Fastest way to list all primes below N
This is the best algorithm I could come up.
def get_primes(n):
numbers = set(range(n, 1, -1))
primes = []
while numbers:
p = numbers.pop()
primes.append(p)
numbers.difference_update(set(range(p*2, n+1, p)))
return primes
>>> timeit.Timer(stmt='get_primes.get_primes(1000000)', setup='import get_primes').timeit(1)
1.1499958793645562
Can it be made even faster?
This code has a flaw: Since numbers is an unordered set, there is no guarantee that numbers.pop() will remove the lowest number from the set. Nevertheless, it works (at least for me) for some input numbers:
>>> sum(get_primes(2000000))
142913828922L
#That's the correct sum of all numbers below 2 million
>>> 529 in get_primes(1000)
False
>>> 529 in get_primes(530)
True
Solution 1:
Warning: timeit results may vary due to differences in hardware or
version of Python.
Below is a script which compares a number of implementations:
- ambi_sieve_plain,
- rwh_primes,
- rwh_primes1,
- rwh_primes2,
- sieveOfAtkin,
- sieveOfEratosthenes,
- sundaram3,
- sieve_wheel_30,
- ambi_sieve (requires numpy)
- primesfrom3to (requires numpy)
- primesfrom2to (requires numpy)
Many thanks to stephan for bringing sieve_wheel_30 to my attention. Credit goes to Robert William Hanks for primesfrom2to, primesfrom3to, rwh_primes, rwh_primes1, and rwh_primes2.
Of the plain Python methods tested, with psyco, for n=1000000, rwh_primes1 was the fastest tested.
+---------------------+-------+
| Method | ms |
+---------------------+-------+
| rwh_primes1 | 43.0 |
| sieveOfAtkin | 46.4 |
| rwh_primes | 57.4 |
| sieve_wheel_30 | 63.0 |
| rwh_primes2 | 67.8 |
| sieveOfEratosthenes | 147.0 |
| ambi_sieve_plain | 152.0 |
| sundaram3 | 194.0 |
+---------------------+-------+
Of the plain Python methods tested, without psyco, for n=1000000, rwh_primes2 was the fastest.
+---------------------+-------+
| Method | ms |
+---------------------+-------+
| rwh_primes2 | 68.1 |
| rwh_primes1 | 93.7 |
| rwh_primes | 94.6 |
| sieve_wheel_30 | 97.4 |
| sieveOfEratosthenes | 178.0 |
| ambi_sieve_plain | 286.0 |
| sieveOfAtkin | 314.0 |
| sundaram3 | 416.0 |
+---------------------+-------+
Of all the methods tested, allowing numpy, for n=1000000, primesfrom2to was the fastest tested.
+---------------------+-------+
| Method | ms |
+---------------------+-------+
| primesfrom2to | 15.9 |
| primesfrom3to | 18.4 |
| ambi_sieve | 29.3 |
+---------------------+-------+
Timings were measured using the command:
python -mtimeit -s"import primes" "primes.{method}(1000000)"
with {method} replaced by each of the method names.
primes.py:
#!/usr/bin/env python
import psyco; psyco.full()
from math import sqrt, ceil
import numpy as np
def rwh_primes(n):
# https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
""" Returns a list of primes < n """
sieve = [True] * n
for i in xrange(3,int(n**0.5)+1,2):
if sieve[i]:
sieve[i*i::2*i]=[False]*((n-i*i-1)/(2*i)+1)
return [2] + [i for i in xrange(3,n,2) if sieve[i]]
def rwh_primes1(n):
# https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
""" Returns a list of primes < n """
sieve = [True] * (n/2)
for i in xrange(3,int(n**0.5)+1,2):
if sieve[i/2]:
sieve[i*i/2::i] = [False] * ((n-i*i-1)/(2*i)+1)
return [2] + [2*i+1 for i in xrange(1,n/2) if sieve[i]]
def rwh_primes2(n):
# https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
""" Input n>=6, Returns a list of primes, 2 <= p < n """
correction = (n%6>1)
n = {0:n,1:n-1,2:n+4,3:n+3,4:n+2,5:n+1}[n%6]
sieve = [True] * (n/3)
sieve[0] = False
for i in xrange(int(n**0.5)/3+1):
if sieve[i]:
k=3*i+1|1
sieve[ ((k*k)/3) ::2*k]=[False]*((n/6-(k*k)/6-1)/k+1)
sieve[(k*k+4*k-2*k*(i&1))/3::2*k]=[False]*((n/6-(k*k+4*k-2*k*(i&1))/6-1)/k+1)
return [2,3] + [3*i+1|1 for i in xrange(1,n/3-correction) if sieve[i]]
def sieve_wheel_30(N):
# http://zerovolt.com/?p=88
''' Returns a list of primes <= N using wheel criterion 2*3*5 = 30
Copyright 2009 by zerovolt.com
This code is free for non-commercial purposes, in which case you can just leave this comment as a credit for my work.
