借用本文，我们了解了如何在多线程、多处理和异步/异步之间进行选择及其用法。

Python 3有一个新的内置库，以实现并发和并行-concurrent.futures

因此，我将通过一个实验演示如何通过线程池运行四个任务（即.sleep（）方法）：

from concurrent.futures import ThreadPoolExecutor, as_completed
from time import sleep, time

def concurrent(max_worker):
    futures = []
    tic = time()
    with ThreadPoolExecutor(max_workers=max_worker) as executor:
        futures.append(executor.submit(sleep, 2))  # Two seconds sleep
        futures.append(executor.submit(sleep, 1))
        futures.append(executor.submit(sleep, 7))
        futures.append(executor.submit(sleep, 3))
        for future in as_completed(futures):
            if future.result() is not None:
                print(future.result())
    print(f'Total elapsed time by {max_worker} workers:', time()-tic)

concurrent(5)
concurrent(4)
concurrent(3)
concurrent(2)
concurrent(1)

输出：

Total elapsed time by 5 workers: 7.007831811904907
Total elapsed time by 4 workers: 7.007944107055664
Total elapsed time by 3 workers: 7.003149509429932
Total elapsed time by 2 workers: 8.004627466201782
Total elapsed time by 1 workers: 13.013478994369507

[注]：

正如您在上面的结果中看到的，最好的情况是这四项任务有3名员工。如果有进程任务而不是I/O绑定或阻塞（多处理而不是线程），则可以将ThreadPoolExecutor更改为ProcessPoolExecutoor。

注意：对于Python中的实际并行化，您应该使用多处理模块来分叉并行执行的多个进程（由于全局解释器锁，Python线程提供了交织，但实际上它们是串行执行的，而不是并行执行的，并且仅在交织I/O操作时有用）。

然而，如果您只是在寻找交错（或者正在执行可以并行化的I/O操作，尽管存在全局解释器锁），那么线程模块就是开始的地方。作为一个非常简单的例子，让我们考虑通过并行对子范围求和来对大范围求和的问题：

import threading

class SummingThread(threading.Thread):
     def __init__(self,low,high):
         super(SummingThread, self).__init__()
         self.low=low
         self.high=high
         self.total=0

     def run(self):
         for i in range(self.low,self.high):
             self.total+=i


thread1 = SummingThread(0,500000)
thread2 = SummingThread(500000,1000000)
thread1.start() # This actually causes the thread to run
thread2.start()
thread1.join()  # This waits until the thread has completed
thread2.join()
# At this point, both threads have completed
result = thread1.total + thread2.total
print result

请注意，以上是一个非常愚蠢的示例，因为它绝对没有I/O，并且由于全局解释器锁，虽然在CPython中交错执行（增加了上下文切换的开销），但仍将串行执行。

对我来说，线程的最佳示例是监视异步事件。看看这个代码。

# thread_test.py
import threading
import time

class Monitor(threading.Thread):
    def __init__(self, mon):
        threading.Thread.__init__(self)
        self.mon = mon

    def run(self):
        while True:
            if self.mon[0] == 2:
                print "Mon = 2"
                self.mon[0] = 3;

您可以通过打开IPython会话并执行以下操作来使用此代码：

>>> from thread_test import Monitor
>>> a = [0]
>>> mon = Monitor(a)
>>> mon.start()
>>> a[0] = 2
Mon = 2
>>>a[0] = 2
Mon = 2

等几分钟

>>> a[0] = 2
Mon = 2

下面的代码可以运行10个线程同时打印0到99之间的数字：

from threading import Thread

def test():
    for i in range(0, 100):
        print(i)

thread_list = []

for _ in range(0, 10):
    thread = Thread(target=test)
    thread_list.append(thread)

for thread in thread_list:
    thread.start()

for thread in thread_list:
    thread.join()

下面的代码是上述代码循环版本的简写，运行10个线程，同时打印0到99之间的数字：

from threading import Thread

def test():
    [print(i) for i in range(0, 100)]

thread_list = [Thread(target=test) for _ in range(0, 10)]

[thread.start() for thread in thread_list]

[thread.join() for thread in thread_list]

结果如下：

...
99
83
97
84
98
99
85
86
87
88
...

使用线程/多处理的最简单方法是使用更多高级库，如autothread。

import autothread
from time import sleep as heavyworkload

@autothread.multithreaded() # <-- This is all you need to add
def example(x: int, y: int):
    heavyworkload(1)
    return x*y

现在，您可以为函数提供int列表。Autothread将为您处理所有事务，并只提供并行计算的结果。

result = example([1, 2, 3, 4, 5], 10)

import threading
import requests

def send():

  r = requests.get('https://www.stackoverlow.com')

thread = []
t = threading.Thread(target=send())
thread.append(t)
t.start()