Python Eratosthenes筛-寻找素数
澄清一下,这不是家庭作业问题:) 我想为我正在构建的数学应用程序寻找素数&偶然发现的方法 我已经用Python编写了它的一个实现。但是速度太慢了。比如说,如果我想找到所有小于200万的素数。需要20分钟以上。(我在这一点上停了下来)。我怎样才能加快速度Python Eratosthenes筛-寻找素数,python,math,primes,sieve-of-eratosthenes,Python,Math,Primes,Sieve Of Eratosthenes,澄清一下,这不是家庭作业问题:) 我想为我正在构建的数学应用程序寻找素数&偶然发现的方法 我已经用Python编写了它的一个实现。但是速度太慢了。比如说,如果我想找到所有小于200万的素数。需要20分钟以上。(我在这一点上停了下来)。我怎样才能加快速度 def primes_sieve(limit): limitn = limit+1 primes = range(2, limitn) for i in primes: factors = range(i
def primes_sieve(limit):
limitn = limit+1
primes = range(2, limitn)
for i in primes:
factors = range(i, limitn, i)
for f in factors[1:]:
if f in primes:
primes.remove(f)
return primes
print primes_sieve(2000)
更新:
最后我对这段代码进行了分析&发现从列表中删除一个元素花费了相当多的时间。考虑到它必须遍历整个列表(最坏的情况)才能找到元素&然后删除它,然后重新调整列表(可能会有副本出现?),这是可以理解的。不管怎样,我扔掉了字典的清单。我的新实施-
def primes_sieve1(limit):
limitn = limit+1
primes = dict()
for i in range(2, limitn): primes[i] = True
for i in primes:
factors = range(i,limitn, i)
for f in factors[1:]:
primes[f] = False
return [i for i in primes if primes[i]==True]
print primes_sieve1(2000000)
从数组(列表)的开头删除需要向下移动数组(列表)后面的所有项。这意味着从前面开始,以这种方式从列表中删除每个元素是一个O(n^2)操作 使用集合可以更高效地执行此操作:
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
您没有完全实现正确的算法: 在第一个示例中,
primes\u sieve
不维护要删除/取消设置的素性标志列表(如在算法中),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
在第二个示例中,primes\u sieve1
维护一个素性标志字典,这是正确的步骤,但它以未定义的顺序在字典中迭代,并冗余地删除因子中的因子(而不是像算法中那样仅删除素数的因子)。您可以通过对键进行排序和跳过非素数(这已经使它快了一个数量级)来解决这个问题,但直接使用列表仍然更有效
正确的算法(使用列表而不是字典)如下所示:
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,这还包括在素数平方(
i*i
)而不是它的双精度处开始非素数标记的算法优化。)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案很有趣:因为python通过生成器提供了惰性求值,所以eratosthenes的筛选可以完全按照如下所述实现:
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块就在那里,因为算法会一直运行,直到它吹出堆栈,并且没有
try block将显示回溯,将实际输出推离屏幕。一个简单的速度破解: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 然后您可以进一步优化,而不是在素数中使用for i,在素数中使用for i[:round(len(素数)**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。更快:
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))
我的实施:
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
通过结合许多爱好者的贡献(包括以上评论中的Glenn Maynard和MrHIDEn),我在python 2中获得了以下代码:
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次方的不同输入所需的时间为:
- 3:0.3毫秒
- 4:2.4毫秒
- 5:23毫秒
- 6:0.26秒
- 7:3.1秒
- 8时33分
- 这是一个内存效率更高的版本(而且是一个合适的筛选,而不是试用版)。基本上,不是保留一个包含所有数字的数组,并划掉那些不是质数的,而是保留一个计数器数组——发现的每一个质数对应一个计数器——并将它们跳到假定质数之前。这样,它使用与素数成比例的存储,而不是最高素数
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))
您会注意到,primes()
是一个生成器,因此您可以将结果保存在列表中,也可以直接使用它们。下面是第一个n
素数:
import itertools
for k in itertools.islice (primes(), n):
print (k)
为了完整起见,这里有一个计时器来测量性能:
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,我还编写了
primes()
作为一个简单的迭代器(使用\uuuu iter\uuuu
和\uuu next\uuuu
),它以几乎相同的速度运行。我也很惊讶 由于速度的原因,我更喜欢NumPy
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)]
在Jupyter笔记本上比较Eratosthenes的筛子速度和蛮力。用539倍于蛮力的速度筛分埃拉托斯坦(Eratosthenes)中的百万元素
