Check if a number is a prime number (Python)

Updated vegaseat 3 Tallied Votes 24K Views Share

This simple isprime(number) function checks if the given integer number is a prime number and returns True or False. The function makes sure that the number is a positive integer, and that 1 is not considered a prime number.

To find out if an integer n is odd one can use n & 1, to check for even one can then use not n & 1 or the more traditional n % 2 == 0 which is about 30% slower.

# prime numbers are only divisible by unity and themselves
# (1 is not considered a prime number by convention)

def isprime(n):
    '''check if integer n is a prime'''
    # make sure n is a positive integer
    n = abs(int(n))
    # 0 and 1 are not primes
    if n < 2:
        return False
    # 2 is the only even prime number
    if n == 2: 
        return True    
    # all other even numbers are not primes
    if not n & 1: 
        return False
    # range starts with 3 and only needs to go up the squareroot of n
    # for all odd numbers
    for x in range(3, int(n**0.5)+1, 2):
        if n % x == 0:
            return False
    return True

# test ...
print isprime(1)       # False
print isprime(2)       # True
print isprime(3)       # True
print isprime(29)      # True
print isprime(345)     # False
print isprime(999979)  # True
print isprime(999981)  # False

# extra test ...
print isprime(49)      # False
print isprime(95)      # False
vegaseat 1,735 DaniWeb's Hypocrite Team Colleague

Filtered out even numbers, except number 2, to speed things up.

theApprentice() 0 Newbie Poster


The code listed above does not seem accurate as numbers like 49 and 95 pass the test and yet they are not prime numbers!

Perhaps a flaw in the algorithm used?

Editor's note: as you can see, 49 and 95 do not pass the test, not sure what's wrong with you?

Toster 0 Newbie Poster

I think the mistake is that you converted the square root to an int, because the code I wrote a couple of years ago works fine, and its almost the same

def isprime(n):
    if n == 2:
        return 1
    if n % 2 == 0:
        return 0

    max = n**0.5+1
    i = 3

    while i <= max:
        if n % i == 0:
            return 0

    return 1

Editors note: the range function in vegaseat's code needs integers, so it's correct.

bishisht commented: could you please explain line 7 m a newbie +0
slate 241 Posting Whiz in Training

Simple RSA primtest. Does not work on universal pseudo primes.

def isprime(n,PROB):
    '''returns if the number is prime. Failure rate: 1/4**PROB '''
    if n==2: return '1'
    if n==1 or n&1==0:return '0'
    while 1&d==0:
    for i in range(PROB):
        if pow(a,d,n)==1:
        if composit:
            for r in xrange(0,s):
                if pow(a,d*2**r,n)==n-1:
        if composit: return False
    return True
coolkirankumar 0 Newbie Poster

daniweb is very useful to beginners to learn IT.i hope i would learn more from u...........bye

kungfujam 0 Newbie Poster

this works fine returning true if prime and false with the smallest number that divides it if it is false.

def isprime(x):
	x = abs(int(x))
	if x < 2:
		return "Less 2", False
	elif x == 2:
		return True
	elif x % 2 == 0:
		return False	
		for n in range(3, int(x**0.5)+2, 2):
			if x % n == 0:
				return n, False
		return True
arindam31 -8 Junior Poster in Training

Pardon my previous post. The indentation was not right.This one works fine.

for n in range(2,10):
  for x in range(2,n):
      if n%x == 0:
        print n,'equal',x,'*',n/x
       print n,'is a prime number'
Gribouillis 1,391 Programming Explorer Team Colleague
zjtpjs4 0 Newbie Poster

Hey thank you for this epic post!!! I used it as a base for checking if a number was prime in an algorithm I made for Project Euler #3.
Thanks again!!!!

gmorcan 0 Newbie Poster

how avout this one?:

def is_prime(n):
    i = 2
    if n < 2:
        return False
    while i < n:        
        if n % i == 0:
            return False
            i += 1
    return True

Editor's note: too slow for large n

Lardmeister 461 Posting Virtuoso

using Python module timeit and a rather small prime number like 999979, your function is 2700 times slower than vegaseat's function.

TrustyTony 888 pyMod Team Colleague Featured Poster

OK, here is a really optimized version for this classic problem. This is not originally mine (but have done very similar).
It utilizes the fact that primes > 4 are 1(mod 6) or 5(mod 6). Actually I editted it a lot now before posting to be more according to format standards of PEP8.

def is_prime(n):
    if n == 2 or n == 3: 
        return True
    elif n < 2 or n % 2 == 0: 
        return False
    elif n < 9:
        return True
    elif n % 3 == 0: 
        return False

    r = int(sqrt(n))
    f = 5

    while f <= r:
        if n % f == 0 or n % (f + 2) == 0: 
            return False
            f += 6

    return True
filtre 0 Newbie Poster

How do you feel witth this?

import re 
def is_prime(num): return not re.match(r"^1?$|^(11+?)\1+$", "1" * num)
def pri(num): return is_prime(num)
TrustyTony 888 pyMod Team Colleague Featured Poster

Real hack, and surely stupendously inefficient. Best as joke or concept. Do not try to test (2**13-1) with it.

