我知道我在这里有点晚了，但我有一个非常类似的问题，但它包括使用tkinter作为GUI，并且主循环使它不可能使用依赖于.join()的任何解决方案。因此，我调整了原问题EDIT中给出的解决方案，但使其更一般，以便于其他人更容易理解。

下面是运行中的新线程类:

import threading
import traceback
import logging


class ExceptionThread(threading.Thread):
    def __init__(self, *args, **kwargs):
        threading.Thread.__init__(self, *args, **kwargs)

    def run(self):
        try:
            if self._target:
                self._target(*self._args, **self._kwargs)
        except Exception:
            logging.error(traceback.format_exc())


def test_function_1(input):
    raise IndexError(input)


if __name__ == "__main__":
    input = 'useful'

    t1 = ExceptionThread(target=test_function_1, args=[input])
    t1.start()

当然，您总是可以让它以日志以外的其他方式处理异常，例如将其打印出来，或将其输出到控制台。

这允许您像使用Thread类一样使用ExceptionThread类，无需任何特殊修改。

我喜欢的一种方法是基于观察者模式。我定义了一个信号类，线程用它向侦听器发出异常。它还可以用于从线程返回值。例子:

import threading

class Signal:
    def __init__(self):
        self._subscribers = list()

    def emit(self, *args, **kwargs):
        for func in self._subscribers:
            func(*args, **kwargs)

    def connect(self, func):
        self._subscribers.append(func)

    def disconnect(self, func):
        try:
            self._subscribers.remove(func)
        except ValueError:
            raise ValueError('Function {0} not removed from {1}'.format(func, self))


class WorkerThread(threading.Thread):

    def __init__(self, *args, **kwargs):
        super(WorkerThread, self).__init__(*args, **kwargs)
        self.Exception = Signal()
        self.Result = Signal()

    def run(self):
        if self._Thread__target is not None:
            try:
                self._return_value = self._Thread__target(*self._Thread__args, **self._Thread__kwargs)
            except Exception as e:
                self.Exception.emit(e)
            else:
                self.Result.emit(self._return_value)

if __name__ == '__main__':
    import time

    def handle_exception(exc):
        print exc.message

    def handle_result(res):
        print res

    def a():
        time.sleep(1)
        raise IOError('a failed')

    def b():
        time.sleep(2)
        return 'b returns'

    t = WorkerThread(target=a)
    t2 = WorkerThread(target=b)
    t.Exception.connect(handle_exception)
    t2.Result.connect(handle_result)
    t.start()
    t2.start()

    print 'Threads started'

    t.join()
    t2.join()
    print 'Done'

我没有足够的使用线程的经验来断言这是一种完全安全的方法。但这对我来说很管用，我喜欢这种灵活性。

我认为其他的解决方案有点复杂，如果你唯一想要的是真正看到某个异常，而不是完全无视和盲目。

解决方案是创建一个自定义线程，从主线程获取记录器并记录任何异常。

class ThreadWithLoggedException(threading.Thread):
    """
    Similar to Thread but will log exceptions to passed logger.

    Args:
        logger: Logger instance used to log any exception in child thread

    Exception is also reachable via <thread>.exception from the main thread.
    """

    def __init__(self, *args, **kwargs):
        try:
            self.logger = kwargs.pop("logger")
        except KeyError:
            raise Exception("Missing 'logger' in kwargs")
        super().__init__(*args, **kwargs)
        self.exception = None

    def run(self):
        try:
            if self._target is not None:
                self._target(*self._args, **self._kwargs)
        except Exception as exception:
            thread = threading.current_thread()
            self.exception = exception
            self.logger.exception(f"Exception in child thread {thread}: {exception}")
        finally:
            del self._target, self._args, self._kwargs

例子:

logger = logging.getLogger(__name__)
logger.setLevel(logging.INFO)
logger.addHandler(logging.StreamHandler())

def serve():
    raise Exception("Earth exploded.")

th = ThreadWithLoggedException(target=serve, logger=logger)
th.start()

主线程输出:

Exception in child thread <ThreadWithLoggedException(Thread-1, started 139922384414464)>: Earth exploded.
Traceback (most recent call last):
  File "/core/utils.py", line 108, in run
    self._target(*self._args, **self._kwargs)
  File "/myapp.py", line 105, in serve
    raise Exception("Earth exploded.")
Exception: Earth exploded.

