看看这个;

http://docs.python.org/library/queue.html

这可能不是正确的方法，但我会这样做;

实际的代码;

from multiprocessing import Process, JoinableQueue as Queue 

class CustomWorker(Process):
    def __init__(self,workQueue, out1,out2,out3):
        Process.__init__(self)
        self.input=workQueue
        self.out1=out1
        self.out2=out2
        self.out3=out3
    def run(self):
            while True:
                try:
                    value = self.input.get()
                    #value modifier
                    temp1,temp2,temp3 = self.calc_stuff(value)
                    self.out1.put(temp1)
                    self.out2.put(temp2)
                    self.out3.put(temp3)
                    self.input.task_done()
                except Queue.Empty:
                    return
                   #Catch things better here
    def calc_stuff(self,param):
        out1 = param * 2
        out2 = param * 4
        out3 = param * 8
        return out1,out2,out3
def Main():
    inputQueue = Queue()
    for i in range(10):
        inputQueue.put(i)
    out1 = Queue()
    out2 = Queue()
    out3 = Queue()
    processes = []
    for x in range(2):
          p = CustomWorker(inputQueue,out1,out2,out3)
          p.daemon = True
          p.start()
          processes.append(p)
    inputQueue.join()
    while(not out1.empty()):
        print out1.get()
        print out2.get()
        print out3.get()
if __name__ == '__main__':
    Main()

希望这能有所帮助。

由于全局解释器锁(GIL)的存在，在CPython上使用多线程并不能为纯python代码提供更好的性能。我建议使用multiprocessing模块:

pool = multiprocessing.Pool(4)
out1, out2, out3 = zip(*pool.map(calc_stuff, range(0, 10 * offset, offset)))

注意，这在交互式解释器中不起作用。

为了避免GIL周围常见的FUD:在本例中使用线程没有任何优势。这里要使用进程，而不是线程，因为它们避免了一大堆问题。

为什么不用线程和一个互斥来保护一个全局列表呢?

import os
import re
import time
import sys
import thread

from threading import Thread

class thread_it(Thread):
    def __init__ (self,param):
        Thread.__init__(self)
        self.param = param
    def run(self):
        mutex.acquire()
        output.append(calc_stuff(self.param))
        mutex.release()   


threads = []
output = []
mutex = thread.allocate_lock()

for j in range(0, 10):
    current = thread_it(j * offset)
    threads.append(current)
    current.start()

for t in threads:
    t.join()

#here you have output list filled with data

请记住，您的速度将与最慢的线程一样快

看看这个;

http://docs.python.org/library/queue.html

这可能不是正确的方法，但我会这样做;

实际的代码;

from multiprocessing import Process, JoinableQueue as Queue 

class CustomWorker(Process):
    def __init__(self,workQueue, out1,out2,out3):
        Process.__init__(self)
        self.input=workQueue
        self.out1=out1
        self.out2=out2
        self.out3=out3
    def run(self):
            while True:
                try:
                    value = self.input.get()
                    #value modifier
                    temp1,temp2,temp3 = self.calc_stuff(value)
                    self.out1.put(temp1)
                    self.out2.put(temp2)
                    self.out3.put(temp3)
                    self.input.task_done()
                except Queue.Empty:
                    return
                   #Catch things better here
    def calc_stuff(self,param):
        out1 = param * 2
        out2 = param * 4
        out3 = param * 8
        return out1,out2,out3
def Main():
    inputQueue = Queue()
    for i in range(10):
        inputQueue.put(i)
    out1 = Queue()
    out2 = Queue()
    out3 = Queue()
    processes = []
    for x in range(2):
          p = CustomWorker(inputQueue,out1,out2,out3)
          p.daemon = True
          p.start()
          processes.append(p)
    inputQueue.join()
    while(not out1.empty()):
        print out1.get()
        print out2.get()
        print out3.get()
if __name__ == '__main__':
    Main()

希望这能有所帮助。

为了并行化一个简单的for循环，joblib为多处理的原始使用带来了很多价值。不仅是简短的语法，还包括在迭代非常快的时候(以消除开销)进行透明的迭代聚集，或者捕获子进程的回溯，以获得更好的错误报告。

免责声明:我是joblib的原作者。

并行化这段代码最简单的方法是什么?

使用concurrent.futures中的PoolExecutor。将原始代码与此代码并排比较。首先，最简洁的方法是使用executor.map:

...
with ProcessPoolExecutor() as executor:
    for out1, out2, out3 in executor.map(calc_stuff, parameters):
        ...

