如何恢复传递给multiprocessing.Process的函数的返回值？

在下面的示例代码中，我想恢复该函数的返回值 worker。我该怎么做呢？此值存储在哪里？

示例代码：

import multiprocessing

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


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

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

输出：

0 represent!
1 represent!
2 represent!
3 represent!
4 represent!
[<Process(Process-1, stopped)>, <Process(Process-2, stopped)>, <Process(Process-3, stopped)>, <Process(Process-4, stopped)>, <Process(Process-5, stopped)>]

我似乎在中存储的对象中找不到相关的属性jobs。

使用共享变量进行通信。例如这样：

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()

我认为@sega_sai建议的方法更好。但这确实需要一个代码示例，因此请按以下步骤进行：

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)))

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]

如果您熟悉map（内置Python 2），这应该不会太有挑战性。否则，请查看sega_Sai的链接。

注意需要很少的代码。（还请注意如何重用流程）。

本示例说明如何使用multiprocessing.Pipe实例列表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!']

该解决方案使用较少的资源比multiprocessing.Queue其用途

• 管道
• 至少一把锁
• 缓冲区
• 一个线程

或multiprocessing.SimpleQueue其用途

• 管道
• 至少一把锁

查看每种类型的源代码非常有启发性。

出于某种原因，我找不到在Queue任何地方执行此操作的一般示例（即使Python的doc示例也不会生成多个进程），所以这是经过10次尝试后我才开始工作的内容：

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是一个线程安全的阻塞队列，可用于存储子进程的返回值。因此，您必须将队列传递给每个进程。一些不太明显的是，你们必须get()从队列你之前join的ProcessES否则队列已满，并且块一切。

针对那些面向对象的人的更新（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)

对于任何人谁是寻求如何从获取值Process使用Queue：

import multiprocessing

ret = {'foo': False}

def worker(queue):
    ret = queue.get()
    ret['foo'] = True
    queue.put(ret)

if __name__ == '__main__':
    queue = multiprocessing.Queue()
    queue.put(ret)
    p = multiprocessing.Process(target=worker, args=(queue,))
    p.start()
    print queue.get()  # Prints {"foo": True}
    p.join()

似乎应该改用multiprocessing.Pool类，并使用.apply（）.apply_async（），map（）方法

http://docs.python.org/library/multiprocessing.html?highlight=pool#multiprocessing.pool.AsyncResult

您可以使用exit内置功能来设置流程的退出代码。可以从exitcode流程的属性中获取：

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]

该石子包有一个很好的抽象杠杆multiprocessing.Pipe，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

以为我会简化上面复制的最简单的示例，为我在Py3.6上工作。最简单的是multiprocessing.Pool：

import multiprocessing
import time

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

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

您可以使用设置池中的进程数Pool(processes=5)。但是，它默认为CPU计数，因此对于与CPU绑定的任务，请将其留空。（I / O绑定的任务无论如何通常都适合线程，因为线程大多在等待，因此可以共享一个CPU内核。）Pool还应用了分块优化。

（请注意，worker方法不能嵌套在一个方法中。我最初在该方法内定义了我的worker方法，该方法对进行调用pool.map，以使其全部独立，但随后这些进程无法导入它，并抛出了“ AttributeError ：无法腌制本地对象“ outer_method..inner_method”。在此处更多。它可以在类内部。）

（赞赏最初指定打印'represent!'而不是的问题time.sleep()，但没有它，我以为某些代码不是在同时运行的。）

Py3的 ProcessPoolExecutor也是两行（.map返回一个生成器，因此您需要list()）：

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

使用纯ProcessES：

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())

SimpleQueue如果您需要的是put和，请使用get。第一个循环开始所有进程，第二个循环进行阻塞之前queue.get调用。我认为也没有任何理由打电话p.join()。

一个简单的解决方案：

import multiprocessing

output=[]
data = range(0,10)

def f(x):
    return x**2

def handler():
    p = multiprocessing.Pool(64)
    r=p.map(f, data)
    return r

if __name__ == '__main__':
    output.append(handler())

print(output[0])

输出：

[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]

如果您使用的是Python 3，则可以concurrent.futures.ProcessPoolExecutor用作方便的抽象：

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]

由于需要从函数中获取错误代码，因此我对vartec的答案做了一些修改。（感谢vertec !!!这是一个很棒的技巧）

也可以使用a来完成此操作，manager.list但我认为最好将它放在字典中并在其中存储列表。这样，由于我们无法确定列表的填充顺序，因此我们保留了函数和结果。

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