带有RPYC的多处理Python“ ValueError：禁用酸洗”

Question

我试图在rpyc服务中使用多处理程序包，但是ValueError: pickling is disabled当我尝试从客户端调用公开函数时会得到提示。我知道该multiprocesing程序包使用酸洗在进程之间传递信息，并且不允许酸洗，rpyc因为这是不安全的协议。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)

Answer 1

您可以在协议配置中启用酸洗。配置存储为字典，您可以修改默认值并将其传递给服务器 ( protocol_config= ) 和客户端 ( config =)。您还需要定义在客户端和服务器端并行化的函数。所以这里是完整的代码server.py：

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()

代码client.py是：

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)