或者通过单独提交每个电话来分解:

...
with ThreadPoolExecutor() as executor:
    futures = []
    for parameter in parameters:
        futures.append(executor.submit(calc_stuff, parameter))

    for future in futures:
        out1, out2, out3 = future.result() # this will block
        ...

离开上下文表示执行程序释放资源

您可以使用线程或进程，并使用完全相同的接口。

一个工作示例

下面是工作示例代码，将演示的价值:

把它放在一个文件futuretest.py中:

from concurrent.futures import ProcessPoolExecutor, ThreadPoolExecutor
from time import time
from http.client import HTTPSConnection

def processor_intensive(arg):
    def fib(n): # recursive, processor intensive calculation (avoid n > 36)
        return fib(n-1) + fib(n-2) if n > 1 else n
    start = time()
    result = fib(arg)
    return time() - start, result

def io_bound(arg):
    start = time()
    con = HTTPSConnection(arg)
    con.request('GET', '/')
    result = con.getresponse().getcode()
    return time() - start, result

def manager(PoolExecutor, calc_stuff):
    if calc_stuff is io_bound:
        inputs = ('python.org', 'stackoverflow.com', 'stackexchange.com',
                  'noaa.gov', 'parler.com', 'aaronhall.dev')
    else:
        inputs = range(25, 32)
    timings, results = list(), list()
    start = time()
    with PoolExecutor() as executor:
        for timing, result in executor.map(calc_stuff, inputs):
            # put results into correct output list:
            timings.append(timing), results.append(result)
    finish = time()
    print(f'{calc_stuff.__name__}, {PoolExecutor.__name__}')
    print(f'wall time to execute: {finish-start}')
    print(f'total of timings for each call: {sum(timings)}')
    print(f'time saved by parallelizing: {sum(timings) - (finish-start)}')
    print(dict(zip(inputs, results)), end = '\n\n')

def main():
    for computation in (processor_intensive, io_bound):
        for pool_executor in (ProcessPoolExecutor, ThreadPoolExecutor):
            manager(pool_executor, calc_stuff=computation)

if __name__ == '__main__':
    main()

下面是python -m futuretest一次运行的输出:

processor_intensive, ProcessPoolExecutor
wall time to execute: 0.7326343059539795
total of timings for each call: 1.8033506870269775
time saved by parallelizing: 1.070716381072998
{25: 75025, 26: 121393, 27: 196418, 28: 317811, 29: 514229, 30: 832040, 31: 1346269}

processor_intensive, ThreadPoolExecutor
wall time to execute: 1.190223217010498
total of timings for each call: 3.3561410903930664
time saved by parallelizing: 2.1659178733825684
{25: 75025, 26: 121393, 27: 196418, 28: 317811, 29: 514229, 30: 832040, 31: 1346269}

io_bound, ProcessPoolExecutor
wall time to execute: 0.533886194229126
total of timings for each call: 1.2977914810180664
time saved by parallelizing: 0.7639052867889404
{'python.org': 301, 'stackoverflow.com': 200, 'stackexchange.com': 200, 'noaa.gov': 301, 'parler.com': 200, 'aaronhall.dev': 200}

io_bound, ThreadPoolExecutor
wall time to execute: 0.38941240310668945
total of timings for each call: 1.6049387454986572
time saved by parallelizing: 1.2155263423919678
{'python.org': 301, 'stackoverflow.com': 200, 'stackexchange.com': 200, 'noaa.gov': 301, 'parler.com': 200, 'aaronhall.dev': 200}

处理器密集型的分析

在Python中执行处理器密集型计算时，期望ProcessPoolExecutor比ThreadPoolExecutor性能更好。

由于全局解释器锁(又名GIL)的存在，线程不能使用多个处理器，因此每次计算的时间和壁时间(实际运行的时间)会更大。

IO-bound分析

另一方面，当执行IO绑定操作时，期望ThreadPoolExecutor比ProcessPoolExecutor性能更好。

Python的线程是真实的，OS，线程。操作系统可以让它们进入睡眠状态，并在信息到达时将它们重新唤醒。

最终的想法

我怀疑在Windows上多处理会更慢，因为Windows不支持分叉，所以每个新进程都要花时间启动。

您可以在多个进程中嵌套多个线程，但建议不要使用多个线程来派生多个进程。

如果在Python中面临一个繁重的处理问题，您可以简单地使用额外的进程来扩展—但不能使用线程。