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tl;博士我产生 3 个线程,每个线程抛出一个异常,最 Pythonic 的方式来引发所有 3 个异常?

下面是一个与我正在做的类似的代码示例。

from multiprocessing.pool import ThreadPool

def fail_func(host):
    raise Exception('{} FAILED!!!'.format(host))

hosts = ['172.1.1.1', '172.1.1.2', '172.1.1.3']
pool = ThreadPool(processes=5)
workers = [pool.apply_async(fail_func(host)) for host in hosts]
# join and close thread pool
pool.join(); pool.close()
# get the exceptions
[worker.get() for worker in workers if not worker.successful()]

它最终做的只是在第一台主机上失败,并带有以下回溯:

Traceback (most recent call last):
  File "thread_exception_example.py", line 8, in <module>
    workers = [pool.apply_async(fail_func(host)) for host in hosts]
  File "thread_exception_example.py", line 4, in fail_func
    raise Exception('{} FAILED!!!'.format(host))
Exception: 172.1.1.1 FAILED!!!

但我想要它做的是为每个失败的线程引发多个异常,如下所示:

Traceback (most recent call last):
  File "thread_exception_example.py", line 8, in <module>
    workers = [pool.apply_async(fail_func(host)) for host in hosts]
  File "thread_exception_example.py", line 4, in fail_func
    raise Exception('{} FAILED!!!'.format(host))
Exception: 172.1.1.1 FAILED!!!

Traceback (most recent call last):
  File "thread_exception_example.py", line 8, in <module>
    workers = [pool.apply_async(fail_func(host)) for host in hosts]
  File "thread_exception_example.py", line 4, in fail_func
    raise Exception('{} FAILED!!!'.format(host))
Exception: 172.1.1.2 FAILED!!!

Traceback (most recent call last):
  File "thread_exception_example.py", line 8, in <module>
    workers = [pool.apply_async(fail_func(host)) for host in hosts]
  File "thread_exception_example.py", line 4, in fail_func
    raise Exception('{} FAILED!!!'.format(host))
Exception: 172.1.1.3 FAILED!!!

有没有这样做的pythonic方式?还是我需要将所有内容包装在 try/except 中,收集所有消息,然后重新引发单个异常?

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1 回答 1

2

没有办法“引发多个异常”。在给定的异常上下文中,要么存在异常,要么没有异常。

所以是的,您将必须创建一个包含所有异常的包装异常,并引发它。但是您几乎已经获得了所需的所有代码:

def get_exception():
    try:
        worker.get()
    except Exception as e:
        return e

现在,而不是:

[worker.get() for worker in workers if not worker.successful()]

......你可以这样做:

[get_exception(worker.get) for worker in workers if not worker.successful()]

这是一个例外列表。


就个人而言,我一直认为AsyncResult应该有一种exception方法,类似于concurrent.futures.Future. 但是我会futures首先在这里使用(如果我被迫使用 Python 2.x,则安装backport )。

于 2013-08-16T22:38:56.610 回答