Python：从multiprocessing.Process获取回溯

Question

我试图从multiprocessing.Process获取一个追溯对象。不幸的是，通过管道传递异常信息不起作用，因为无法腌制回溯对象：

def foo(pipe_to_parent):
    try:
        raise Exception('xxx')
    except:
        pipe_to_parent.send(sys.exc_info())

to_child, to_self = multiprocessing.Pipe()
process = multiprocessing.Process(target = foo, args = (to_self,))
process.start()
exc_info = to_child.recv()
process.join()
print traceback.format_exception(*exc_info)
to_child.close()
to_self.close()

追溯：

Traceback (most recent call last):
  File "/usr/lib/python2.6/multiprocessing/process.py", line 231, in _bootstrap
    self.run()
  File "/usr/lib/python2.6/multiprocessing/process.py", line 88, in run
    self._target(*self._args, **self._kwargs)
  File "foo", line 7, in foo
    to_parent.send(sys.exc_info())
PicklingError: Can't pickle <type 'traceback'>: attribute lookup __builtin__.traceback failed

还有另一种访问异常信息的方法吗？我想避免传递格式化的字符串。

Answer 1

使用tblib您可以传递包装的异常并在以后重新引发它们：

import tblib.pickling_support
tblib.pickling_support.install()

from multiprocessing import Pool
import sys


class ExceptionWrapper(object):

    def __init__(self, ee):
        self.ee = ee
        __, __, self.tb = sys.exc_info()

    def re_raise(self):
        raise self.ee.with_traceback(self.tb)
        # for Python 2 replace the previous line by:
        # raise self.ee, None, self.tb


# example of how to use ExceptionWrapper

def inverse(i):
    """ will fail for i == 0 """
    try:
        return 1.0 / i
    except Exception as e:
        return ExceptionWrapper(e)


def main():
    p = Pool(1)
    results = p.map(inverse, [0, 1, 2, 3])
    for result in results:
        if isinstance(result, ExceptionWrapper):
            result.re_raise()


if __name__ == "__main__":
    main()

因此，如果您在远程进程中捕获到异常，则将其包装ExceptionWrapper，然后将其传递回去。调用re_raise()主流程即可完成工作。