%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)
我认为必须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
limit=100
ints=列表(范围(2,限制))#将以空结束
当len(ints)>0时:
素数=整数[0]
打印素数
整数删除(素数)
i=2
倍数=素数*i
当多重导入数学时
def筛(n):
素数=[True]*n
素数[0]=假
素数[1]=假
对于范围(2,int(数学sqrt(n))+1)内的i:
j=i*i
而j
我能想到的最快实现:
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
我认为这是用eratosthenes方法寻找素数的最短代码
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
,我可以在2秒钟多一点的时间内找到所有低于1亿的素数
有两个关键特性需要注意
- 仅对
i
至n
- 使用
x[2*i::i]=False将i
的倍数设置为False
比显式python for循环快得多
这两种方法大大加快了代码的速度。对于低于一百万的限制,没有明显的运行时间
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)))
不确定我的代码是否有效,有人愿意评论吗
from math import isqrt
def isPrime(n):
if n >= 2: # cheating the 2, is 2 even prime?
for i in range(3, int(n / 2 + 1),2): # dont waste time with even numbers
if n % i == 0:
return False
return True
def primesTo(n):
x = [2] if n >= 2 else [] # cheat the only even prime
if n >= 2:
for i in range(3, n + 1,2): # dont waste time with even numbers
if isPrime(i):
x.append(i)
return x
def primes2(n): # trying to do this using set methods and the "Sieve of Eratosthenes"
base = {2} # again cheating the 2
base.update(set(range(3, n + 1, 2))) # build the base of odd numbers
for i in range(3, isqrt(n) + 1, 2): # apply the sieve
base.difference_update(set(range(2 * i, n + 1 , i)))
return list(base)
print(primesTo(10000)) # 2 different methods for comparison
print(primes2(10000))
我只是
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)))
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)))
from math import isqrt
def isPrime(n):
if n >= 2: # cheating the 2, is 2 even prime?
for i in range(3, int(n / 2 + 1),2): # dont waste time with even numbers
if n % i == 0:
return False
return True
def primesTo(n):
x = [2] if n >= 2 else [] # cheat the only even prime
if n >= 2:
for i in range(3, n + 1,2): # dont waste time with even numbers
if isPrime(i):
x.append(i)
return x
def primes2(n): # trying to do this using set methods and the "Sieve of Eratosthenes"
base = {2} # again cheating the 2
base.update(set(range(3, n + 1, 2))) # build the base of odd numbers
for i in range(3, isqrt(n) + 1, 2): # apply the sieve
base.difference_update(set(range(2 * i, n + 1 , i)))
return list(base)
print(primesTo(10000)) # 2 different methods for comparison
print(primes2(10000))
def nondivsby2():
j = 1
while True:
j += 2
yield j
def nondivsbyk(k, nondivs):
j = 0
for i in nondivs:
while j < i:
j += k
if j > i:
yield i
def primes():
nd = nondivsby2()
while True:
p = next(nd)
nd = nondivsbyk(p, nd)
yield p
def main():
for p in primes():
print(p)
import sympy
list(sympy.primerange(lower, upper+1))
def prime_factors(n):
for i in range(2, int(n ** 0.5) + 1):
if (q_r := divmod(n, i))[1] == 0:
return [i] + factor_list(q_r[0])
return [n]
sieve = lambda j: [print(x) for x in filter(lambda n: 0 not in map(lambda i: n % i, range(2, n)) and (n!=1)&(n!=0), range(j + 1))]