我怎样才能恢复传递给multiprocessing.Process函数的返回值？我怎样才能恢复传递给

2019-05-13 19:07发布

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在下面的示例代码，我想恢复函数的返回值worker 。我怎么能去这样做？当存储在此值？

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)>]

我似乎无法找到存储在对象的相关属性jobs 。

在此先感谢，BLZ

Answer 1:

使用共享变量进行通信。例如像这样：

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()

Answer 2:

我认为@sega_sai建议的方法是最好的一个。但它确实需要一个代码示例，所以这里有云：

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))

这将打印的返回值：

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的2个内置），这应该不会太困难。否则，看看sega_Sai的链接。

注意代码是如何一点是必要的。（还要注意如何处理被再次使用）。

Answer 3:

这个例子说明如何使用列表multiprocessing.Pipe实例从进程任意数量的返回字符串：

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它使用

管道
至少一个锁
缓冲区
一个线程

或multiprocessing.SimpleQueue其使用

管道
至少一个锁

这是非常有启发看看源每种类型的。

Answer 4:

看来你应该使用multiprocessing.Pool类，而不是和。适用使用方法（）.apply_async（），地图（）

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

Answer 5:

您可以使用exit内置设置进程的退出代码。它可以从获得exitcode的方法的属性：

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]

Answer 6:

出于某种原因，我无法找到如何做到这一点与一般的例子Queue的任何地方（甚至Python的文档例子不产卵多个进程），所以这里我得到了像10次尝试后的工作是什么：

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是阻塞，线程安全的队列中，你可以用它来返回值从子进程存储。所以，你必须排队传递给每一个过程。一些不太明显的是，你们必须get()你以前从队列中join的Process ES否则队列已满，并且块一切。

更新对于那些谁是面向对象（在Python 3.4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)

Answer 7:

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()

Answer 8:

我修改vartec的回答有点，因为我需要得到从函数的错误代码。（感谢VERTEC！它的真棒招）

这也可以用做manager.list但我认为最好是有它在一个字典，并在其中存储的列表。这样一来，这样我们保持功能和结果，因为我们不能确定，其中，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

Answer 9:

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：