If you need this code for commercial purposes, please contact me by sending an email to: info [at] zerovolt [dot] com.'''
__smallp = ( 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59,
61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139,
149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227,
229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311,
313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401,
409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491,
499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599,
601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683,
691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797,
809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887,
907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997)
wheel = (2, 3, 5)
const = 30
if N < 2:
return []
if N <= const:
pos = 0
while __smallp[pos] <= N:
pos += 1
return list(__smallp[:pos])
# make the offsets list
offsets = (7, 11, 13, 17, 19, 23, 29, 1)
# prepare the list
p = [2, 3, 5]
dim = 2 + N // const
tk1 = [True] * dim
tk7 = [True] * dim
tk11 = [True] * dim
tk13 = [True] * dim
tk17 = [True] * dim
tk19 = [True] * dim
tk23 = [True] * dim
tk29 = [True] * dim
tk1[0] = False
# help dictionary d
# d[a , b] = c ==> if I want to find the smallest useful multiple of (30*pos)+a
# on tkc, then I need the index given by the product of [(30*pos)+a][(30*pos)+b]
# in general. If b < a, I need [(30*pos)+a][(30*(pos+1))+b]
d = {}
for x in offsets:
for y in offsets:
res = (x*y) % const
if res in offsets:
d[(x, res)] = y
# another help dictionary: gives tkx calling tmptk[x]
tmptk = {1:tk1, 7:tk7, 11:tk11, 13:tk13, 17:tk17, 19:tk19, 23:tk23, 29:tk29}
pos, prime, lastadded, stop = 0, 0, 0, int(ceil(sqrt(N)))
# inner functions definition
def del_mult(tk, start, step):
for k in xrange(start, len(tk), step):
tk[k] = False
# end of inner functions definition
cpos = const * pos
while prime < stop:
# 30k + 7
if tk7[pos]:
prime = cpos + 7
p.append(prime)
lastadded = 7
for off in offsets:
tmp = d[(7, off)]
start = (pos + prime) if off == 7 else (prime * (const * (pos + 1 if tmp < 7 else 0) + tmp) )//const
del_mult(tmptk[off], start, prime)
# 30k + 11
if tk11[pos]:
prime = cpos + 11
p.append(prime)
lastadded = 11
for off in offsets:
tmp = d[(11, off)]
start = (pos + prime) if off == 11 else (prime * (const * (pos + 1 if tmp < 11 else 0) + tmp) )//const
del_mult(tmptk[off], start, prime)
# 30k + 13
if tk13[pos]:
prime = cpos + 13
p.append(prime)
lastadded = 13
for off in offsets:
tmp = d[(13, off)]
start = (pos + prime) if off == 13 else (prime * (const * (pos + 1 if tmp < 13 else 0) + tmp) )//const
