使用Sundaram的Sieve，我想我打破了pure-Python的记录:

def sundaram3(max_n):
    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)

Comparasion:

C:\USERS>python -m timeit -n10 -s "import get_primes" "get_primes.get_primes_erat(1000000)"
10 loops, best of 3: 710 msec per loop

C:\USERS>python -m timeit -n10 -s "import get_primes" "get_primes.daniel_sieve_2(1000000)"
10 loops, best of 3: 435 msec per loop

C:\USERS>python -m timeit -n10 -s "import get_primes" "get_primes.sundaram3(1000000)"
10 loops, best of 3: 327 msec per loop

第一次使用python，所以我在这里使用的一些方法可能看起来有点麻烦。我只是直接将我的c++代码转换为python，这就是我所拥有的(尽管在python中有点慢)

#!/usr/bin/env python
import time

def GetPrimes(n):

    Sieve = [1 for x in xrange(n)]

    Done = False
    w = 3

    while not Done:

        for q in xrange (3, n, 2):
            Prod = w*q
            if Prod < n:
                Sieve[Prod] = 0
            else:
                break

        if w > (n/2):
            Done = True
        w += 2

    return Sieve



start = time.clock()

d = 10000000
Primes = GetPrimes(d)

count = 1 #This is for 2

for x in xrange (3, d, 2):
    if Primes[x]:
        count+=1

elapsed = (time.clock() - start)
print "\nFound", count, "primes in", elapsed, "seconds!\n"

pythonw Primes.py 在12.799119秒内找到664579个质数!

#!/usr/bin/env python
import time

def GetPrimes2(n):

    Sieve = [1 for x in xrange(n)]

    for q in xrange (3, n, 2):
        k = q
        for y in xrange(k*3, n, k*2):
            Sieve[y] = 0

    return Sieve



start = time.clock()

d = 10000000
Primes = GetPrimes2(d)

count = 1 #This is for 2

for x in xrange (3, d, 2):
    if Primes[x]:
        count+=1

elapsed = (time.clock() - start)
print "\nFound", count, "primes in", elapsed, "seconds!\n"

pythonw Primes2.py 在10.230172秒内找到664579个质数!

#!/usr/bin/env python
import time

def GetPrimes3(n):

    Sieve = [1 for x in xrange(n)]

    for q in xrange (3, n, 2):
        k = q
        for y in xrange(k*k, n, k << 1):
            Sieve[y] = 0

    return Sieve



start = time.clock()

d = 10000000
Primes = GetPrimes3(d)

count = 1 #This is for 2

for x in xrange (3, d, 2):
    if Primes[x]:
        count+=1

elapsed = (time.clock() - start)
print "\nFound", count, "primes in", elapsed, "seconds!\n"

python Primes2.py 在7.113776秒内找到664579个质数!

对于Python 3

def rwh_primes2(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 range(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 range(1,n//3-correction) if sieve[i]]

下面是Eratosthenes的一个numpy版本，具有良好的复杂度(低于排序长度为n的数组)和向量化。与@unutbu相比，用46微秒就可以找到100万以下的所有质数。

import numpy as np 
def generate_primes(n):
    is_prime = np.ones(n+1,dtype=bool)
    is_prime[0:2] = False
    for i in range(int(n**0.5)+1):
        if is_prime[i]:
            is_prime[i**2::i]=False
    return np.where(is_prime)[0]

计时:

import time    
for i in range(2,10):
    timer =time.time()
    generate_primes(10**i)
    print('n = 10^',i,' time =', round(time.time()-timer,6))

>> n = 10^ 2  time = 5.6e-05
>> n = 10^ 3  time = 6.4e-05
>> n = 10^ 4  time = 0.000114
>> n = 10^ 5  time = 0.000593
>> n = 10^ 6  time = 0.00467
>> n = 10^ 7  time = 0.177758
>> n = 10^ 8  time = 1.701312
>> n = 10^ 9  time = 19.322478

这个算法很快，但它有一个严重的缺陷:

>>> sorted(get_primes(530))
[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, 527, 529]
>>> 17*31
527
>>> 23*23
529

您假设numbers.pop()将返回集合中最小的数字，但这根本不能保证。集合是无序的，pop()删除并返回任意元素，因此不能使用它从剩余数字中选择下一个质数。

编写自己的质数查找代码很有指导意义，但手边有一个快速可靠的库也很有用。我围绕c++库primesieve编写了一个包装器，命名为primesieve-python

试试pip install primesieve吧

import primesieve
primes = primesieve.generate_primes(10**8)

我很好奇对比一下速度。