V2EX = way to explore
V2EX 是一个关于分享和探索的地方
现在注册
已注册用户请  登录
推荐学习书目
Learn Python the Hard Way
Python Sites
PyPI - Python Package Index
http://diveintopython.org/toc/index.html
Pocoo
值得关注的项目
PyPy
Celery
Jinja2
Read the Docs
gevent
pyenv
virtualenv
Stackless Python
Beautiful Soup
结巴中文分词
Green Unicorn
Sentry
Shovel
Pyflakes
pytest
Python 编程
pep8 Checker
Styles
PEP 8
Google Python Style Guide
Code Style from The Hitchhiker's Guide
wuwukai007
V2EX  ›  Python

Python 捕获多进程 OR 多线程 error 的最佳实现是什么?

  •  
  •   wuwukai007 · 2020-01-07 14:50:42 +08:00 · 2644 次点击
    这是一个创建于 1812 天前的主题,其中的信息可能已经有所发展或是发生改变。
    def test():
        a=[2]
        print(a[3])
    
    def run():
        test()
    
    def err(error):
        logger.exception(error)
    
    if __name__ == '__main__':
        pool = Pool(2)
        t1 = pool.apply_async(run,error_callback=err)
        pool.close()
        pool.join()
    

    日志记录 :

    list index out of range
    
    • 显然并没有拿到堆栈信息,在 callback 里再加上 try
    def test():
        a=[2]
        print(a[3])
    
    def run():
        test()
    
    def err(error):
        #这边已经拿到 error 了但是还要 try,要不然日志只记录 list index out of range
        try:
            raise error
        except Exception as e:
            logger.exception(e)
    
    if __name__ == '__main__':
        pool = Pool(2)
        t1 = pool.apply_async(run,error_callback=err)
        pool.close()
        pool.join()
    

    日志输出 :

    list index out of range
    multiprocessing.pool.RemoteTraceback: 
    """
    Traceback (most recent call last):
      File "/usr/lib/python3.5/multiprocessing/pool.py", line 119, in worker
        result = (True, func(*args, **kwds))
      File "/home/work_dir/test/timer.py", line 13, in run
        test()
      File "/home/work_dir/test/timer.py", line 10, in test
        print(a[3])
    IndexError: list index out of range
    """
    
    The above exception was the direct cause of the following exception:
    
    Traceback (most recent call last):
      File "/home/work_dir/test/timer.py", line 17, in err
        raise error
    IndexError: list index out of range
    
    

    难道要在回调函数里再次捕获输出?

    第 1 条附言  ·  2020-01-07 17:32:29 +08:00

    用了个简单的办法

    from multiprocessing.pool import rebuild_exc
    from multiprocessing import pool
    
    class My_Except(pool.ExceptionWithTraceback):
        def __init__(self, exc, tb):
            logger.exception(tb)
            return super().__init__(exc,tb)
    
    pool.ExceptionWithTraceback = My_Except
    

    在包的__init__.py中 打一个猴子补丁,这样就可以抓到堆栈,然后写日志,或刷新状态.

    然后在代码中导入__init__.py,也不需要回调函数直接就拿到堆栈了

    from multiprocessing import Pool
    import py_mutilprocessing
    
    def test():
        a=[2]
        print(a[3])
    
    def run():
        test()
    
    if __name__ == '__main__':
        pool = Pool(2)
        t1 = pool.apply_async(run,)
        pool.close()
        pool.join()
    

    日志:

    <traceback object at 0x7ff050695988>
    Traceback (most recent call last):
      File "/usr/local/lib/python3.6/multiprocessing/pool.py", line 119, in worker
        result = (True, func(*args, **kwds))
      File "/home/tarena/work_dir/py_mutilprocessing/timer.py", line 9, in run
        test()
      File "/home/tarena/work_dir/py_mutilprocessing/timer.py", line 6, in test
        print(a[3])
    IndexError: list index out of range
    

