两个并行流程的顺序流程

时间:2017-03-06 03:47:56

标签: python python-2.7

我有几个进程,比如A_step1,A_step2,B_step1,B_step2 ......它们必须以step1必须在step2开始运行之前完成的方式运行。这就是我所做的:

from subprocess import check_call
check_call(A_step1)
check_call(A_step2)
check_call(B_step1)
check_call(B_step2)

但是,我希望A和B进程并行运行。有没有在Python中实现这一点?

非常感谢

2 个答案:

答案 0 :(得分:0)

您可能可以将相关进程放在函数中,然后异步运行它们。对于异步部分,我建议使用multiprocessing模块

答案 1 :(得分:0)

一个常见的策略是使用队列作为一种机制,允许协调员(通常是您的主要流程)发放工作,并允许工作人员在完成某事时告诉协调员。

这是一个简化的例子。您可以尝试随机睡眠时间来说服自己,在两名工人完成第一步工作之前,第二步工作都不会开始。

from multiprocessing import Process, Manager
from time import sleep
from random import randint

def main():

    # Some queues so that we can tell the workers to advance
    # to the next step, and so that the workers to tell
    # us when they have completed a step.
    workQA = Manager().Queue()
    workQB = Manager().Queue()
    progQ = Manager().Queue()

    # Start the worker processes.
    pA = Process(target = workerA, args = (workQA, progQ))
    pB = Process(target = workerB, args = (workQB, progQ))
    pA.start()
    pB.start()

    # Step through some work.
    for step in (1, 2):
        workQA.put(step)
        workQB.put(step)
        done = []
        while True:
            item_done = progQ.get()
            print item_done
            done.append(item_done)
            if len(done) == 2:
                break

    # Tell the workers to stop and wait for everything to wrap up.
    workQA.put('stop')
    workQB.put('stop')
    pA.join()
    pB.join()

def workerA(workQ, progQ):
    do_work('A', workQ, progQ)

def workerB(workQ, progQ):
    do_work('B', workQ, progQ)

def do_work(worker, workQ, progQ):
    # Of course, in your real code the two workers won't
    # be doing the same thing.
    while True:
        step = workQ.get()
        if step == 1:
            do_step(worker, step, progQ)
        elif step == 2:
            do_step(worker, step, progQ)
        else:
            return

def do_step(worker, step, progQ):
    n = randint(1, 5)
    msg = 'worker={} step={} sleep={}'.format(worker, step, n)
    sleep(n)
    progQ.put(msg)   

main()

示例输出:

worker=B step=1 sleep=2
worker=A step=1 sleep=4
worker=A step=2 sleep=1
worker=B step=2 sleep=3