出于某种原因，我找不到一个通用的例子，如何做到这一点与队列在任何地方(甚至Python的文档示例不会产生多个进程)，所以这是我得到的工作后，10次尝试:

from multiprocessing import Process, Queue

def add_helper(queue, arg1, arg2): # the func called in child processes
    ret = arg1 + arg2
    queue.put(ret)

def multi_add(): # spawns child processes
    q = Queue()
    processes = []
    rets = []
    for _ in range(0, 100):
        p = Process(target=add_helper, args=(q, 1, 2))
        processes.append(p)
        p.start()
    for p in processes:
        ret = q.get() # will block
        rets.append(ret)
    for p in processes:
        p.join()
    return rets

Queue是一个阻塞的、线程安全的队列，您可以使用它来存储来自子进程的返回值。因此，您必须将队列传递给每个进程。这里不太明显的一点是，您必须在加入进程之前从队列中获取()，否则队列将被填满并阻塞所有内容。

面向对象的更新(在Python 3.4中测试):

from multiprocessing import Process, Queue

class Multiprocessor():

    def __init__(self):
        self.processes = []
        self.queue = Queue()

    @staticmethod
    def _wrapper(func, queue, args, kwargs):
        ret = func(*args, **kwargs)
        queue.put(ret)

    def run(self, func, *args, **kwargs):
        args2 = [func, self.queue, args, kwargs]
        p = Process(target=self._wrapper, args=args2)
        self.processes.append(p)
        p.start()

    def wait(self):
        rets = []
        for p in self.processes:
            ret = self.queue.get()
            rets.append(ret)
        for p in self.processes:
            p.join()
        return rets

# tester
if __name__ == "__main__":
    mp = Multiprocessor()
    num_proc = 64
    for _ in range(num_proc): # queue up multiple tasks running `sum`
        mp.run(sum, [1, 2, 3, 4, 5])
    ret = mp.wait() # get all results
    print(ret)
    assert len(ret) == num_proc and all(r == 15 for r in ret)

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

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)。

pebble包有一个很好的利用多处理的抽象。管道，这使得这个非常简单:

from pebble import concurrent

@concurrent.process
def function(arg, kwarg=0):
    return arg + kwarg

future = function(1, kwarg=1)

print(future.result())

示例来自:https://pythonhosted.org/Pebble/#concurrent-decorators

这个例子展示了如何使用multiprocessing的列表。管道实例从任意数量的进程中返回字符串:

import multiprocessing

def worker(procnum, send_end):
    '''worker function'''
    result = str(procnum) + ' represent!'
    print result
    send_end.send(result)

def main():
    jobs = []
    pipe_list = []
    for i in range(5):
        recv_end, send_end = multiprocessing.Pipe(False)
        p = multiprocessing.Process(target=worker, args=(i, send_end))
        jobs.append(p)
        pipe_list.append(recv_end)
        p.start()

    for proc in jobs:
        proc.join()
    result_list = [x.recv() for x in pipe_list]
    print result_list

if __name__ == '__main__':
    main()

输出:

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

这种解决方案使用的资源比多进程少。使用

一个管道 至少一个锁 一个缓冲区 一个线程

或者多处理。SimpleQueue使用

一个管道 至少一个锁

查看这些类型的源代码是非常有指导意义的。