一尘不染

Python中的简单Prime生成器

python

有人可以告诉我这段代码在做什么吗?无论如何,它只是打印“计数”。我只想要一个非常简单的素数生成器(没什么花哨的)。

import math

def main():
    count = 3
    one = 1
    while one == 1:
        for x in range(2, int(math.sqrt(count) + 1)):
            if count % x == 0: 
                continue
            if count % x != 0:
                print count

        count += 1

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2020-02-17

共1个答案

一尘不染

有一些问题:

  • 当计数不除以x时,为什么要打印计数?这并不意味着它是素数,仅意味着该特定x不会将其除
  • continue 移至下一个循环迭代-但你确实想使用停止它 break
    这是你的代码,其中包含一些修复程序,它仅输出质数:
import math

def main():
    count = 3

    while True:
        isprime = True

        for x in range(2, int(math.sqrt(count) + 1)):
            if count % x == 0: 
                isprime = False
                break

        if isprime:
            print count

        count += 1

有关更有效的质子生成,请参见其他人的建议,参见戊二烯筛。这是一个不错的,经过优化的实现,带有很多注释:

# Sieve of Eratosthenes
# Code by David Eppstein, UC Irvine, 28 Feb 2002
# http://code.activestate.com/recipes/117119/

def gen_primes():
    """ Generate an infinite sequence of prime numbers.
    """
    # Maps composites to primes witnessing their compositeness.
    # This is memory efficient, as the sieve is not "run forward"
    # indefinitely, but only as long as required by the current
    # number being tested.
    #
    D = {}

    # The running integer that's checked for primeness
    q = 2

    while True:
        if q not in D:
            # q is a new prime.
            # Yield it and mark its first multiple that isn't
            # already marked in previous iterations
            # 
            yield q
            D[q * q] = [q]
        else:
            # q is composite. D[q] is the list of primes that
            # divide it. Since we've reached q, we no longer
            # need it in the map, but we'll mark the next 
            # multiples of its witnesses to prepare for larger
            # numbers
            # 
            for p in D[q]:
                D.setdefault(p + q, []).append(p)
            del D[q]

        q += 1

请注意,它返回一个生成器。

2020-02-17