HiHe 174 Junior Poster

Super fast for very large numbers:

def miller_rabin_isprime(a, i, n):
    Miller-Rabin primality test
    returns a 1 if n is a prime
    usually i = n - 1
    does not test prime 2
    if i == 0:
        return 1
    x = miller_rabin_isprime(a, i // 2, n)
    if x == 0:
        return 0
    y = (x * x) % n
    if ((y == 1) and (x != 1) and (x != (n - 1))):
        return 0
    if (i % 2) != 0:
        y = (a * y) % n
    return y
Bikash1990 0 Newbie Poster
i = 0
while(i <100):
    j = 2
    while (j<=(i/j)):
        if not (i%j):break
    if (j > i/j):print i ," is prime number"
    i = i+1

Editor's note:
Let's say you want to know if 999979 is a prime number. How long would that take?

woooee 814 Nearly a Posting Maven

You have made the classic mistake of including even numbers = testing twice as many numbers. Start out with i=3 and add two on each pass.

Eyeteeorg 0 Newbie Poster

I just got an automated email from Daniweb suggesting that I read this thread based on my recent activity. The code posts in this thread are fine, but most of the comment posts are curmudgeonly, rude, standoffish, exclusivist, and socially inept. People seem to think it's a violation of some natural law if a beginner programmer joins a discussion and tries to make a positive contribution by posting code. I'll definitely be putting some of you on my ignore list.

Gribouillis 1,391 Programming Explorer Team Colleague

People seem to think it's a violation of some natural law if a beginner programmer joins a discussion and tries to make a positive contribution by posting code.

On the contrary, we answer beginner programmer's questions every day in the python forum and try to help them as much as we can. Correct errors and inefficiencies in code is not rude or socially inept: we all learn by our own mistakes.

I would advise beginner programmers to start their own thread with well described issues to solve. They will probably get better answers than by adding to a code snippet thread.

They may also find many ideas in Vegaseat's thread projects-for-the-beginner.

Eyeteeorg 0 Newbie Poster

Hi, Grib,

Thanks for the link to that thread. I'll definitely be checking it out. What I'm really hungry for is some real-world problems to try to provide solutions for, but I realize it might be a bit too early for me to do that. Doing those kinds of projects could be a good alternative.

If there are similar threads available for html and javascript, I'd love to see links for them as well.


long5001 0 Newbie Poster

I don't understand what's going on with line 15 in vegaseat's code. What does not n & 1 do?

Gribouillis 1,391 Programming Explorer Team Colleague

What does not n & 1 do?

& is the integer bitwise and operator. When n is an integer, n & 1 is the value of its first bit in base 2. Therefore, n & 1 has value 0 if n is even and 1 if n is odd. not n & 1 has the same meaning as n is even (which is less cryptic but is not python). Another way to say it is n % 2 == 0, but vegaseat thinks he gains some speed.

vegaseat 1,735 DaniWeb's Hypocrite Team Colleague
andrew.jhonson.71 0 Newbie Poster
memilanuk 14 Newbie Poster

Found this... thought it might be of help to folks here - the main module is pretty well commented, with lots of links to references:

Gribouillis commented: interesting +14
diego.rivarola 0 Newbie Poster


def is_prime(x):
    prime = False
    if x > 1:
        for i in range(2, x+1):
            if x % i == 0 and i != x:
                prime = False
                prime = True
    return prime


Gu1tar1st 0 Newbie Poster

Simpler than above examples, and creates a list of primes (to 10,000 in this example). Also, you only have to test x up to a half of n, as anything over a half of the number is not worth testing. (not worth testing if 17 can be dividied by 9, 10, 11, 12 etc.!) Also only tests odd numbers with the range (1,n,2):

primes = []
for n in range(1,10000,2):
  for x in range(2,n/2):
      if n%x == 0:
print primes
ddanbe 2,724 Professional Procrastinator Featured Poster

@Gu1tar1st: As the square root of 17 is 4.123 you don't even have to test for divisibility from 5,6,7,8,9....

justin.thong.39 0 Newbie Poster

what does "not n & 1" do specifically?
Does it return False for all the even numbers above 2?

Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.