(2, num)])


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


def simple_is_prime(num):
    """Simple check if number is prime
    More efficient than oneliner_is_prime as it breaks the loop
    """
    for x in range(2, num):
        if num % x == 0:
            return False
    return num > 1


def simple_gen_primes(limit=None):
    """Prime number generator based on simple gen"""
    num = 2
    while limit is None or num <= limit:
        if simple_is_prime(num):
            yield num
        num += 1


if __name__ == "__main__":
    less1000primes = list(gen_primes(limit=1000))
    assert less1000primes == list(simple_gen_primes(limit=1000))
    for num in range(1000):
        assert (
            (num in less1000primes)
            == is_prime(num)
            == oneliner_is_prime(num)
            == regex_is_prime(num)
            == simple_is_prime(num)
        )
    print("Primes less than 1000:")
    print(less1000primes)

    from timeit import timeit

    print("\nTimeit:")
    print(
        "gen_primes:",
        timeit(
            "list(gen_primes(limit=1000))",
            setup="from __main__ import gen_primes",
            number=1000,
        ),
    )
    print(
        "simple_gen_primes:",
        timeit(
            "list(simple_gen_primes(limit=1000))",
            setup="from __main__ import simple_gen_primes",
            number=1000,
        ),
    )
    print(
        "is_prime:",
        timeit(
            "[is_prime(num) for num in range(2, 1000)]",
            setup="from __main__ import is_prime",
            number=100,
        ),
    )
    print(
        "oneliner_is_prime:",
        timeit(
            "[oneliner_is_prime(num) for num in range(2, 1000)]",
            setup="from __main__ import oneliner_is_prime",
            number=100,
        ),
    )
    print(
        "regex_is_prime:",
        timeit(
            "[regex_is_prime(num) for num in range(2, 1000)]",
            setup="from __main__ import regex_is_prime",
            number=100,
        ),
    )
    print(
        "simple_is_prime:",
        timeit(
            "[simple_is_prime(num) for num in range(2, 1000)]",
            setup="from __main__ import simple_is_prime",
            number=100,
        ),
    )

The result of running this code show interesting results:

$ python prime_time.py
Primes less than 1000:
[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, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997]

Timeit:
gen_primes: 0.6738066330144648
simple_gen_primes: 4.738092333020177
is_prime: 31.83770858097705
oneliner_is_prime: 3.3708438930043485
regex_is_prime: 8.692703998007346
simple_is_prime: 0.4686249239894096

So I can see that we have right answers for different questions here; for a prime number generator gen_primes looks like the right answer; but for a prime number check, the simple_is_prime function is better suited.

This works, but I am always open to better ways to make is_prime function.

Answer 14

Here is what I have:

def is_prime(num):
    if num < 2:         return False
    elif num < 4:       return True
    elif not num % 2:   return False
    elif num < 9:       return True
    elif not num % 3:   return False
    else:
        for n in range(5, int(math.sqrt(num) + 1), 6):
            if not num % n:
                return False
            elif not num % (n + 2):
                return False

    return True

It's pretty fast for large numbers, as it only checks against already prime numbers for divisors of a number.

Now if you want to generate a list of primes, you can do:

# primes up to 'max'
def primes_max(max):
    yield 2
    for n in range(3, max, 2):
        if is_prime(n):
            yield n

# the first 'count' primes
def primes_count(count):
    counter = 0
    num = 3

    yield 2

    while counter < count:
        if is_prime(num):
            yield num
            counter += 1
        num += 2

using generators here might be desired for efficiency.

And just for reference, instead of saying:

one = 1
while one == 1:
    # do stuff

you can simply say:

while 1:
    #do stuff

Answer 15

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 16

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 17

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

Answer 18

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 19

You need to make sure that all possible divisors don't evenly divide the number you're checking. In this case you'll print the number you're checking any time just one of the possible divisors doesn't evenly divide the number.

Also you don't want to use a continue statement because a continue will just cause it to check the next possible divisor when you've already found out that the number is not a prime.

Answer 20

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 21

You can create a list of primes using list comprehensions in a fairly elegant manner. Taken from here:

>>> noprimes = [j for i in range(2, 8) for j in range(i*2, 50, i)]
>>> primes = [x for x in range(2, 50) if x not in noprimes]
>>> print primes
>>> [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]

Answer 22

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 23

def genPrimes():
    primes = []   # primes generated so far
    last = 1      # last number tried
    while True:
        last += 1
        for p in primes:
            if last % p == 0:
                break
        else:
            primes.append(last)
            yield last

Answer 24

For me, the below solution looks simple and easy to follow.

import math

def is_prime(num):

    if num < 2:
        return False

    for i in range(2, int(math.sqrt(num) + 1)):
        if num % i == 0:
            return False

return True

Answer 25

def check_prime(x):
    if (x < 2): 
       return 0
    elif (x == 2): 
       return 1
    t = range(x)
    for i in t[2:]:
       if (x % i == 0):
            return 0
    return 1

Answer 26

• The continue statement looks wrong.

• You want to start at 2 because 2 is the first prime number.

• You can write "while True:" to get an infinite loop.