Python中的简单Prime生成器
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 mathdef 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
回答:
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 mathdef 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。
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