这里有一个简单的示例：您需要尝试一些替代URL，并返回第一个URL的内容以进行响应。

import Queue
import threading
import urllib2

# Called by each thread
def get_url(q, url):
    q.put(urllib2.urlopen(url).read())

theurls = ["http://google.com", "http://yahoo.com"]

q = Queue.Queue()

for u in theurls:
    t = threading.Thread(target=get_url, args = (q,u))
    t.daemon = True
    t.start()

s = q.get()
print s

在这种情况下，线程被用作一种简单的优化：每个子线程都在等待URL解析和响应，以将其内容放入队列；每个线程都是一个守护进程（如果主线程结束，则不会保持进程运行——这比不结束更常见）；主线程启动所有子线程，在队列中执行get以等待其中一个线程完成put，然后发出结果并终止（这将删除所有可能仍在运行的子线程，因为它们是守护进程线程）。

Python中线程的正确使用总是与I/O操作相关（因为CPython无论如何都不使用多个内核来运行CPU绑定的任务，线程的唯一原因是在等待一些I/O时不会阻塞进程）。顺便说一句，队列几乎总是将工作分配给线程和/或收集工作结果的最佳方式，而且它们本质上是线程安全的，因此它们使您不用担心锁、条件、事件、信号量和其他线程间协调/通信概念。

大多数文档和教程都使用Python的“线程和队列”模块，对于初学者来说，它们可能会让人不知所措。

也许可以考虑Python 3的concurrent.futures.ThreadPoolExecutor模块。

结合子句和列表理解，这可能是一个真正的魅力。

from concurrent.futures import ThreadPoolExecutor, as_completed

def get_url(url):
    # Your actual program here. Using threading.Lock() if necessary
    return ""

# List of URLs to fetch
urls = ["url1", "url2"]

with ThreadPoolExecutor(max_workers = 5) as executor:

    # Create threads
    futures = {executor.submit(get_url, url) for url in urls}

    # as_completed() gives you the threads once finished
    for f in as_completed(futures):
        # Get the results
        rs = f.result()

我在这里看到了很多没有执行实际工作的示例，它们大多是CPU限制的。这里是一个CPU绑定任务的示例，它计算1000万到1005万之间的所有素数。我在这里使用了所有四种方法：

import math
import timeit
import threading
import multiprocessing
from concurrent.futures import ThreadPoolExecutor, ProcessPoolExecutor


def time_stuff(fn):
    """
    Measure time of execution of a function
    """
    def wrapper(*args, **kwargs):
        t0 = timeit.default_timer()
        fn(*args, **kwargs)
        t1 = timeit.default_timer()
        print("{} seconds".format(t1 - t0))
    return wrapper

def find_primes_in(nmin, nmax):
    """
    Compute a list of prime numbers between the given minimum and maximum arguments
    """
    primes = []

    # Loop from minimum to maximum
    for current in range(nmin, nmax + 1):

        # Take the square root of the current number
        sqrt_n = int(math.sqrt(current))
        found = False

        # Check if the any number from 2 to the square root + 1 divides the current numnber under consideration
        for number in range(2, sqrt_n + 1):

            # If divisible we have found a factor, hence this is not a prime number, lets move to the next one
            if current % number == 0:
                found = True
                break

        # If not divisible, add this number to the list of primes that we have found so far
        if not found:
            primes.append(current)

    # I am merely printing the length of the array containing all the primes, but feel free to do what you want
    print(len(primes))

@time_stuff
def sequential_prime_finder(nmin, nmax):
    """
    Use the main process and main thread to compute everything in this case
    """
    find_primes_in(nmin, nmax)

@time_stuff
def threading_prime_finder(nmin, nmax):
    """
    If the minimum is 1000 and the maximum is 2000 and we have four workers,
    1000 - 1250 to worker 1
    1250 - 1500 to worker 2
    1500 - 1750 to worker 3
    1750 - 2000 to worker 4
    so let’s split the minimum and maximum values according to the number of workers
    """
    nrange = nmax - nmin
    threads = []
    for i in range(8):
        start = int(nmin + i * nrange/8)
        end = int(nmin + (i + 1) * nrange/8)

        # Start the thread with the minimum and maximum split up to compute
        # Parallel computation will not work here due to the GIL since this is a CPU-bound task
        t = threading.Thread(target = find_primes_in, args = (start, end))
        threads.append(t)
        t.start()

