埃拉托舍内斯的筛子-寻找素数巨蟒

您没有完全实现正确的算法:

primes_sieve 不维护要删除/取消设置的素性标志列表(如算法中所示),而是连续调整整数列表的大小,这非常昂贵:从列表中删除项需要将所有后续项下移一个。

在第二个例子中, primes_sieve1 保持 素性标志,这是朝着正确方向迈出的一步,但它以未定义的顺序在字典上迭代,并冗余地删除因子(而不是仅删除素数的因子,如在算法中)。您可以通过对键进行排序并跳过非素数(这已经使它的速度提高了一个数量级)来解决这个问题,但是直接使用列表仍然有效得多。

正确的算法(使用列表而不是字典)类似于:

def primes_sieve2(limit):
    a = [True] * limit                          # Initialize the primality list
    a[0] = a[1] = False

    for (i, isprime) in enumerate(a):
        if isprime:
            yield i
            for n in range(i*i, limit, i):     # Mark factors non-prime
                a[n] = False

(请注意,这还包括在素数的平方开始非素数标记的算法优化( i*i

def eratosthenes(n):
    multiples = []
    for i in range(2, n+1):
        if i not in multiples:
            print (i)
            for j in range(i*i, n+1, i):
                multiples.append(j)

eratosthenes(100)

要从数组(列表)的开头删除,需要向下移动该数组之后的所有项。这意味着以这种方式从前面开始删除列表中的每个元素是一个O(n^2)操作。

使用集合可以更有效地执行此操作:

def primes_sieve(limit):
    limitn = limit+1
    not_prime = set()
    primes = []

    for i in range(2, limitn):
        if i in not_prime:
            continue

        for f in range(i*2, limitn, i):
            not_prime.add(f)

        primes.append(i)

    return primes

print primes_sieve(1000000)

... 或者,避免重新排列列表:

def primes_sieve(limit):
    limitn = limit+1
    not_prime = [False] * limitn
    primes = []

    for i in range(2, limitn):
        if not_prime[i]:
            continue
        for f in xrange(i*2, limitn, i):
            not_prime[f] = True

        primes.append(i)

    return primes

import time
def get_primes(n):
  m = n+1
  #numbers = [True for i in range(m)]
  numbers = [True] * m #EDIT: faster
  for i in range(2, int(n**0.5 + 1)):
    if numbers[i]:
      for j in range(i*i, m, i):
        numbers[j] = False
  primes = []
  for i in range(2, m):
    if numbers[i]:
      primes.append(i)
  return primes

start = time.time()
primes = get_primes(10000)
print(time.time() - start)
print(get_primes(100))

我意识到这并不能真正回答如何快速生成素数的问题,但也许有些人会发现这个替代方法很有意思:因为python通过生成器提供了惰性的计算,所以可以按照下面的说明来实现eratosthenes的sieve:

def intsfrom(n):
    while True:
        yield n
        n += 1

def sieve(ilist):
    p = next(ilist)
    yield p
    for q in sieve(n for n in ilist if n%p != 0):
        yield q


try:
    for p in sieve(intsfrom(2)):
        print p,

    print ''
except RuntimeError as e:
    print e

try块在那里是因为算法一直运行到它吹出堆栈,而没有 try block将在屏幕外显示backtrace并推送您要查看的实际输出。

def simpleSieve(sieveSize):
    #creating Sieve.
    sieve = [True] * (sieveSize+1)
    # 0 and 1 are not considered prime.
    sieve[0] = False
    sieve[1] = False
    for i in xrange(2,int(math.sqrt(sieveSize))+1):
        if sieve[i] == False:
            continue
        for pointer in xrange(i**2, sieveSize+1, i):
            sieve[pointer] = False
    # Sieve is left with prime numbers == True
    primes = []
    for i in xrange(sieveSize+1):
        if sieve[i] == True:
            primes.append(i)
    return primes

sieveSize = input()
primes = simpleSieve(sieveSize)

在我的机器上计算10次方不同输入所用的时间是:

• 3:0.3毫秒
• 4:2.4毫秒
• 6:0.26秒
• 7:3.1秒

一个简单的速度技巧:当你定义变量“primes”时,将步骤设置为2以自动跳过所有偶数,并将起点设置为1。

然后,可以进一步优化by,而不是for i in primes,使用for i in primes[:round(len(primes)**0.5]。这将大大提高性能。此外,您可以消除以5结尾的数字,以进一步提高速度。

