Hi Gregor,
I had done the same thing. I also noted that assigning (or inserting) an element into a list is faster than creating a new list: l.insert(0,2) is faster than l = [2]+l.
###
def sieve (maximum):
if maximum < 2: return []
limit = int(maximum**0.5)
nums = range(1,maximum+1,2)
nums[0] = None
for p in nums:
if p:
if p > limit: break
nums[(p*p)//2::p] = [False]*(1+(maximum//p- p)//2)
nums[0] = 2
return filter(None, nums)
###
/cHi Sean!
Thanks for your measurements. In the meantime I did another amendment, leaving out the even numbers from the sieve. It goes like this:
def sieve(maximum): nums = range(3, maximum+1, 2) for p in nums: if p: if p*p > maximum: break start = (p*p-2)//2 nums[start::p] = [False]*(1+((maximum-3)//2-start)//p) return [2] + filter(None, nums)
Perhaps not very elegant. But approximately twice as fast as the former version.
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