python - 是否可以在 Python 的线程内生成进程？

Question

我正在编写一个生成进程并在特定条件下重新启动进程的程序。比如子进程不再给母进程发送数据，一段时间后，我想让母进程终止子进程，重新启动。我以为我可以使用线程从子进程接收数据并重新启动子进程，但它并不像我想的那样工作。

import numpy as np
import multiprocessing as mp
import threading
import time
from apscheduler.schedulers.background import BackgroundScheduler

pipe_in, pipe_out = mp.Pipe()

class Mother():
    def __init__(self):
        self.pipe_out = pipe_out

        self.proc = mp.Process(target = self.test_func, args=(pipe_in, ))
        self.proc.start()

        self.thread = threading.Thread(target=self.thread_reciever, args=(self.pipe_out, ))
        self.thread.start()

    def thread_reciever(self, pipe_out):
        while True:
            value = pipe_out.recv()

            print(value)
            if value == 5:
                self.proc.terminate()
                time.sleep(2)
                self.proc = mp.Process(target = self.test_func)
                self.proc.start()

    def test_func(self, pipe_in):
        for i in range(10):
            pipe_in.send(i)
            time.sleep(1)


if __name__ == '__main__':
    r = Mother()

它打印出这个错误。

D:\>d:\python36-32\python.exe temp06.py
    0
1
2
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5
Exception in thread Thread-1:
Traceback (most recent call last):
File "d:\python36-32\lib\threading.py", line 916, in _bootstrap_inner
    self.run()
File "d:\python36-32\lib\threading.py", line 864, in run
    self._target(*self._args, **self._kwargs)
File "temp06.py", line 28, in thread_reciever
    self.proc.start()
File "d:\python36-32\lib\multiprocessing\process.py", line 105, in start
    self._popen = self._Popen(self)
File "d:\python36-32\lib\multiprocessing\context.py", line 223, in _Popen
    return _default_context.get_context().Process._Popen(process_obj)
File "d:\python36-32\lib\multiprocessing\context.py", line 322, in _Popen
    return Popen(process_obj)
File "d:\python36-32\lib\multiprocessing\popen_spawn_win32.py", line 65, in __init__
    reduction.dump(process_obj, to_child)
File "d:\python36-32\lib\multiprocessing\reduction.py", line 60, in dump
    ForkingPickler(file, protocol).dump(obj)
TypeError: can't pickle _thread.lock objects


D:\>Traceback (most recent call last):
File "<string>", line 1, in <module>
File "d:\python36-32\lib\multiprocessing\spawn.py", line 99, in spawn_main
    new_handle = reduction.steal_handle(parent_pid, pipe_handle)
File "d:\python36-32\lib\multiprocessing\reduction.py", line 82, in steal_handle
    _winapi.PROCESS_DUP_HANDLE, False, source_pid)
OSError: [WinError 87]

如何在线程内启动和终止进程？ (我正在使用一个线程，因为它可以同步接收来自不同进程的数据)或者还有其他方法可以完成这项工作吗？

作为全局函数的 test_func

import numpy as np
import multiprocessing as mp
import threading
import time
from apscheduler.schedulers.background import BackgroundScheduler

pipe_in, pipe_out = mp.Pipe()  

def test_func( pipe_in):
    for i in range(10):
        pipe_in.send(i)
        time.sleep(1)

class Mother():
    def __init__(self):
        self.pipe_out = pipe_out
        mp.freeze_support()
        self.proc = mp.Process(target = test_func, args=(pipe_in, ))
        self.proc.start()

        self.thread = threading.Thread(target=self.thread_reciever, args=(self.pipe_out, ))
        self.thread.start()

    def thread_reciever(self, pipe_out):
        while True:
            value = pipe_out.recv()

            print(value)
            if value == 5:
                self.proc.terminate()
                time.sleep(2)
                mp.freeze_support()
                self.proc = mp.Process(target = test_func, args=(pipe_in,))
                self.proc.start()


if __name__ == '__main__':

    r = Mother()

输出

D:\> d:\python36-32\python.exe temp06.py
0
1
2
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5
Traceback (most recent call last):
File "<string>", line 1, in <module>
File "d:\python36-32\lib\multiprocessing\spawn.py", line 105, in spawn_main
    exitcode = _main(fd)
File "d:\python36-32\lib\multiprocessing\spawn.py", line 115, in _main
    self = reduction.pickle.load(from_parent)
AttributeError: Can't get attribute 'test_func' on <module '__main__' (built-in)>

Answer 1

在windows下，由于没有fork系统调用，python启动一个新的解释器实例，使用pickle/unpickle重构执行上下文，但是 thread.Lock 不可 picklable。在 pickling self.test_func 时，self.thread 引用了一个 thread.Lock 对象，使其不可 picklable。

您可以简单地将 test_func 更改为普通的全局函数，而无需线程对象引用:

self.proc = mp.Process(target = test_func, args=(pipe_in,))
...
def test_func(pipe_in):
    for i in range(10):
        pipe_in.send(i)
        time.sleep(1)