April 2023
Intermediate to advanced
769 pages
25h 16m
Chinese
Content preview from 流畅的Python（第2版）
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552
｜
第
19
章
➊
每
50 000
次迭代（因为
range
的步幅是
2
）休眠一次。
asyncio
仓库中的
284
号工单详细讨论了
asyncio.sleep(0)
的用法。
然而，需要注意的是，这将拖慢
is_prime
，以及事件循环和整个程序的速度——后二
者才是重点。换成每
10
万次迭代执行一次
await asyncio.sleep(0)
之后，在我的设备
中，旋转指针依然顺滑，只是用时达到了
4.9
秒，比原来的
primes.is_prime
函数慢了
将近
50%
（参数同为
5_000_111_000_222_021
）。
使用
await asyncio.sleep(0)
只是权宜之计，更稳妥的办法是重构异步代码，把
CPU
密集型计算委托给另一个进程。第
21
章将给出一个使用
asyncio.loop.run_in_
executor
的方案。另外，还可以使用任务队列（详见
19.7.5
节 ）。
目前，我们所做的实验只调用了一个
CPU
密集型函数。
19.6
节说明如何并发执行多个
CPU
密集型调用。
19.6
自建进程池
我写这一节是为了展示如何使用多个进程处理
CPU
密集型任务，以及使用
队列分配任务和收集结果的常见模式。第
20
章将给出一种为进程分配任务
更简单的方式：
concurrent.futures
包中的
ProcessPoolExecutor
（内部使用
队列）。
本节将编写一个程序，计算
2
和
9 999 999 999 999 999
（
10
16
–
1
）之间或大于
2
53
的
20
个整
数样本是否为素数。样本中的素数有大有小，还有可以分解为大小素数的合数。 ...
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I'm always learning. So when I got on to O'Reilly, I was like a kid in a candy store. There are playlists. There are answers. There's on-demand training. It's worth its weight in gold, in terms of what it allows me to do.