您的版本有几个小的优化。通过颠倒 True 和 False 的角色，您可以将“ if flags[i] is False:”更改为“ if flags[i]:”。并且第二条语句的起始值range可以i*i代替i*3. 您的原始版本在我的系统上需要 0.166 秒。通过这些更改，以下版本在我的系统上需要 0.156 秒。
def prime_numbers(limit=1000000):
'''Prime number generator. Yields the series
2, 3, 5, 7, 11, 13, 17, 19, 23, 29 ...
using Sieve of Eratosthenes.
'''
yield 2
sub_limit = int(limit**0.5)
flags = [True, True] + [False] * (limit - 2)
# Step through all the odd numbers
for i in range(3, limit, 2):
if flags[i]:
continue
yield i
# Exclude further multiples of the current prime number
if i <= sub_limit:
for j in range(i*i, limit, i<<1):
flags[j] = True
不过，这对您的记忆问题没有帮助。
进入 C 扩展的世界，我使用了gmpy的开发版本。（免责声明：我是维护者之一。）开发版本称为 gmpy2 并支持称为 xmpz 的可变整数。使用 gmpy2 和以下代码，我的运行时间为 0.140 秒。限制为 1,000,000,000 的运行时间为 158 秒。
import gmpy2
def prime_numbers(limit=1000000):
'''Prime number generator. Yields the series
2, 3, 5, 7, 11, 13, 17, 19, 23, 29 ...
using Sieve of Eratosthenes.
'''
yield 2
sub_limit = int(limit**0.5)
# Actual number is 2*bit_position + 1.
oddnums = gmpy2.xmpz(1)
current = 0
while True:
current += 1
current = oddnums.bit_scan0(current)
prime = 2 * current + 1
if prime > limit:
break
yield prime
# Exclude further multiples of the current prime number
if prime <= sub_limit:
for j in range(2*current*(current+1), limit>>1, prime):
oddnums.bit_set(j)
推动优化并牺牲清晰度，我使用以下代码获得了 0.107 和 123 秒的运行时间：
import gmpy2
def prime_numbers(limit=1000000):
'''Prime number generator. Yields the series
2, 3, 5, 7, 11, 13, 17, 19, 23, 29 ...
using Sieve of Eratosthenes.
'''
yield 2
sub_limit = int(limit**0.5)
# Actual number is 2*bit_position + 1.
oddnums = gmpy2.xmpz(1)
f_set = oddnums.bit_set
f_scan0 = oddnums.bit_scan0
current = 0
while True:
current += 1
current = f_scan0(current)
prime = 2 * current + 1
if prime > limit:
break
yield prime
# Exclude further multiples of the current prime number
if prime <= sub_limit:
list(map(f_set,range(2*current*(current+1), limit>>1, prime)))
编辑：基于这个练习，我将 gmpy2 修改为 accept xmpz.bit_set(iterator)。使用以下代码，所有小于 1,000,000,000 的素数的运行时间对于 Python 2.7 为 56 秒，对于 Python 3.2 为 74 秒。（如评论中所述，xrange比 快range。）
import gmpy2
try:
range = xrange
except NameError:
pass
def prime_numbers(limit=1000000):
'''Prime number generator. Yields the series
2, 3, 5, 7, 11, 13, 17, 19, 23, 29 ...
using Sieve of Eratosthenes.
'''
yield 2
sub_limit = int(limit**0.5)
oddnums = gmpy2.xmpz(1)
f_scan0 = oddnums.bit_scan0
current = 0
while True:
current += 1
current = f_scan0(current)
prime = 2 * current + 1
if prime > limit:
break
yield prime
if prime <= sub_limit:
oddnums.bit_set(iter(range(2*current*(current+1), limit>>1, prime)))
编辑#2：再试一次！我修改了 gmpy2 以接受xmpz.bit_set(slice). 使用以下代码，对于 Python 2.7 和 Python 3.2，所有小于 1,000,000,000 的素数的运行时间约为 40 秒。
from __future__ import print_function
import time
import gmpy2
def prime_numbers(limit=1000000):
'''Prime number generator. Yields the series
2, 3, 5, 7, 11, 13, 17, 19, 23, 29 ...
using Sieve of Eratosthenes.