del_mult(tmptk[off], start, prime)
# 30k + 17
if tk17[pos]:
prime = cpos + 17
p.append(prime)
lastadded = 17
for off in offsets:
tmp = d[(17, off)]
start = (pos + prime) if off == 17 else (prime * (const * (pos + 1 if tmp < 17 else 0) + tmp) )//const
del_mult(tmptk[off], start, prime)
# 30k + 19
if tk19[pos]:
prime = cpos + 19
p.append(prime)
lastadded = 19
for off in offsets:
tmp = d[(19, off)]
start = (pos + prime) if off == 19 else (prime * (const * (pos + 1 if tmp < 19 else 0) + tmp) )//const
del_mult(tmptk[off], start, prime)
# 30k + 23
if tk23[pos]:
prime = cpos + 23
p.append(prime)
lastadded = 23
for off in offsets:
tmp = d[(23, off)]
start = (pos + prime) if off == 23 else (prime * (const * (pos + 1 if tmp < 23 else 0) + tmp) )//const
del_mult(tmptk[off], start, prime)
# 30k + 29
if tk29[pos]:
prime = cpos + 29
p.append(prime)
lastadded = 29
for off in offsets:
tmp = d[(29, off)]
start = (pos + prime) if off == 29 else (prime * (const * (pos + 1 if tmp < 29 else 0) + tmp) )//const
del_mult(tmptk[off], start, prime)
# now we go back to top tk1, so we need to increase pos by 1
pos += 1
cpos = const * pos
# 30k + 1
if tk1[pos]:
prime = cpos + 1
p.append(prime)
lastadded = 1
for off in offsets:
tmp = d[(1, off)]
start = (pos + prime) if off == 1 else (prime * (const * pos + tmp) )//const
del_mult(tmptk[off], start, prime)
# time to add remaining primes
# if lastadded == 1, remove last element and start adding them from tk1
# this way we don't need an "if" within the last while
if lastadded == 1:
p.pop()
# now complete for every other possible prime
while pos < len(tk1):
cpos = const * pos
if tk1[pos]: p.append(cpos + 1)
if tk7[pos]: p.append(cpos + 7)
if tk11[pos]: p.append(cpos + 11)
if tk13[pos]: p.append(cpos + 13)
if tk17[pos]: p.append(cpos + 17)
if tk19[pos]: p.append(cpos + 19)
if tk23[pos]: p.append(cpos + 23)
if tk29[pos]: p.append(cpos + 29)
pos += 1
# remove exceeding if present
pos = len(p) - 1
while p[pos] > N:
pos -= 1
if pos < len(p) - 1:
del p[pos+1:]
# return p list
return p
def sieveOfEratosthenes(n):
"""sieveOfEratosthenes(n): return the list of the primes < n."""
# Code from: <[email protected]>, Nov 30 2006
# http://groups.google.com/group/comp.lang.python/msg/f1f10ced88c68c2d
if n <= 2:
return []
sieve = range(3, n, 2)
top = len(sieve)
for si in sieve:
if si:
bottom = (si*si - 3) // 2
if bottom >= top:
break
sieve[bottom::si] = [0] * -((bottom - top) // si)
return [2] + [el for el in sieve if el]
def sieveOfAtkin(end):
"""sieveOfAtkin(end): return a list of all the prime numbers <end
using the Sieve of Atkin."""
