# Another Python prime-listing algorithm

I wrote a small program in Python that finds primes (how original) up to some limit. I tried to produce very clean code, so I'd be vary grateful if you could point out any improvement I could do to make it more pythonic and more readable.

I'm also interested in knowing whether or not it can be sped up (with the same method I used here). I tried to use generator, but since I want to read my list of primes multiple times it didn't work... I'm also wondering if a set or some other structure could be better than a list here.

Thank you very much !

from __future__ import print_function
import time

def measure_time(func):
def wrapper(*args, **kw):
starting_time = time.clock()
res = func(*args, **kw)
ending_time = time.clock()
print('Duration : %fs' % (ending_time - starting_time))
return res
return wrapper

def ask_user_input():
"""Relentlessly ask user to enter an integer >1."""
while True:
try:
lim = int(raw_input('What\'s the limit?   '))
if lim == int(lim) and lim > 2:
break
else:
print('Must be an integer larger than 1. Try again.')
except (NameError, ValueError):
print('Not a number. Try again.')
return lim

def print_primes(primes):
for p in primes:
print(p)

@measure_time
def construct_primes_list(lim):
"""Builds the list of primes smaller than lim."""
prime_divisors = [2]

for num in xrange(2, lim):
for d1 in prime_divisors:
if num % d1 == 0:
break
else:
prime_divisors.append(num)

return prime_divisors

def main():
lim = ask_user_input()
primes = construct_primes_list(lim)
# print_primes(primes)

if __name__ == '__main__':
main()


Here is the benchmarking on my computer:

lim      | time
-----------------------
100      | 0.000143s
1000     | 0.003231s
10000    | 0.071654s
100000   | 2.598098s
1000000  | 169.052574s


EDIT :

As ChatterOne's comment suggested, I tried to compute the square root of num because I know I couldn't find any divisors after that. But it gets way slower. Here is the code :

for num in xrange(3, lim):
sqrt_num = int(sqrt(num)+1)
for d1 in [p for p in prime_divisors if p < sqrt_num]:
if num % d1 == 0:
break
else:
prime_divisors.append(num)


Any thoughts?

• If you want to keep your algorithm (that is, using the modulo), you should at least stop at the square root of lim, without going all the way to lim – ChatterOne Aug 2 '16 at 14:45
• @ChatterOne If lim is 100, I want to be able to get 83, 89, and 97 too... I think you mean the square root of num, in which case I have tried it, and when computing the square root it gets way slower... Maybe I've done this wrong ? I will edit so that you can see that piece of code too – BusyAnt Aug 3 '16 at 7:40

## 2 Answers

Regarding the code:

for num in xrange(3, lim):
sqrt_num = int(sqrt(num)+1)
for d1 in [p for p in prime_divisors if p < sqrt_num]:
if num % d1 == 0:
break
else:
prime_divisors.append(num)


One small improvement would be to either change the squarebrackets in the for statement into parenthesis or just move the test into the loop. Something like:

for num in xrange(3, lim):
for d1 in prime_divisors:
if d1 > sqrt_num:
break
if num % d1 == 0:
break
else:
prime_divisors.append(num)


As written, the code first makes loops over prime_divisors to make a temporary list of the ones < sqrt_num, and then loops over that temporary list to run the modulus test. Changing the square brackets to parenthesis would make it a generator instead of a list so that performance hit would go away. Actually, the whole thing could be rewritten as:

for num in xrange(3, lim):
prime_divisors.extend([] if any(p for p in prime_divisors if p < sqrt_num and num % p == 0) else [num])


..but that may be taking things a bit far.

• This is indeed way faster, and keeps the same algorithm. Thank you for this review ! You had a small mistake in your code, sometimes primes weren't added to the list, I fixed it. As for a comparison, with your loop I can calculate primes below 1 million in just 1.88" ! The one-liner is not as efficient though – BusyAnt Aug 4 '16 at 8:21

You will probably get better performance from the sieve of Eratosthenes. The key difference is that the sieve never performs modulo. Instead it assumes primes, and then crosses them off as you find multiples of primes.

def sieve(n):
"""Return a list of the primes below n."""
prime = [True] * n
result = [2]
append = result.append
sqrt_n = (int(n ** .5) + 1) | 1    # ensure it's odd
for p in range(3, sqrt_n, 2):
if prime[p]:
append(p)
prime[p*p::2*p] = [False] * ((n - p*p - 1) // (2*p) + 1)
for p in range(sqrt_n, n, 2):
if prime[p]:
append(p)
return result

• What improvements brings this to the OPs code ? Here, we try to improve the code already written, then recommend other solutions (if necessary). More, the OP clearly specified: "with the same method I used here" or "... point out any improvement I could do to make it more pythonic and more readable". – Grajdeanu Alex. Aug 2 '16 at 15:04
• This is the same algorithm, just changed slightly to use multiplication rather than division. – Oscar Smith Aug 2 '16 at 15:06
• I would agree with you if I had suggested something like the General number sieve, but op's sieve is basically the same as this one. – Oscar Smith Aug 2 '16 at 15:07
• I'm not familiar with the lines where you assign sqrt_n and where you do prime[p*p::2*p]. Could you please explain these lines a bit more? Thank you for this answer anyway, it doesn't look like my algorithm at first sight, but I'll take it as long as I can learn new stuff. I'd be interested in an actual review of my code if possible. – BusyAnt Aug 3 '16 at 7:25
• Kudos on getting the Sieve right. However, OP uses trial division, which is something different. Eyeballing says it has complexity around O(n log n) instead of Sieve's O(n log log n). Also, OP's algo is trivially adapted to a generator, whereas with a Sieve things are a little bit trickier. – Daerdemandt Aug 3 '16 at 18:44