我认为@sega_sai建议的方法更好。但它确实需要一个代码示例，所以如下:

import multiprocessing
from os import getpid

def worker(procnum):
    print('I am number %d in process %d' % (procnum, getpid()))
    return getpid()

if __name__ == '__main__':
    pool = multiprocessing.Pool(processes = 3)
    print(pool.map(worker, range(5)))

它将打印返回值:

I am number 0 in process 19139
I am number 1 in process 19138
I am number 2 in process 19140
I am number 3 in process 19139
I am number 4 in process 19140
[19139, 19138, 19140, 19139, 19140]

如果你熟悉map (Python 2内置的)，这应该不是太有挑战性。否则，请查看sega_Sai的链接。

注意，只需要很少的代码。(还要注意如何重用流程)。

您可以使用内置的exit来设置进程的退出代码。它可以从进程的exitcode属性中获得:

import multiprocessing

def worker(procnum):
    print str(procnum) + ' represent!'
    exit(procnum)

if __name__ == '__main__':
    jobs = []
    for i in range(5):
        p = multiprocessing.Process(target=worker, args=(i,))
        jobs.append(p)
        p.start()

    result = []
    for proc in jobs:
        proc.join()
        result.append(proc.exitcode)
    print result

输出:

0 represent!
1 represent!
2 represent!
3 represent!
4 represent!
[0, 1, 2, 3, 4]

我想我应该简化上面复制的最简单的例子，在Py3.6上为我工作。最简单的是多处理。池:

import multiprocessing
import time

def worker(x):
    time.sleep(1)
    return x

pool = multiprocessing.Pool()
print(pool.map(worker, range(10)))

您可以设置池中的进程数，例如，pool (processes=5)。但是，它默认为CPU计数，因此对于CPU受限的任务，请将其保留为空。(I/ o绑定的任务通常适合线程，因为线程大部分都在等待，所以可以共享一个CPU内核。)Pool还应用了分块优化。

(注意，工作方法不能嵌套在方法中。我最初在调用池的方法中定义了工作方法。map，以保持它完全自包含，但随后进程无法导入它，并抛出“AttributeError: Can't pickle local object outer_method..inner_method”。更多的在这里。它可以在类内部。)

(欣赏原文问题指定打印’代表!'而不是time.sleep()，但没有它，我认为一些代码是并发运行时，它不是。)

Py3的ProcessPoolExecutor也是两行(。Map返回一个生成器，所以你需要list()):

from concurrent.futures import ProcessPoolExecutor
with ProcessPoolExecutor() as executor:
    print(list(executor.map(worker, range(10))))

使用普通流程:

import multiprocessing
import time

def worker(x, queue):
    time.sleep(1)
    queue.put(x)

queue = multiprocessing.SimpleQueue()
tasks = range(10)

for task in tasks:
    multiprocessing.Process(target=worker, args=(task, queue,)).start()

for _ in tasks:
    print(queue.get())

如果您所需要的只是放置和获取，请使用SimpleQueue。在第二个循环进入阻塞队列之前，第一个循环启动所有进程。得到调用。我认为没有任何理由也调用p.join()。

我修改了vartec的答案，因为我需要从函数中获得错误代码。(由于vertec ! !这是一个很棒的技巧)

这也可以通过经理来实现。列表，但我认为最好是在字典中，并在其中存储一个列表。这样，我们就可以保留函数和结果，因为我们不能确定列表将被填充的顺序。

from multiprocessing import Process
import time
import datetime
import multiprocessing


def func1(fn, m_list):
    print 'func1: starting'
    time.sleep(1)
    m_list[fn] = "this is the first function"
    print 'func1: finishing'
    # return "func1"  # no need for return since Multiprocess doesnt return it =(

def func2(fn, m_list):
    print 'func2: starting'
    time.sleep(3)
    m_list[fn] = "this is function 2"
    print 'func2: finishing'
    # return "func2"

def func3(fn, m_list):
    print 'func3: starting'
    time.sleep(9)
    # if fail wont join the rest because it never populate the dict
    # or do a try/except to get something in return.
    raise ValueError("failed here")
    # if we want to get the error in the manager dict we can catch the error
    try:
        raise ValueError("failed here")
        m_list[fn] = "this is third"
    except:
        m_list[fn] = "this is third and it fail horrible"
        # print 'func3: finishing'
        # return "func3"


def runInParallel(*fns):  # * is to accept any input in list
    start_time = datetime.datetime.now()
    proc = []
    manager = multiprocessing.Manager()
    m_list = manager.dict()
    for fn in fns:
        # print fn
        # print dir(fn)
        p = Process(target=fn, name=fn.func_name, args=(fn, m_list))
        p.start()
        proc.append(p)
    for p in proc:
        p.join()  # 5 is the time out

    print datetime.datetime.now() - start_time
    return m_list, proc

if __name__ == '__main__':
    manager, proc = runInParallel(func1, func2, func3)
    # print dir(proc[0])
    # print proc[0]._name
    # print proc[0].name
    # print proc[0].exitcode

    # here you can check what did fail
    for i in proc:
        print i.name, i.exitcode  # name was set up in the Process line 53

    # here will only show the function that worked and where able to populate the 
    # manager dict
    for i, j in manager.items():
        print dir(i)  # things you can do to the function
        print i, j

如果你正在使用Python 3，你可以使用concurrent.futures.ProcessPoolExecutor作为一个方便的抽象:

from concurrent.futures import ProcessPoolExecutor

def worker(procnum):
    '''worker function'''
    print(str(procnum) + ' represent!')
    return procnum


if __name__ == '__main__':
    with ProcessPoolExecutor() as executor:
        print(list(executor.map(worker, range(5))))

输出:

0 represent!
1 represent!
2 represent!
3 represent!
4 represent!
[0, 1, 2, 3, 4]

你可以使用ProcessPoolExecutor从函数中获取一个返回值，如下所示:

from concurrent.futures import ProcessPoolExecutor

def test(num1, num2):
    return num1 + num2

with ProcessPoolExecutor() as executor:
    feature = executor.submit(test, 2, 3)
    print(feature.result()) # 5