I did totally forget about negative numbers, probably because no one cares about negative prime numbers and I wasn’t worried about completeness (hence my ADD-fueled additional posts in what was the middle of the night for me). If I were writing a library, I’d have tests to prevent this sort of thing.

As for the question of whether you need to search higher than the sqrt of the number, every factor greater than the sqrt of the number will have a match that is less than the sqrt of the number.

Also, your revised program still fails for 2.

If I were to write a ‘proper’ version, it would be:
def prime_test(n):
.. if n < 2 or n % 2 == 0:
…. return False
.. elif n == 2:
…. return True
.. else:
…. for i in range(3, int(math.sqrt(n))+1, 2):
…… if n % i == 0:
…….. return False
…. return True

If you want speed, however, find a library where someone else has worried about such things, such as sympy.

%timeit assert sympy.isprime(2**61-1)==True # 9th Mersenne prime
1000 loops, best of 3: 743 µs per loop

%timeit assert sympy.isprime(2**89-1)==True # 10th.
1000 loops, best of 3: 1.22 ms per loop

%timeit assert sympy.isprime(2**(2**6)+1)==False # 6th Fermat number
1000 loops, best of 3: 324 µs per loop

And for reference to my previous post, the redefined prime_test vs. sympy:
%timeit -n1000 -r10 assert prime_test(982451653)==True # Mine
1000 loops, best of 10: 2.51 ms per loop
%timeit -n1000 -r10 assert sympy.isprime(982451653)==True # sympy
1000 loops, best of 10: 133 µs per loop

http://www.tiawichiresearch.com/?p=44

The main speed advantage in your program over mine is provided by the limit (sqrt(number)) which is something I was looking into but I wanted to find a proof for it before using it.

Your program also doesn’t handle numbers below 1. I didn’t realise range had an optional third parameter for steps, this will be very useful for me, thanks!

One thing I don’t understand is why you’d need to use else so much. I.E.
….if num is 2:
……..return True
….else:

as you will return if num is 2, there is no need to use an else statement and it just takes up room. And I’d change this

…………if num % n:
…………….continue
…………else:
…………….return False

to

…………if not num % n:
…………….return False

and I’d change my original code to:

def isprime(number):
	if number< =1 or number%2==0:
		return 0
	for check in range(3,int(math.sqrt(number))+1,2):
		if number%check==0:
			return 0
	return 1

I’m sorry to sound so critical, but, in this example in particular, your code is, frankly, terrible python. And it’s slow.

Try this:

import math

def prime_test(num):
….if num is 2:
……..return True
….else:
……..for n in range(3, int(math.sqrt(num)), 2):
…………if num % n:
…………….continue
…………else:
…………….return False
……..return True

In my ipython shell:

%timeit -n1000 -r10 isprime(982451653) #improved version
1000 loops, best of 10: 6.25 ms per loop

%timeit -n1000 -r10 prime_test(982451653)
1000 loops, best of 10: 2.69 ms per loop

Python has many high level constructs. They are what make the language so powerful.