    但是觉得这么写怪怪的~~~~

    10 条回复    2020-05-19 16:29:07 +08:00
    BingoXuan
        1
    BingoXuan  
       2020-01-07 14:58:58 +08:00
    多进程和多线程通讯的问题,内置有 queue 之类的数据类型可以使用。
    wuwukai007
        2
    wuwukai007  
    OP
       2020-01-07 15:01:43 +08:00
    @BingoXuan 不是通信,是抓到子进程子线程错误,输出到日志或者写状态等,但是拿不到堆栈信息。我现在是再次 try 或者是用装饰器来做
    zyqzyq08
        3
    zyqzyq08  
       2020-01-07 15:07:23 +08:00
    可以考虑直接重写 sys.stderr
    BingoXuan
        4
    BingoXuan  
       2020-01-07 15:17:04 +08:00
    你试图捕抓子进程或子线程信息本质就是通讯。为什么一定要主进程去干着活呢?你完全可以让子进程或子线程运行时候去捕捉啊。装饰器的其中一个广泛用途就是捕获并处理错误信息。如果子进程是一个 worker,那么子进程 worker 接受主进程的请求运行指定函数,记录运行结果或捕获运行错误,并通过 socket 或文件通知或保存信息。
    Vegetable
        5
    Vegetable  
       2020-01-07 15:18:59 +08:00
    traceback.format_exc()是你想要的?
    oahebky
        6
    oahebky  
       2020-01-07 15:19:26 +08:00
    了解一下 traceback 库或许你就有答案了。
    wuwukai007
        7
    wuwukai007  
    OP
       2020-01-07 15:22:47 +08:00
    @Vegetable traceback.format_exc() 是主进程中错误会抛出,用的进程池或线程池,除非用装饰器包起来,要不然子进程错误不会抛出来的,比如我现在用的回调,用这个直接输出 NoneType
    todd7zhang
        8
    todd7zhang  
       2020-01-07 17:42:25 +08:00
    子进程的错误, 就让子进程抛呗
    wuwukai007
        9
    wuwukai007  
    OP
       2020-01-07 17:49:30 +08:00
    @todd7zhang 子进程要捕获,之前用的装饰器把子进程整个包起来可以拿到 error,但是这样感觉很浪费,应为进程池本身就能拿到 error,这样多此一举了
    hiyoi
        10
    hiyoi  
       2020-05-19 16:29:07 +08:00
    最近刚好碰到这个情况,我是这么解决的,把要子线程和日志对象封装到一起,使用 logger.exception()来 traceback 错误


    ```
    from concurrent.futures import ThreadPoolExecutor
    import logging


    class Main(object):
    def __init__(self, logger=None):
    self.pool = ThreadPoolExecutor
    self.logger = logger or self._get_logger(__name__)

    def _get_logger(self, name):
    logger = logging.getLogger(name)
    handler = logging.StreamHandler()
    formatter = logging.Formatter(
    "%(asctime)s - %(levelname)s - %(threadName)s - %(message)s")
    handler.setFormatter(formatter)
    logger.addHandler(handler)
    logger.setLevel(logging.DEBUG)
    return logger

    def do_something(self):
    try:
    a = 1 / 0
    except Exception as e:
    self.logger.exception(e)

    def async_task(self):
    with self.pool() as p:
    for i in range(3):
    p.submit(self.do_something)


    if __name__ == '__main__':
    m = Main()
    m.async_task()
    ```

    日志:
    ```
    2020-05-19 16:23:38,378 - ERROR - ThreadPoolExecutor-0_0 - division by zero
    Traceback (most recent call last):
    File "thread.py", line 22, in do_something
    a = 1 / 0
    ZeroDivisionError: division by zero
    2020-05-19 16:23:38,378 - ERROR - ThreadPoolExecutor-0_0 - division by zero
    Traceback (most recent call last):
    File "thread.py", line 22, in do_something
    a = 1 / 0
    ZeroDivisionError: division by zero
    2020-05-19 16:23:38,378 - ERROR - ThreadPoolExecutor-0_2 - division by zero
    Traceback (most recent call last):
    File "thread.py", line 22, in do_something
    a = 1 / 0
    ZeroDivisionError: division by zero
    ```
    关于   ·   帮助文档   ·   博客   ·   API   ·   FAQ   ·   实用小工具   ·   5172 人在线   最高记录 6679   ·     Select Language
    创意工作者们的社区
    World is powered by solitude
    VERSION: 3.9.8.5 · 28ms · UTC 05:43 · PVG 13:43 · LAX 21:43 · JFK 00:43
    Developed with CodeLauncher
    ♥ Do have faith in what you're doing.