AWS Lambda中的Python：多处理，一个完成时终止其他进程

Question

我从this other question了解到AWS Lambda不支持multiprocessing.Pool和multiprocessing.Queue。

我还在 AWS Lambda 中研究 Python 多处理。但我的问题是：当第一个子进程返回时，我们如何终止主进程？ （所有子进程将以不同的执行时间返回）

我这里有什么：

import time
from multiprocessing import Process, Pipe

class run_func():
    
    number = 0

    def __init__(self, number):
        self.number = number

    def subrun(self, input, conn):
         # subprocess function with different execution time based on input.
        response = subprocess(input)
        conn.send([input, response])
        conn.close()

    def run(self):
        number = self.number
        processes = []
        parent_connections = []
        for i in range(0, number):
            parent_conn, child_conn = Pipe()
            parent_connections.append(parent_conn)
            process = Process(target=self.subrun, args=(i, child_conn,))
            processes.append(process)

        for process in processes:
            process.start()
        for process in processes:
            process.join()

        results = []
        for parent_connection in parent_connections:
            resp = parent_connection.recv()
            print(resp)
            results.append((resp[0],resp[1]))
        return results

def lambda_handler(event, context):
    starttime = time.time()
    results = []
    work = run_func(int(event['number']))
    results = work.run()
    print("Results : {}".format(results))
    print('Time: {} seconds'.format(time.time() - starttime))
    return output

当前程序将返回，直到所有子进程完成（for parent_connection in parent_connections）。但我想知道如何在第一个子进程完成时终止？ （至少终止主进程，其他子进程 - 可以让它继续运行）

添加： 明确地说，我的意思是第一个返回的子进程（可能不是第一个创建的子进程）。

Answer 1

所以 join() 循环是等待所有子进程完成的循环。

如果我们在完成第一个子进程后打破它并强制终止所有其他进程，它将为您工作

class run_func():
    
    number = 0

    def __init__(self, number):
        self.number = number

    def subrun(self, input, conn):
         # subprocess function with different execution time based on input.
        response = subprocess(input)
        conn.send([input, response])
        conn.close()

    def run(self):
        number = self.number
        processes = []
        parent_connections = []
        for i in range(0, number):
            parent_conn, child_conn = Pipe()
            parent_connections.append(parent_conn)
            process = Process(target=self.subrun, args=(i, child_conn,))
            processes.append(process)

        for process in processes:
            process.start()
        for process in processes:
            process.join()
            break
        results = []
        for parent_connection in parent_connections:
            resp = parent_connection.recv()
            print(resp)