# Code by Steve Krenzel, <[email protected]>, improved
# Code: https://web.archive.org/web/20080324064651/http://krenzel.info/?p=83
# Info: http://en.wikipedia.org/wiki/Sieve_of_Atkin
assert end > 0
lng = ((end-1) // 2)
sieve = [False] * (lng + 1)
x_max, x2, xd = int(sqrt((end-1)/4.0)), 0, 4
for xd in xrange(4, 8*x_max + 2, 8):
x2 += xd
y_max = int(sqrt(end-x2))
n, n_diff = x2 + y_max*y_max, (y_max << 1) - 1
if not (n & 1):
n -= n_diff
n_diff -= 2
for d in xrange((n_diff - 1) << 1, -1, -8):
m = n % 12
if m == 1 or m == 5:
m = n >> 1
sieve[m] = not sieve[m]
n -= d
x_max, x2, xd = int(sqrt((end-1) / 3.0)), 0, 3
for xd in xrange(3, 6 * x_max + 2, 6):
x2 += xd
y_max = int(sqrt(end-x2))
n, n_diff = x2 + y_max*y_max, (y_max << 1) - 1
if not(n & 1):
n -= n_diff
n_diff -= 2
for d in xrange((n_diff - 1) << 1, -1, -8):
if n % 12 == 7:
m = n >> 1
sieve[m] = not sieve[m]
n -= d
x_max, y_min, x2, xd = int((2 + sqrt(4-8*(1-end)))/4), -1, 0, 3
for x in xrange(1, x_max + 1):
x2 += xd
xd += 6
if x2 >= end: y_min = (((int(ceil(sqrt(x2 - end))) - 1) << 1) - 2) << 1
n, n_diff = ((x*x + x) << 1) - 1, (((x-1) << 1) - 2) << 1
for d in xrange(n_diff, y_min, -8):
if n % 12 == 11:
m = n >> 1
sieve[m] = not sieve[m]
n += d
primes = [2, 3]
if end <= 3:
return primes[:max(0,end-2)]
for n in xrange(5 >> 1, (int(sqrt(end))+1) >> 1):
if sieve[n]:
primes.append((n << 1) + 1)
aux = (n << 1) + 1
aux *= aux
for k in xrange(aux, end, 2 * aux):
sieve[k >> 1] = False
s = int(sqrt(end)) + 1
if s % 2 == 0:
s += 1
primes.extend([i for i in xrange(s, end, 2) if sieve[i >> 1]])
return primes
def ambi_sieve_plain(n):
s = range(3, n, 2)
for m in xrange(3, int(n**0.5)+1, 2):
if s[(m-3)/2]:
for t in xrange((m*m-3)/2,(n>>1)-1,m):
s[t]=0
return [2]+[t for t in s if t>0]
def sundaram3(max_n):
# https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/2073279#2073279
numbers = range(3, max_n+1, 2)
half = (max_n)//2
initial = 4
for step in xrange(3, max_n+1, 2):
for i in xrange(initial, half, step):
numbers[i-1] = 0
initial += 2*(step+1)
if initial > half:
return [2] + filter(None, numbers)
################################################################################
# Using Numpy:
def ambi_sieve(n):
# http://tommih.blogspot.com/2009/04/fast-prime-number-generator.html
s = np.arange(3, n, 2)
for m in xrange(3, int(n ** 0.5)+1, 2):
if s[(m-3)/2]:
s[(m*m-3)/2::m]=0
return np.r_[2, s[s>0]]
def primesfrom3to(n):
# https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
""" Returns a array of primes, p < n """
assert n>=2
sieve = np.ones(n/2, dtype=np.bool)
for i in xrange(3,int(n**0.5)+1,2):
if sieve[i/2]:
sieve[i*i/2::i] = False
return np.r_[2, 2*np.nonzero(sieve)[0][1::]+1]
def primesfrom2to(n):
# https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
""" Input n>=6, Returns a array of primes, 2 <= p < n """
sieve = np.ones(n/3 + (n%6==2), dtype=np.bool)
sieve[0] = False
for i in xrange(int(n**0.5)/3+1):
if sieve[i]:
k=3*i+1|1
sieve[ ((k*k)/3) ::2*k] = False
sieve[(k*k+4*k-2*k*(i&1))/3::2*k] = False
return np.r_[2,3,((3*np.nonzero(sieve)[0]+1)|1)]
if __name__=='__main__':
import itertools
import sys
def test(f1,f2,num):
print('Testing {f1} and {f2} return same results'.format(
f1=f1.func_name,
f2=f2.func_name))
if not all([a==b for a,b in itertools.izip_longest(f1(num),f2(num))]):
sys.exit("Error: %s(%s) != %s(%s)"%(f1.func_name,num,f2.func_name,num))
n=1000000
test(sieveOfAtkin,sieveOfEratosthenes,n)
test(sieveOfAtkin,ambi_sieve,n)
test(sieveOfAtkin,ambi_sieve_plain,n)
test(sieveOfAtkin,sundaram3,n)
test(sieveOfAtkin,sieve_wheel_30,n)
test(sieveOfAtkin,primesfrom3to,n)
test(sieveOfAtkin,primesfrom2to,n)
test(sieveOfAtkin,rwh_primes,n)
test(sieveOfAtkin,rwh_primes1,n)
test(sieveOfAtkin,rwh_primes2,n)
Running the script tests that all implementations give the same result.
