兩個順序進程並行流程

Question

我有幾個進程，比如說A_step1，A_step2，B_step1，B_step2 ......它們必須以step1必須在step2開始運行之前完成的方式運行。這是我所做的：

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進程並行運行。無論如何要在Python中實現這一點？

非常感謝

Answer 1

你或許可以把相關的過程中的功能，然後異步運行它們。對於異步部分，我會推薦multiprocessing模塊

Answer 2

一個常見的策略是使用隊列作爲機制來允許協調員（通常是您的主進程）完成工作，並且允許工作人員告訴協調者當他們完成了一些事情。

這裏是一個簡化的例子。你可以嘗試隨機的睡眠時間來說服自己，直到兩名工作人員完成第一步的工作後，第二步的工作纔會開始。

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() 

輸出示例：

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