python - 带有RPYC "ValueError: pickling is disabled"的多处理Python

我正在尝试在rpyc服务中使用多处理程序包，但是当我尝试从客户端调用公开函数时会得到ValueError: pickling is disabled。我知道multiprocesing包使用酸洗在进程之间传递信息，并且rpyc不允许酸洗，因为它是不安全的协议(protocol)。因此，我不确定将多处理与rpyc一起使用的最佳方法(或者是否存在)。如何在rpyc服务中使用多重处理？这是服务器端代码:

import rpyc
from multiprocessing import Pool

class MyService(rpyc.Service):

    def exposed_RemotePool(self, function, arglist):

        pool = Pool(processes = 8)
        result = pool.map(function, arglist)
        pool.close()
        return result


if __name__ == "__main__":
    from rpyc.utils.server import ThreadedServer
    t = ThreadedServer(MyService, port = 18861)
    t.start()

这是产生错误的客户端代码:

import rpyc

def square(x):
    return x*x

c = rpyc.connect("localhost", 18861)
result = c.root.exposed_RemotePool(square, [1,2,3,4])
print(result)

最佳答案

您可以在协议(protocol)配置中启用酸洗。配置存储为字典，您可以修改default并将其传递到服务器(protocol_config =)和客户端(config =)。您还需要定义要在客户端和服务器端并行化的功能。因此，这是server.py的完整代码:

import rpyc
from multiprocessing import Pool
rpyc.core.protocol.DEFAULT_CONFIG['allow_pickle'] = True

def square(x):
    return x*x


class MyService(rpyc.Service):

    def exposed_RemotePool(self, function, arglist):

        pool = Pool(processes = 8)
        result = pool.map(function, arglist)
        pool.close()
        return result



if __name__ == "__main__":
    from rpyc.utils.server import ThreadedServer
    t = ThreadedServer(MyService, port = 18861, protocol_config = rpyc.core.protocol.DEFAULT_CONFIG)
    t.start()

对于client.py，代码为:

import rpyc

rpyc.core.protocol.DEFAULT_CONFIG['allow_pickle'] = True

def square(x):
    return x*x

c = rpyc.connect("localhost", port = 18861, config = rpyc.core.protocol.DEFAULT_CONFIG)
result = c.root.exposed_RemotePool(square, [1,2,3,4])
print(result)