Python-列出N以下所有素数的最快方法

这是我能想到的最好的算法。

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

可以使它更快吗?

此代码有一个缺陷:由于numbers是无序集合,因此不能保证numbers.pop()从集合中删除最低的数字。但是,它对某些输入数字有效(至少对我而言):

>>> 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

回答:

警告: timeit由于硬件或Python版本的差异,结果可能会有所不同。

下面是一个脚本,它比较了许多实现:

  • ambi_sieve_plain,
  • rwh_primes,
  • rwh_primes1,
  • rwh_primes2,
  • sieveOfAtkin,
  • 埃拉托斯特尼筛法,
  • 孙达拉姆3,
  • sieve_wheel_30,
  • ambi_sieve(需要numpy)
  • primesfrom3to(需要numpy)
  • primesfrom2to(需要numpy)

在使用psyco测试的简单Python方法中,对于n = 1000000, rwh_primes1是测试最快的方法。

+---------------------+-------+

| 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 |

+---------------------+-------+

在没有psyco的情况下,经过测试的普通Python方法中n = 1000000, rwh_primes2是最快的。

+---------------------+-------+

| 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 |

+---------------------+-------+

在所有测试的方法中,允许numpy,对于n = 1000000, primesfrom2to是测试最快的方法。

+---------------------+-------+

| Method | ms |

+---------------------+-------+

| primesfrom2to | 15.9 |

| primesfrom3to | 18.4 |

| ambi_sieve | 29.3 |

+---------------------+-------+

使用以下命令测量时间:

python -mtimeit -s"import primes" "primes.{method}(1000000)"

{method}每个方法名称替换。

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: <dickinsm@gmail.com>, 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, <Sgk284@gmail.com>, 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)

运行脚本会测试所有实现都给出相同的结果。

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