I have multiple functions in my script which does a REST API api requests.As i need to handle the error scenarios i have put a retry mechanism as below.
no_of_retries = 3
def check_status():
for i in range(0,no_of_retries):
url = "http://something/something"
try:
result = requests.get(url, auth=HTTPBasicAuth(COMMON_USERNAME, COMMON_PASSWORD)).json()
if 'error' not in result:
return result
else:
continue
except Exception as e:
continue
return None
I have several different methods which does similar operation. How can we do it better way to avoid duplication may be using decorators.
You can use a decorator like this and handle your own exception.
def retry(times, exceptions):
"""
Retry Decorator
Retries the wrapped function/method `times` times if the exceptions listed
in ``exceptions`` are thrown
:param times: The number of times to repeat the wrapped function/method
:type times: Int
:param Exceptions: Lists of exceptions that trigger a retry attempt
:type Exceptions: Tuple of Exceptions
"""
def decorator(func):
def newfn(*args, **kwargs):
attempt = 0
while attempt < times:
try:
return func(*args, **kwargs)
except exceptions:
print(
'Exception thrown when attempting to run %s, attempt '
'%d of %d' % (func, attempt, times)
)
attempt += 1
return func(*args, **kwargs)
return newfn
return decorator
#retry(times=3, exceptions=(ValueError, TypeError))
def foo1():
print('Some code here ....')
print('Oh no, we have exception')
raise ValueError('Some error')
foo1()
if you do not mind installing a library you could use the tenacity (github.com/jd/tenacity) module. one of their examples:
import random
from tenacity import retry, stop_after_attempt
# #retry # retry forever
#retry(stop=stop_after_attempt(3))
def do_something_unreliable():
if random.randint(0, 10) > 1:
raise IOError("Broken sauce, everything is hosed!!!111one")
else:
return "Awesome sauce!"
print(do_something_unreliable())
this also allows you to specify the number of tries or seconds you want to keep retrying.
for your case this might look something like this (not tested!):
#retry(stop=stop_after_attempt(3))
def retry_get():
result = requests.get(
url, auth=HTTPBasicAuth(COMMON_USERNAME, COMMON_PASSWORD)).json()
if 'error' not in result:
raise RequestException(result)
The third-party retry module is now widely accepted for this. You can also pass the list of exceptions to retry for, number of retries, delays, maximum delay, exponential back-off, etc.
$ pip install retry
Example usage:
from retry import retry
#retry(ZeroDivisionError, tries=3, delay=2)
def make_trouble():
'''Retry on ZeroDivisionError, raise error after 3 attempts, sleep 2 seconds between attempts.'''
Production level example
import logging
import time
import functools
import traceback
LOG_FORMAT = "%(asctime)s - %(levelname)s - %(pathname)s - %(funcName)s - %(lineno)d -msg: %(message)s"
logging.basicConfig(level=logging.INFO, format=LOG_FORMAT)
def retry(retry_num, retry_sleep_sec):
"""
retry help decorator.
:param retry_num: the retry num; retry sleep sec
:return: decorator
"""
def decorator(func):
"""decorator"""
# preserve information about the original function, or the func name will be "wrapper" not "func"
#functools.wraps(func)
def wrapper(*args, **kwargs):
"""wrapper"""
for attempt in range(retry_num):
try:
return func(*args, **kwargs) # should return the raw function's return value
except Exception as err: # pylint: disable=broad-except
logging.error(err)
logging.error(traceback.format_exc())
time.sleep(retry_sleep_sec)
logging.error("Trying attempt %s of %s.", attempt + 1, retry_num)
logging.error("func %s retry failed", func)
raise Exception('Exceed max retry num: {} failed'.format(retry_num))
return wrapper
return decorator
usage
# this means try your function 5 times, each time sleep 60 seconds
#retry(5, 60)
def your_func():
pass
Formal reference: https://peps.python.org/pep-0318/
I would recommend using the retry library like #MohitC mentioned. However, if you're restricted to import 3rd party libraries like I was, you're welcome to try my version:
import time
def retry(tries= -1, delay=0, max_delay=None, backoff=1, exceptions=Exception, log=False):
"""Retry Decorator with arguments
Args:
tries (int): The maximum number of attempts. Defaults to -1 (infinite)
delay (int, optional): Delay between attempts (seconds). Defaults to 0
max_delay (int, optional): The maximum value of delay (seconds). Defaults to None (Unlimited)
backoff (int, optional): Multiplier applied to delay between attempts (seconds). Defaults to 1 (No backoff)
exceptions (tuple, optional): Types of exceptions to catch. Defaults to Exception (all)
log (bool, optional): Print debug logs. Defaults to False
"""
def retry_decorator(func):
def retry_wrapper(*args, **kwargs):
nonlocal tries, delay, max_delay, backoff, exceptions, log
while tries:
try:
return func(*args, **kwargs)
except exceptions:
tries -= 1
# Reached to maximum tries
if not tries:
raise
# Log the retry logs for the given function
if log:
print(f"Retrying {func.__name__} in {delay} seconds")
# Apply delay between requests
time.sleep(delay)
# Adjust the next delay according to backoff
delay *= backoff
# Adjust maximum delay duration
if max_delay is not None:
delay = min(delay, max_delay)
return retry_wrapper
return retry_decorator
Example Uses:
Simple:
#retry(10, delay=5)
def do_something(params):
# Example func to retry
pass
Advanced:
#retry(10, delay=1, backoff=2, max_delay=10, exceptions=(TimeoutError), log=True)
def do_something(params):
# Example func to retry only for TimeoutErrors
pass
Instead of using decorators, the probably better solution is to move the request to its own function, arriving at a structure similar to this:
no_of_retries = 3
def make_request(url):
for i in range(0,no_of_retries):
try:
result = requests.get(url, auth=HTTPBasicAuth(COMMON_USERNAME, COMMON_PASSWORD)).json()
if 'error' not in result:
return result
else:
continue
except Exception as e:
continue
return result
def check_status():
result = make_request("http://something/status")
def load_file():
result = make_request("http://something/file")
This way, you avoid duplicate code while encapsulating the request. If you were to use a decorator, you would need to wrap the whole load_file() method which would prevent you from further processing the request's result within this function.
