我对Python和多线程编程非常陌生。基本上,我有一个脚本,将文件复制到另一个位置。我想把这个放在另一个线程,这样我就可以输出....表示脚本仍在运行。
我遇到的问题是,如果文件不能复制,它将抛出异常。如果在主线程中运行,这是可以的;但是,使用以下代码是无效的:
try:
threadClass = TheThread(param1, param2, etc.)
threadClass.start() ##### **Exception takes place here**
except:
print "Caught an exception"
在线程类本身中,我试图重新抛出异常,但它不起作用。我在这里看到有人问类似的问题,但他们似乎都在做一些比我试图做的更具体的事情(我不太理解所提供的解决方案)。我看到有人提到sys.exc_info()的用法,但我不知道在哪里或如何使用它。
编辑:线程类的代码如下:
class TheThread(threading.Thread):
def __init__(self, sourceFolder, destFolder):
threading.Thread.__init__(self)
self.sourceFolder = sourceFolder
self.destFolder = destFolder
def run(self):
try:
shul.copytree(self.sourceFolder, self.destFolder)
except:
raise
并发。Futures模块使得在单独的线程(或进程)中工作并处理任何由此产生的异常变得简单:
import concurrent.futures
import shutil
def copytree_with_dots(src_path, dst_path):
with concurrent.futures.ThreadPoolExecutor(max_workers=1) as executor:
# Execute the copy on a separate thread,
# creating a future object to track progress.
future = executor.submit(shutil.copytree, src_path, dst_path)
while future.running():
# Print pretty dots here.
pass
# Return the value returned by shutil.copytree(), None.
# Raise any exceptions raised during the copy process.
return future.result()
并发。futures包含在Python 3.2中,并可作为早期版本的反向移植futures模块使用。
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.
并发。Futures模块使得在单独的线程(或进程)中工作并处理任何由此产生的异常变得简单:
import concurrent.futures
import shutil
def copytree_with_dots(src_path, dst_path):
with concurrent.futures.ThreadPoolExecutor(max_workers=1) as executor:
# Execute the copy on a separate thread,
# creating a future object to track progress.
future = executor.submit(shutil.copytree, src_path, dst_path)
while future.running():
# Print pretty dots here.
pass
# Return the value returned by shutil.copytree(), None.
# Raise any exceptions raised during the copy process.
return future.result()
并发。futures包含在Python 3.2中,并可作为早期版本的反向移植futures模块使用。
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。
如果在线程中发生异常,最好的方法是在连接期间在调用线程中重新引发它。您可以使用sys.exc_info()函数获取当前正在处理的异常的信息。此信息可以简单地存储为线程对象的属性,直到调用join,此时可以重新引发它。
注意,队列。队列(在其他回答中建议)在这个简单的情况下是不必要的,因为线程最多抛出1个异常,并且在抛出一个异常后立即完成。我们通过简单地等待线程完成来避免竞争条件。
例如,扩展ExcThread(如下),覆盖excRun(而不是run)。
Python 2. x:
import threading
class ExcThread(threading.Thread):
def excRun(self):
pass
def run(self):
self.exc = None
try:
# Possibly throws an exception
self.excRun()
except:
import sys
self.exc = sys.exc_info()
# Save details of the exception thrown but don't rethrow,
# just complete the function
def join(self):
threading.Thread.join(self)
if self.exc:
msg = "Thread '%s' threw an exception: %s" % (self.getName(), self.exc[1])
new_exc = Exception(msg)
raise new_exc.__class__, new_exc, self.exc[2]
Python 3. x:
在Python 3中,raise的参数形式为3,因此将最后一行更改为:
raise new_exc.with_traceback(self.exc[2])
我做的是,简单的覆盖连接和运行线程的方法:
class RaisingThread(threading.Thread):
def run(self):
self._exc = None
try:
super().run()
except Exception as e:
self._exc = e
def join(self, timeout=None):
super().join(timeout=timeout)
if self._exc:
raise self._exc
用途如下:
def foo():
time.sleep(2)
print('hi, from foo!')
raise Exception('exception from foo')
t = RaisingThread(target=foo)
t.start()
try:
t.join()
except Exception as e:
print(e)
结果:
hi, from foo!
exception from foo!