'''
yield 2
sub_limit = int(limit**0.5)
flags = gmpy2.xmpz(1)
# pre-allocate the total length
flags.bit_set((limit>>1)+1)
f_scan0 = flags.bit_scan0
current = 0
while True:
current += 1
current = f_scan0(current)
prime = 2 * current + 1
if prime > limit:
break
yield prime
if prime <= sub_limit:
flags.bit_set(slice(2*current*(current+1), limit>>1, prime))
start = time.time()
result = list(prime_numbers(1000000000))
print(time.time() - start)
编辑#3：我已更新 gmpy2 以正确支持 xmpz 位级别的切片。性能没有变化，但 API 非常好。我做了一些调整，我把时间缩短到了大约 37 秒。（请参阅编辑 #4 以了解 gmpy2 2.0.0b1 中的更改。）
from __future__ import print_function
import time
import gmpy2
def prime_numbers(limit=1000000):
'''Prime number generator. Yields the series
2, 3, 5, 7, 11, 13, 17, 19, 23, 29 ...
using Sieve of Eratosthenes.
'''
sub_limit = int(limit**0.5)
flags = gmpy2.xmpz(1)
flags[(limit>>1)+1] = True
f_scan0 = flags.bit_scan0
current = 0
prime = 2
while prime <= sub_limit:
yield prime
current += 1
current = f_scan0(current)
prime = 2 * current + 1
flags[2*current*(current+1):limit>>1:prime] = True
while prime <= limit:
yield prime
current += 1
current = f_scan0(current)
prime = 2 * current + 1
start = time.time()
result = list(prime_numbers(1000000000))
print(time.time() - start)
编辑#4：我在 gmpy2 2.0.0b1 中做了一些改变，打破了前面的例子。gmpy2 不再将 True 视为提供无限源的 1 位的特殊值。-1 应改为使用。
from __future__ import print_function
import time
import gmpy2
def prime_numbers(limit=1000000):
'''Prime number generator. Yields the series
2, 3, 5, 7, 11, 13, 17, 19, 23, 29 ...
using Sieve of Eratosthenes.
'''
sub_limit = int(limit**0.5)
flags = gmpy2.xmpz(1)
flags[(limit>>1)+1] = 1
f_scan0 = flags.bit_scan0
current = 0
prime = 2
while prime <= sub_limit:
yield prime
current += 1
current = f_scan0(current)
prime = 2 * current + 1
flags[2*current*(current+1):limit>>1:prime] = -1
while prime <= limit:
yield prime
current += 1
current = f_scan0(current)
prime = 2 * current + 1
start = time.time()
result = list(prime_numbers(1000000000))
print(time.time() - start)
编辑#5：我对 gmpy2 2.0.0b2 做了一些改进。您现在可以遍历所有设置或清除的位。运行时间提高了约 30%。
from __future__ import print_function
import time
import gmpy2
def sieve(limit=1000000):
'''Returns a generator that yields the prime numbers up to limit.'''
# Increment by 1 to account for the fact that slices do not include
# the last index value but we do want to include the last value for
# calculating a list of primes.
sieve_limit = gmpy2.isqrt(limit) + 1
limit += 1
# Mark bit positions 0 and 1 as not prime.
bitmap = gmpy2.xmpz(3)
# Process 2 separately. This allows us to use p+p for the step size
# when sieving the remaining primes.
bitmap[4 : limit : 2] = -1
# Sieve the remaining primes.
for p in bitmap.iter_clear(3, sieve_limit):
bitmap[p*p : limit : p+p] = -1
return bitmap.iter_clear(2, limit)
if __name__ == "__main__":
start = time.time()
result = list(sieve(1000000000))
print(time.time() - start)
print(len(result))