AttributeError：无法使用多处理队列来腌制本地对象的计算..function1

Question

我有以下使用调度程序和多处理模块的代码：

def computation():
    def function1(q):
        while True:
            daydate = datetime.now()
            number = random.randrange(1, 215)
            print('Sent to function2: ({}, {})'.format(daydate, number))
            q.put((daydate, number))
            time.sleep(2)

    def function2(q):
        while True:
            date, number = q.get()
            print("Recevied values from function1: ({}, {})".format(date, number))
            time.sleep(2)

    if __name__ == "__main__":
        q = Queue()
        a = Process(target=function1, args=(q,))
        a.start()
        b = Process(target=function2, args=(q,))
        b.start()
        a.join()
        b.join()

schedule.every().monday.at("08:45").do(computation)
schedule.every().tuesday.at("08:45").do(computation)

while True:
    schedule.run_pending()
    time.sleep(1)

但是，在执行代码时会出现以下错误：

AttributeError：无法腌制本地对象的计算。

和：

OSError：[WinError 87]参数不正确

如何解决这个问题？我已经尝试通过在文档中声明的模块顶层定义函数来解决此问题（https://docs.python.org/2/library/pickle.html#what-c​​an-
be-pickled-和未腌制
），但仍会给出相同的错误。

Answer 1

嵌套函数不是在顶级定义的函数，因此这就是您收到错误的原因。您需要重新定位计算的定义function1以及function2计算之外。

编写方式时，您的流程将立即启动，而不是在安排它们运行的​​日期开始。那可能会达到您的预期目的：

import os
import time
import random
from multiprocessing import Process, Queue
from threading import Thread
from datetime import datetime
import schedule


def function1(q):
    while True:
        daydate = datetime.now()
        number = random.randrange(1, 215)
        fmt = '(pid: {}) Sent to function2: ({}, {})'
        print(fmt.format(os.getpid(), daydate, number))
        q.put((daydate, number))
        time.sleep(2)


def function2(q):
    while True:
        date, number = q.get()
        fmt = "(pid: {}) Received values from function1: ({}, {})"
        print(fmt.format(os.getpid(), date, number))
        # time.sleep(2) no need to sleep here because q.get will block until
        # new items are available


def computation():
    q = Queue()
    a = Process(target=function1, args=(q,))
    a.start()
    b = Process(target=function2, args=(q,))
    b.start()
    a.join()
    b.join()


if __name__ == "__main__":

    # We are spawning new threads as a launching platform for
    # computation. Without it, the next job couldn't start before the last
    # one has finished. If your jobs always end before the next one should 
    # start, you don't need this construct and you can just pass 
    # ...do(computation)
    schedule.every().friday.at("01:02").do(
        Thread(target=computation).start
    )
    schedule.every().friday.at("01:03").do(
        Thread(target=computation).start
    )

    while True:
        schedule.run_pending()
        time.sleep(1)

就像现在一样，一旦启动，您的进程将永远运行。如果这不是您想要的，则必须考虑实现一些停止条件。