273

在下面的示例代码中，我想恢复 function 的返回值worker。我该怎么做呢？这个值存储在哪里？

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)>]

我似乎无法在存储在jobs.

4

12 回答 12

280

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())
于 2012-05-02T13:55:08.283 回答
82

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

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（Python 2 内置），这应该不会太具有挑战性。否则，请查看sega_Sai 的链接。

请注意需要多少代码。（还要注意如何重用流程）。

于 2015-03-01T21:09:38.773 回答
47

对于其他寻求如何从Processusing中获取价值的人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()
    p.join()
    print(queue.get())  # Prints {"foo": True}

请注意，在 Windows 或 Jupyter Notebook 中，multithreading您必须将其保存为文件并执行该文件。如果您在命令提示符下执行此操作，您将看到如下错误：

 AttributeError: Can't get attribute 'worker' on <module '__main__' (built-in)>
于 2016-06-09T21:41:44.947 回答
41

出于某种原因，我找不到如何在Queue任何地方执行此操作的一般示例（即使 Python 的文档示例也不会产生多个进程），所以这是我在尝试 10 次后得到的工作：

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队列，可用于存储子进程的返回值。所以你必须将队列传递给每个进程。这里不太明显的是，您必须在esget()之前从队列中退出，否则队列会填满并阻塞所有内容。joinProcess

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)
于 2017-08-23T02:47:23.907 回答
36

此示例显示如何使用multiprocessing.Pipe实例列表从任意数量的进程返回字符串：

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!']

此解决方案使用的资源少于使用的multiprocessing.Queue

  • 管道
  • 至少一个锁
  • 缓冲区
  • 一根线

或使用multiprocessing.SimpleQueue

  • 管道
  • 至少一个锁

查看每种类型的来源非常有启发性。

于 2016-06-09T23:56:07.450 回答
16

看来您应该改用multiprocessing.Pool类并使用方法 .apply() .apply_async(), map()

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

于 2012-05-02T14:05:58.337 回答
16

您可以使用exit内置设置进程的退出代码。可以从exitcode进程的属性中获取：

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]
于 2016-04-06T17:15:28.007 回答
11

pebble包有一个很好的抽象利用multiprocessing.Pipe，这使得这非常简单：

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

于 2019-06-19T21:53:38.620 回答
8

以为我会简化从上面复制的最简单的示例，在 Py3.6 上为我工作。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，例如，设置池中的进程数Pool(processes=5)。但是它默认为 CPU 计数，因此对于 CPU 密集型任务将其留空。（无论如何，I/O-bound 任务通常都适合线程，因为线程大部分都在等待，因此可以共享一个 CPU 内核。）Pool也适用于分块优化。

（请注意，工作方法不能嵌套在方法中。我最初在调用 的方法中定义了工作方法pool.map，以使其完全独立，但随后进程无法导入它，并抛出“AttributeError ：不能腌制本地对象outer_method..inner_method”。更多在这里。它可以在一个类中。）

（欣赏原始问题指定打印'represent!'而不是time.sleep()，但没有它，我认为某些代码不是同时运行的。）

Py3ProcessPoolExecutor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如果您只需要put和 ，请使用get。第一个循环启动所有进程，然后第二个循环进行阻塞queue.get调用。我也不认为有任何理由打电话p.join()。

于 2019-12-26T09:41:45.353 回答
2

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

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]
于 2019-08-27T09:51:46.520 回答
2

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]
于 2018-12-17T22:58:09.280 回答
0

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

这也可以用 a 来完成，manager.list但我认为最好将它放在一个 dict 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，因为我们无法确定列表的填充顺序。

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
于 2017-07-11T21:43:31.140 回答