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#!/usr/bin/env python2
''' Eratosthenes sieve
from __future__ import print_function
import array
import time
start = time.time()
limit = 2*10**6 # Four million
sieve = array.array('l',range(limit))
primes = [2] # Start with just first, then append to it
n = 0 # Index into current prime being eliminated
while n*n < limit:
p = primes[n]
# Eliminate all multiples of primes[n]:
for i in range(p+p,limit, p):
sieve[i] = 0
# Find remaining numbers up to new upperbound (current prime squared)
for i in range(primes[-1]+1,min(limit, p*p)): # min() to avoid out of bounds access.
if sieve[i]:
elapsed = start - time.time()
print(', '.join(['%s' % x for x in primes[:20]]), "...",
', '.join(['%s' % x for x in primes[-10:]]))
print("%d primes found in %g seconds" % (len(primes), elapsed))
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JimDennis commented Jan 5, 2018

For this particular case the use of array.array over just using a plain list seems to give less than a 2% performance benefit.

A quick and dirty test with the limit set at 20 million gives me 10 seconds for the array.array version and 13 seconds for the list version. So that's about a 25% advantage at that scale.

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JimDennis commented Jan 5, 2018 for those who want to know more about this algorithm.

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