Solution 2:
Faster & more memory-wise pure Python code:
def primes(n):
""" Returns a list of primes < n """
sieve = [True] * n
for i in range(3,int(n**0.5)+1,2):
if sieve[i]:
sieve[i*i::2*i]=[False]*((n-i*i-1)//(2*i)+1)
return [2] + [i for i in range(3,n,2) if sieve[i]]
or starting with half sieve
def primes1(n):
""" Returns a list of primes < n """
sieve = [True] * (n//2)
for i in range(3,int(n**0.5)+1,2):
if sieve[i//2]:
sieve[i*i//2::i] = [False] * ((n-i*i-1)//(2*i)+1)
return [2] + [2*i+1 for i in range(1,n//2) if sieve[i]]
Faster & more memory-wise numpy code:
import numpy
def primesfrom3to(n):
""" Returns a array of primes, 3 <= p < n """
sieve = numpy.ones(n//2, dtype=bool)
for i in range(3,int(n**0.5)+1,2):
if sieve[i//2]:
sieve[i*i//2::i] = False
return 2*numpy.nonzero(sieve)[0][1::]+1
a faster variation starting with a third of a sieve:
import numpy
def primesfrom2to(n):
""" Input n>=6, Returns a array of primes, 2 <= p < n """
sieve = numpy.ones(n//3 + (n%6==2), dtype=bool)
for i in range(1,int(n**0.5)//3+1):
if sieve[i]:
k=3*i+1|1
sieve[ k*k//3 ::2*k] = False
sieve[k*(k-2*(i&1)+4)//3::2*k] = False
return numpy.r_[2,3,((3*numpy.nonzero(sieve)[0][1:]+1)|1)]
A (hard-to-code) pure-python version of the above code would be:
def primes2(n):
""" Input n>=6, Returns a list of primes, 2 <= p < n """
n, correction = n-n%6+6, 2-(n%6>1)
sieve = [True] * (n//3)
for i in range(1,int(n**0.5)//3+1):
if sieve[i]:
k=3*i+1|1
sieve[ k*k//3 ::2*k] = [False] * ((n//6-k*k//6-1)//k+1)
sieve[k*(k-2*(i&1)+4)//3::2*k] = [False] * ((n//6-k*(k-2*(i&1)+4)//6-1)//k+1)
return [2,3] + [3*i+1|1 for i in range(1,n//3-correction) if sieve[i]]
Unfortunately pure-python don't adopt the simpler and faster numpy way of doing assignment, and calling len() inside the loop as in [False]*len(sieve[((k*k)//3)::2*k]) is too slow. So I had to improvise to correct input (& avoid more math) and do some extreme (& painful) math-magic.
Personally I think it is a shame that numpy (which is so widely used) is not part of Python standard library, and that the improvements in syntax and speed seem to be completely overlooked by Python developers.
Solution 3:
There's a pretty neat sample from the Python Cookbook here -- the fastest version proposed on that URL is:
import itertools
def erat2( ):
D = { }
yield 2
for q in itertools.islice(itertools.count(3), 0, None, 2):
p = D.pop(q, None)
if p is None:
D[q*q] = q
yield q
else:
x = p + q
while x in D or not (x&1):
x += p
D[x] = p
so that would give
def get_primes_erat(n):
return list(itertools.takewhile(lambda p: p<n, erat2()))
Measuring at the shell prompt (as I prefer to do) with this code in pri.py, I observe:
$ python2.5 -mtimeit -s'import pri' 'pri.get_primes(1000000)'
10 loops, best of 3: 1.69 sec per loop
$ python2.5 -mtimeit -s'import pri' 'pri.get_primes_erat(1000000)'
10 loops, best of 3: 673 msec per loop
so it looks like the Cookbook solution is over twice as fast.