Python 多处理和处理工人中的异常

我使用 python 多处理库来实现一种算法，其中有许多工作人员处理某些数据并将结果返回给父进程。我使用 multiprocessing.Queue 将作业传递给工作人员，然后收集结果。

一切都运行得很好，直到工作人员无法处理某些数据块。在下面的简化示例中，每个工作人员都有两个阶段：

• 初始化 - 可能会失败，在这种情况下，worker 应该被销毁
• 数据处理 - 处理一块数据可能会失败，在这种情况下，工作人员应该跳过该块并继续处理下一个数据。

当这两个阶段中的任何一个失败时，我都会在脚本完成后陷入僵局。这段代码模拟了我的问题：

import multiprocessing as mp
import random

workers_count = 5
# Probability of failure, change to simulate failures
fail_init_p = 0.2
fail_job_p = 0.3


#========= Worker =========
def do_work(job_state, arg):
    if random.random() < fail_job_p:
        raise Exception("Job failed")
    return "job %d processed %d" % (job_state, arg)

def init(args):
    if random.random() < fail_init_p:
        raise Exception("Worker init failed")
    return args

def worker_function(args, jobs_queue, result_queue):
    # INIT
    # What to do when init() fails?
    try:
        state = init(args)
    except:
        print "!Worker %d init fail" % args
        return
    # DO WORK
    # Process data in the jobs queue
    for job in iter(jobs_queue.get, None):
        try:
            # Can throw an exception!
            result = do_work(state, job)
            result_queue.put(result)
        except:
            print "!Job %d failed, skip..." % job
        finally:
            jobs_queue.task_done()
    # Telling that we are done with processing stop token
    jobs_queue.task_done()



#========= Parent =========
jobs = mp.JoinableQueue()
results = mp.Queue()
for i in range(workers_count):
    mp.Process(target=worker_function, args=(i, jobs, results)).start()

# Populate jobs queue
results_to_expect = 0
for j in range(30):
    jobs.put(j)
    results_to_expect += 1

# Collecting the results
# What if some workers failed to process the job and we have
# less results than expected
for r in range(results_to_expect):
    result = results.get()
    print result

#Signal all workers to finish
for i in range(workers_count):
    jobs.put(None)

#Wait for them to finish
jobs.join()

我对这段代码有两个问题：

1. When init()失败，如何检测该工作人员无效而不是等待它完成？
2. When do_work()失败，如何通知父进程结果队列中应该有更少的结果？

谢谢你的帮助！

我稍微更改了您的代码以使其正常工作（请参阅下面的说明）。

import multiprocessing as mp
import random

workers_count = 5
# Probability of failure, change to simulate failures
fail_init_p = 0.5
fail_job_p = 0.4


#========= Worker =========
def do_work(job_state, arg):
    if random.random() < fail_job_p:
        raise Exception("Job failed")
    return "job %d processed %d" % (job_state, arg)

def init(args):
    if random.random() < fail_init_p:
        raise Exception("Worker init failed")
    return args

def worker_function(args, jobs_queue, result_queue):
    # INIT
    # What to do when init() fails?
    try:
        state = init(args)
    except:
        print "!Worker %d init fail" % args
        result_queue.put('init failed')
        return
    # DO WORK
    # Process data in the jobs queue
    for job in iter(jobs_queue.get, None):
        try:
            # Can throw an exception!
            result = do_work(state, job)
            result_queue.put(result)
        except:
            print "!Job %d failed, skip..." % job
            result_queue.put('job failed')


#========= Parent =========
jobs = mp.Queue()
results = mp.Queue()
for i in range(workers_count):
    mp.Process(target=worker_function, args=(i, jobs, results)).start()

# Populate jobs queue
results_to_expect = 0
for j in range(30):
    jobs.put(j)
    results_to_expect += 1

init_failures = 0
job_failures = 0
successes = 0
while job_failures + successes < 30 and init_failures < workers_count:
    result = results.get()
    init_failures += int(result == 'init failed')
    job_failures += int(result == 'job failed')
    successes += int(result != 'init failed' and result != 'job failed')
    #print init_failures, job_failures, successes

for ii in range(workers_count):
    jobs.put(None)

我的改变：

1. Changed jobs成为一个正常人Queue（代替JoinableQueue).
2. 现在，工作人员会返回特殊结果字符串“init failed”和“job failed”。
3. 只要特定条件有效，主进程就会监视所述特殊结果。
4. 最后，提出“停止”请求（即None工作），无论你有多少工人，无论如何。请注意，并非所有这些都可以从队列中提取（以防工作线程初始化失败）。

顺便说一句，您的原始代码很好并且易于使用。随机概率位非常酷。