只需注意：线程不需要队列。

这是我可以想象的最简单的例子，它显示了10个并发运行的进程。

import threading
from random import randint
from time import sleep


def print_number(number):

    # Sleeps a random 1 to 10 seconds
    rand_int_var = randint(1, 10)
    sleep(rand_int_var)
    print "Thread " + str(number) + " slept for " + str(rand_int_var) + " seconds"

thread_list = []

for i in range(1, 10):

    # Instantiates the thread
    # (i) does not make a sequence, so (i,)
    t = threading.Thread(target=print_number, args=(i,))
    # Sticks the thread in a list so that it remains accessible
    thread_list.append(t)

# Starts threads
for thread in thread_list:
    thread.start()

# This blocks the calling thread until the thread whose join() method is called is terminated.
# From http://docs.python.org/2/library/threading.html#thread-objects
for thread in thread_list:
    thread.join()

# Demonstrates that the main process waited for threads to complete
print "Done"

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。

以前的解决方案都没有在我的GNU/Linux服务器上使用多个内核（我没有管理员权限）。他们只是在一个核心上跑步。

我使用较低级别的os.fork接口来派生多个进程。这是对我有用的代码：

from os import fork

values = ['different', 'values', 'for', 'threads']

for i in range(len(values)):
    p = fork()
    if p == 0:
        my_function(values[i])
        break

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

import threading
import requests

def send():

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

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

我想提供一个简单的例子，以及我在自己解决这个问题时发现有用的解释。

在这个答案中，您将找到一些关于Python的GIL（全局解释器锁）的信息，以及一个使用multiprocessing.dummy编写的简单日常示例，以及一些简单的基准测试。

全局解释器锁（GIL）

Python不允许真正意义上的多线程。它有一个多线程包，但是如果你想多线程来加快你的代码，那么使用它通常不是一个好主意。

Python有一个称为全局解释器锁（GIL）的构造。GIL确保在任何时候只能执行一个“线程”。一个线程获取GIL，做一些工作，然后将GIL传递给下一个线程。

这种情况发生得很快，因此在人眼看来，您的线程似乎是并行执行的，但它们实际上只是轮流使用相同的CPU内核。

所有这些GIL传递都增加了执行开销。这意味着如果你想让你的代码运行得更快，那么使用线程打包通常不是个好主意。

使用Python的线程包是有原因的。如果你想同时运行一些事情，而效率不是一个问题，那就很好，也很方便。或者，如果您运行的代码需要等待一些东西（比如一些I/O），那么这可能很有意义。但是线程库不允许您使用额外的CPU内核。

多线程可以外包给操作系统（通过执行多线程处理），以及一些调用Python代码的外部应用程序（例如，Spark或Hadoop），或者Python代码调用的一些代码（例如：您可以让Python代码调用一个C函数来完成昂贵的多线程任务）。

为什么这很重要

因为很多人在了解GIL是什么之前，会花很多时间在他们的Python多线程代码中寻找瓶颈。

一旦这些信息清楚，下面是我的代码：

#!/bin/python
from multiprocessing.dummy import Pool
from subprocess import PIPE,Popen
import time
import os

# In the variable pool_size we define the "parallelness".
# For CPU-bound tasks, it doesn't make sense to create more Pool processes
# than you have cores to run them on.
#
# On the other hand, if you are using I/O-bound tasks, it may make sense
# to create a quite a few more Pool processes than cores, since the processes
# will probably spend most their time blocked (waiting for I/O to complete).
pool_size = 8

def do_ping(ip):
    if os.name == 'nt':
        print ("Using Windows Ping to " + ip)
        proc = Popen(['ping', ip], stdout=PIPE)
        return proc.communicate()[0]
    else:
        print ("Using Linux / Unix Ping to " + ip)
        proc = Popen(['ping', ip, '-c', '4'], stdout=PIPE)
        return proc.communicate()[0]


os.system('cls' if os.name=='nt' else 'clear')
print ("Running using threads\n")
start_time = time.time()
pool = Pool(pool_size)
website_names = ["www.google.com","www.facebook.com","www.pinterest.com","www.microsoft.com"]
result = {}
for website_name in website_names:
    result[website_name] = pool.apply_async(do_ping, args=(website_name,))
pool.close()
pool.join()
print ("\n--- Execution took {} seconds ---".format((time.time() - start_time)))

# Now we do the same without threading, just to compare time
print ("\nRunning NOT using threads\n")
start_time = time.time()
for website_name in website_names:
    do_ping(website_name)
print ("\n--- Execution took {} seconds ---".format((time.time() - start_time)))

# Here's one way to print the final output from the threads
output = {}
for key, value in result.items():
    output[key] = value.get()
print ("\nOutput aggregated in a Dictionary:")
print (output)
print ("\n")

print ("\nPretty printed output: ")
for key, value in output.items():
    print (key + "\n")
    print (value)