Python 中的简单素数生成器

本文介绍了Python 中的简单素数生成器的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

有人能告诉我这段代码有什么问题吗?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(没什么特别的).

Could someone please tell me what I'm doing wrong with this code? It is just printing 'count' anyway. I just want a very simple prime generator (nothing fancy).

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

推荐答案

存在一些问题:

• 当 count 没有除以 x 时，为什么要打印出来?这并不意味着它是素数，它只是意味着这个特定的 x 不能整除它
• continue 移动到下一个循环迭代 - 但您真的想使用 break
停止它

• Why do you print out count when it didn't divide by x? It doesn't mean it's prime, it means only that this particular x doesn't divide it
• continue moves to the next loop iteration - but you really want to stop it using break

这是你的代码，有一些修正，它只打印出素数:

Here's your code with a few fixes, it prints out only primes:

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

如其他人所建议的那样，要获得更高效的素数生成，请参阅埃拉托色尼筛.这是一个很好的优化实现，有很多评论:

For much more efficient prime generation, see the Sieve of Eratosthenes, as others have suggested. Here's a nice, optimized implementation with many comments:

# Sieve of Eratosthenes # Code by David Eppstein, UC Irvine, 28 Feb 2002 # code.activestate/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

注意它返回一个生成器.

Note that it returns a generator.