我的实现:

import math
n = 100
marked = {}
for i in range(2, int(math.sqrt(n))):
    if not marked.get(i):
        for x in range(i * i, n, i):
            marked[x] = True

for i in range(2, n):
    if not marked.get(i):
        print i

import itertools

def primes():

    class counter:
        def __init__ (this,  n): this.n, this.current,  this.isVirgin = n, n*n,  True
            # isVirgin means it's never been incremented
        def advancePast (this,  n): # return true if the counter advanced
            if this.current > n:
                if this.isVirgin: raise StopIteration # if this is virgin, then so will be all the subsequent counters.  Don't need to iterate further.
                return False
            this.current += this.n # pre: this.current == n; post: this.current > n.
            this.isVirgin = False # when it's gone, it's gone
            return True

    yield 1
    multiples = []
    for n in itertools.count(2):
        isPrime = True
        for p in (m.advancePast(n) for m in multiples):
            if p: isPrime = False
        if isPrime:
            yield n
            multiples.append (counter (n))

你会注意到的 primes() n 素数:

import itertools

for k in itertools.islice (primes(),  n):
    print (k)

import time

def timer ():
    t,  k = time.process_time(),  10
    for p in primes():
        if p>k:
            print (time.process_time()-t,  " to ",  p,  "\n")
            k *= 10
            if k>100000: return

如果你想知道的话,我也写了 素数() __iter__ 和 __next__ ),它以几乎相同的速度运行。我也很惊讶!

我喜欢裸体是因为速度。

import numpy as np

# Find all prime numbers using Sieve of Eratosthenes
def get_primes1(n):
    m = int(np.sqrt(n))
    is_prime = np.ones(n, dtype=bool)
    is_prime[:2] = False  # 0 and 1 are not primes

    for i in range(2, m):
        if is_prime[i] == False:
            continue
        is_prime[i*i::i] = False

    return np.nonzero(is_prime)[0]

# Find all prime numbers using brute-force.
def isprime(n):
    ''' Check if integer n is a prime '''
    n = abs(int(n))  # n is a positive integer
    if n < 2:  # 0 and 1 are not primes
        return False
    if n == 2:  # 2 is the only even prime number
        return True
    if not n & 1:  # all other even numbers are not primes
        return False
    # Range starts with 3 and only needs to go up the square root
    # of n for all odd numbers
    for x in range(3, int(n**0.5)+1, 2):
        if n % x == 0:
            return False
    return True

# To apply a function to a numpy array, one have to vectorize the function
def get_primes2(n):
    vectorized_isprime = np.vectorize(isprime)
    a = np.arange(n)
    return a[vectorized_isprime(a)]

检查输出:

n = 100
print(get_primes1(n))
print(get_primes2(n))    
    [ 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]
    [ 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]

%timeit get_primes1(1000000)
%timeit get_primes2(1000000)
4.79 ms ± 90.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
2.58 s ± 31.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

我想应该可以简单地使用空列表作为循环的终止条件,并得出以下结论:

limit = 100
ints = list(range(2, limit))   # Will end up empty

while len(ints) > 0:
    prime = ints[0]
    print prime
    ints.remove(prime)
    i = 2
    multiple = prime * i
    while multiple <= limit:
        if multiple in ints:
            ints.remove(multiple)
        i += 1
        multiple = prime * i

import math
def sieve(n):
    primes = [True]*n
    primes[0] = False
    primes[1] = False
    for i in range(2,int(math.sqrt(n))+1):
            j = i*i
            while j < n:
                    primes[j] = False
                    j = j+i
    return [x for x in range(n) if primes[x] == True]

def prime(r):
    n = range(2,r)
    while len(n)>0:
        yield n[0]
        n = [x for x in n if x not in range(n[0],r,n[0])]


print(list(prime(r)))

使用一点 numpy

有两个关键特性值得注意

• 切出倍数 i 只为 直到根 n
• 设置倍数 我 False x[2*i::i] = False 比显式python for循环快得多。

这两个明显加快了代码的速度。对于100万以下的限制,没有可感知的运行时间。

import numpy as np

def primes(n):
    x = np.ones((n+1,), dtype=np.bool)
    x[0] = False
    x[1] = False
    for i in range(2, int(n**0.5)+1):
        if x[i]:
            x[2*i::i] = False

    primes = np.where(x == True)[0]
    return primes

print(len(primes(100_000_000)))

我能想到的最快的实现:

isprime = [True]*N
isprime[0] = isprime[1] = False
for i in range(4, N, 2):
    isprime[i] = False
for i in range(3, N, 2):
    if isprime[i]:
        for j in range(i*i, N, 2*i):
            isprime[j] = False