    # Don’t forget to wait for the threads to finish
    for t in threads:
        t.join()

@time_stuff
def processing_prime_finder(nmin, nmax):
    """
    Split the minimum, maximum interval similar to the threading method above, but use processes this time
    """
    nrange = nmax - nmin
    processes = []
    for i in range(8):
        start = int(nmin + i * nrange/8)
        end = int(nmin + (i + 1) * nrange/8)
        p = multiprocessing.Process(target = find_primes_in, args = (start, end))
        processes.append(p)
        p.start()

    for p in processes:
        p.join()

@time_stuff
def thread_executor_prime_finder(nmin, nmax):
    """
    Split the min max interval similar to the threading method, but use a thread pool executor this time.
    This method is slightly faster than using pure threading as the pools manage threads more efficiently.
    This method is still slow due to the GIL limitations since we are doing a CPU-bound task.
    """
    nrange = nmax - nmin
    with ThreadPoolExecutor(max_workers = 8) as e:
        for i in range(8):
            start = int(nmin + i * nrange/8)
            end = int(nmin + (i + 1) * nrange/8)
            e.submit(find_primes_in, start, end)

@time_stuff
def process_executor_prime_finder(nmin, nmax):
    """
    Split the min max interval similar to the threading method, but use the process pool executor.
    This is the fastest method recorded so far as it manages process efficiently + overcomes GIL limitations.
    RECOMMENDED METHOD FOR CPU-BOUND TASKS
    """
    nrange = nmax - nmin
    with ProcessPoolExecutor(max_workers = 8) as e:
        for i in range(8):
            start = int(nmin + i * nrange/8)
            end = int(nmin + (i + 1) * nrange/8)
            e.submit(find_primes_in, start, end)

def main():
    nmin = int(1e7)
    nmax = int(1.05e7)
    print("Sequential Prime Finder Starting")
    sequential_prime_finder(nmin, nmax)
    print("Threading Prime Finder Starting")
    threading_prime_finder(nmin, nmax)
    print("Processing Prime Finder Starting")
    processing_prime_finder(nmin, nmax)
    print("Thread Executor Prime Finder Starting")
    thread_executor_prime_finder(nmin, nmax)
    print("Process Executor Finder Starting")
    process_executor_prime_finder(nmin, nmax)
if __name__ == "__main__":
    main()

以下是我的Mac OS X四核计算机的结果

Sequential Prime Finder Starting
9.708213827005238 seconds
Threading Prime Finder Starting
9.81836523200036 seconds
Processing Prime Finder Starting
3.2467174359990167 seconds
Thread Executor Prime Finder Starting
10.228896902000997 seconds
Process Executor Finder Starting
2.656402041000547 seconds

Python 3具有启动并行任务的功能。这使我们的工作更容易。

它有线程池和进程池。

以下内容提供了一个见解：

ThreadPoolExecutor示例（源代码）

import concurrent.futures
import urllib.request

URLS = ['http://www.foxnews.com/',
        'http://www.cnn.com/',
        'http://europe.wsj.com/',
        'http://www.bbc.co.uk/',
        'http://some-made-up-domain.com/']

# Retrieve a single page and report the URL and contents
def load_url(url, timeout):
    with urllib.request.urlopen(url, timeout=timeout) as conn:
        return conn.read()

# We can use a with statement to ensure threads are cleaned up promptly
with concurrent.futures.ThreadPoolExecutor(max_workers=5) as executor:
    # Start the load operations and mark each future with its URL
    future_to_url = {executor.submit(load_url, url, 60): url for url in URLS}
    for future in concurrent.futures.as_completed(future_to_url):
        url = future_to_url[future]
        try:
            data = future.result()
        except Exception as exc:
            print('%r generated an exception: %s' % (url, exc))
        else:
            print('%r page is %d bytes' % (url, len(data)))

ProcessPoolExecutor（源）

import concurrent.futures
import math

PRIMES = [
    112272535095293,
    112582705942171,
    112272535095293,
    115280095190773,
    115797848077099,
    1099726899285419]

def is_prime(n):
    if n % 2 == 0:
        return False

    sqrt_n = int(math.floor(math.sqrt(n)))
    for i in range(3, sqrt_n + 1, 2):
        if n % i == 0:
            return False
    return True

def main():
    with concurrent.futures.ProcessPoolExecutor() as executor:
        for number, prime in zip(PRIMES, executor.map(is_prime, PRIMES)):
            print('%d is prime: %s' % (number, prime))

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。

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