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

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。

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()。

使用共享变量进行通信。比如这样:

import multiprocessing


def worker(procnum, return_dict):
    """worker function"""
    print(str(procnum) + " represent!")
    return_dict[procnum] = procnum


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

    for proc in jobs:
        proc.join()
    print(return_dict.values())

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]

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)