这是一个棘手的小问题，我想提出我的解决方案。我发现了一些其他的解决方案(异步。例如IO)看起来很有前途，但也呈现出一些黑盒子。队列/事件循环方法将您与某个实现联系在一起。然而，并发期货的源代码只有大约1000行，很容易理解。它让我很容易地解决了我的问题:创建临时的工作线程，而不需要太多的设置，并且能够在主线程中捕获异常。

我的解决方案使用并发期货API和线程API。它允许你创建一个worker，给你线程和未来。这样，你就可以加入线程来等待结果:

worker = Worker(test)
thread = worker.start()
thread.join()
print(worker.future.result())

.．.或者你可以让worker在完成时发送一个回调:

worker = Worker(test)
thread = worker.start(lambda x: print('callback', x))

.．.或者你可以循环直到事件完成:

worker = Worker(test)
thread = worker.start()

while True:
    print("waiting")
    if worker.future.done():
        exc = worker.future.exception()
        print('exception?', exc)
        result = worker.future.result()
        print('result', result)           
        break
    time.sleep(0.25)

代码如下:

from concurrent.futures import Future
import threading
import time

class Worker(object):
    def __init__(self, fn, args=()):
        self.future = Future()
        self._fn = fn
        self._args = args

    def start(self, cb=None):
        self._cb = cb
        self.future.set_running_or_notify_cancel()
        thread = threading.Thread(target=self.run, args=())
        thread.daemon = True #this will continue thread execution after the main thread runs out of code - you can still ctrl + c or kill the process
        thread.start()
        return thread

    def run(self):
        try:
            self.future.set_result(self._fn(*self._args))
        except BaseException as e:
            self.future.set_exception(e)

        if(self._cb):
            self._cb(self.future.result())

.．.和测试函数:

def test(*args):
    print('args are', args)
    time.sleep(2)
    raise Exception('foo')

concurrent.futures.as_completed

https://docs.python.org/3.7/library/concurrent.futures.html#concurrent.futures.as_completed

解决方案如下:

当调用异常时，立即返回主线程 不需要额外的用户定义类，因为它不需要: 显式队列 在工作线程周围添加except else

来源:

#!/usr/bin/env python3

import concurrent.futures
import time

def func_that_raises(do_raise):
    for i in range(3):
        print(i)
        time.sleep(0.1)
    if do_raise:
        raise Exception()
    for i in range(3):
        print(i)
        time.sleep(0.1)
    
with concurrent.futures.ThreadPoolExecutor(max_workers=2) as executor:
    futures = []
    futures.append(executor.submit(func_that_raises, False))
    futures.append(executor.submit(func_that_raises, True))
    for future in concurrent.futures.as_completed(futures):
        print(repr(future.exception()))

可能的输出:

0
0
1
1
2
2
0
Exception()
1
2
None

不幸的是，当一个期货失效时，不可能通过终止期货来取消其他期货:

concurrent.futures;Python:并发。如何使其可取消? 线程:有办法杀死一个线程吗? C pthreads:在Pthread库中杀死线程

如果你这样做:

for future in concurrent.futures.as_completed(futures):
    if future.exception() is not None:
        raise future.exception()

然后with语句捕获它，等待第二个线程完成，然后继续。以下行为类似:

for future in concurrent.futures.as_completed(futures):
    future.result()

因为future.result()在发生异常时重新引发异常。

如果你想退出整个Python进程，你可以使用os._exit(0)，但这可能意味着你需要重构。

具有完美异常语义的自定义类

我最终为自己编写了一个完美的接口:限制一次运行的最大线程数的正确方法?部分“错误处理的队列示例”。该类的目标是既方便，又让您完全控制提交和结果/错误处理。

在Python 3.6.7, Ubuntu 18.04上测试。

我喜欢这门课:

https://gist.github.com/earonesty/b88d60cb256b71443e42c4f1d949163e

import threading
from typing import Any


class PropagatingThread(threading.Thread):
    """A Threading Class that raises errors it caught, and returns the return value of the target on join."""

    def __init__(self, *args, **kwargs):
        self._target = None
        self._args = ()
        self._kwargs = {}
        super().__init__(*args, **kwargs)
        self.exception = None
        self.return_value = None
        assert self._target

    def run(self):
        """Don't override this if you want the behavior of this class, use target instead."""
        try:
            if self._target:
                self.return_value = self._target(*self._args, **self._kwargs)
        except Exception as e:
            self.exception = e
        finally:
            # see super().run() for why this is necessary
            del self._target, self._args, self._kwargs

    def join(self, timeout=None) -> Any:
        super().join(timeout)
        if self.exception:
            raise self.exception
        return self.return_value