这很容易理解。这里有两种简单的线程处理方法。

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

def a(a=1, b=2):
    print(a)
    time.sleep(5)
    print(b)
    return a+b

def b(**kwargs):
    if "a" in kwargs:
        print("am b")
    else:
        print("nothing")
        
to_do=[]
executor = ThreadPoolExecutor(max_workers=4)
ex1=executor.submit(a)
to_do.append(ex1)
ex2=executor.submit(b, **{"a":1})
to_do.append(ex2)

for future in as_completed(to_do):
    print("Future {} and Future Return is {}\n".format(future, future.result()))

print("threading")

to_do=[]
to_do.append(threading.Thread(target=a))
to_do.append(threading.Thread(target=b, kwargs={"a":1}))

for threads in to_do:
    threads.start()
    
for threads in to_do:
    threads.join()

Alex Martelli的回答对我有所帮助。不过，这里有一个我认为更有用的修改版本（至少对我来说）。

更新：可在Python 2和Python 3中使用

try:
    # For Python 3
    import queue
    from urllib.request import urlopen
except:
    # For Python 2 
    import Queue as queue
    from urllib2 import urlopen

import threading

worker_data = ['http://google.com', 'http://yahoo.com', 'http://bing.com']

# Load up a queue with your data. This will handle locking
q = queue.Queue()
for url in worker_data:
    q.put(url)

# Define a worker function
def worker(url_queue):
    queue_full = True
    while queue_full:
        try:
            # Get your data off the queue, and do some work
            url = url_queue.get(False)
            data = urlopen(url).read()
            print(len(data))

        except queue.Empty:
            queue_full = False

# Create as many threads as you want
thread_count = 5
for i in range(thread_count):
    t = threading.Thread(target=worker, args = (q,))
    t.start()

给定函数f，如下所示：

import threading
threading.Thread(target=f).start()

向f传递参数

threading.Thread(target=f, args=(a,b,c)).start()

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。

这里是多线程，有一个简单的例子会很有帮助。您可以运行它并轻松了解多线程在Python中的工作方式。我使用了一个锁来防止访问其他线程，直到前面的线程完成它们的工作。通过使用这行代码，

t锁定=线程。有界信号量（值=4）

您可以一次允许多个进程，并保留将在稍后或完成之前的进程后运行的其余线程。

import threading
import time

#tLock = threading.Lock()
tLock = threading.BoundedSemaphore(value=4)
def timer(name, delay, repeat):
    print  "\r\nTimer: ", name, " Started"
    tLock.acquire()
    print "\r\n", name, " has the acquired the lock"
    while repeat > 0:
        time.sleep(delay)
        print "\r\n", name, ": ", str(time.ctime(time.time()))
        repeat -= 1

    print "\r\n", name, " is releaseing the lock"
    tLock.release()
    print "\r\nTimer: ", name, " Completed"

def Main():
    t1 = threading.Thread(target=timer, args=("Timer1", 2, 5))
    t2 = threading.Thread(target=timer, args=("Timer2", 3, 5))
    t3 = threading.Thread(target=timer, args=("Timer3", 4, 5))
    t4 = threading.Thread(target=timer, args=("Timer4", 5, 5))
    t5 = threading.Thread(target=timer, args=("Timer5", 0.1, 5))

    t1.start()
    t2.start()
    t3.start()
    t4.start()
    t5.start()

    print "\r\nMain Complete"

if __name__ == "__main__":
    Main()

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。