I have created a custom retry function. The function will retry if the first number is less than the second.
CODE:
import time
# CUSTOM EXCEPTION
class custom_error(Exception):
pass
# RETRY FUNCTION.
def retry(func, retries=3):
print(func)
def retry_wrapper(*args, **kwargs):
print(args)
n = args[0]
u = args[1]
print(n, u)
attempts = 0
while attempts < retries:
try:
if n > u:
return func(*args, **kwargs)
else:
raise custom_error
except custom_error:
print("error")
time.sleep(2)
attempts += 1
return retry_wrapper
#retry
def akash(a, b):
c = a / b
return c
# CALLING THE FUNCTION
a = akash(1, 2)
print(a)
OUTPUT:
<function akash at 0x00000187C3A66B00>
(1, 2)
1 2
error
error
error
Using functools on top of mrkiril's answer:
from functools import wraps, partial
def retry(f=None, times=10):
if f is None:
return partial(retry, times=times)
#wraps(f)
def wrap(*args, **kwargs):
attempt = 0
while attempt < times:
try:
return f(*args, **kwargs)
except:
print(f"{f.__name__}, attempt {attempt} of {times}")
attempt += 1
return f(*args, **kwargs)
return wrap
Then, wrap your function like the following:
import random
#retry
def foo():
if random.randint(0, 5) != 0:
raise Exception
Related
I'm calling a function in Python which I know may stall and force me to restart the script.
How do I call the function or what do I wrap it in so that if it takes longer than 5 seconds the script cancels it and does something else?
You may use the signal package if you are running on UNIX:
In [1]: import signal
# Register an handler for the timeout
In [2]: def handler(signum, frame):
...: print("Forever is over!")
...: raise Exception("end of time")
...:
# This function *may* run for an indetermined time...
In [3]: def loop_forever():
...: import time
...: while 1:
...: print("sec")
...: time.sleep(1)
...:
...:
# Register the signal function handler
In [4]: signal.signal(signal.SIGALRM, handler)
Out[4]: 0
# Define a timeout for your function
In [5]: signal.alarm(10)
Out[5]: 0
In [6]: try:
...: loop_forever()
...: except Exception, exc:
...: print(exc)
....:
sec
sec
sec
sec
sec
sec
sec
sec
Forever is over!
end of time
# Cancel the timer if the function returned before timeout
# (ok, mine won't but yours maybe will :)
In [7]: signal.alarm(0)
Out[7]: 0
10 seconds after the call signal.alarm(10), the handler is called. This raises an exception that you can intercept from the regular Python code.
This module doesn't play well with threads (but then, who does?)
Note that since we raise an exception when timeout happens, it may end up caught and ignored inside the function, for example of one such function:
def loop_forever():
while 1:
print('sec')
try:
time.sleep(10)
except:
continue
You can use multiprocessing.Process to do exactly that.
Code
import multiprocessing
import time
# bar
def bar():
for i in range(100):
print "Tick"
time.sleep(1)
if __name__ == '__main__':
# Start bar as a process
p = multiprocessing.Process(target=bar)
p.start()
# Wait for 10 seconds or until process finishes
p.join(10)
# If thread is still active
if p.is_alive():
print "running... let's kill it..."
# Terminate - may not work if process is stuck for good
p.terminate()
# OR Kill - will work for sure, no chance for process to finish nicely however
# p.kill()
p.join()
How do I call the function or what do I wrap it in so that if it takes longer than 5 seconds the script cancels it?
I posted a gist that solves this question/problem with a decorator and a threading.Timer. Here it is with a breakdown.
Imports and setups for compatibility
It was tested with Python 2 and 3. It should also work under Unix/Linux and Windows.
First the imports. These attempt to keep the code consistent regardless of the Python version:
from __future__ import print_function
import sys
import threading
from time import sleep
try:
import thread
except ImportError:
import _thread as thread
Use version independent code:
try:
range, _print = xrange, print
def print(*args, **kwargs):
flush = kwargs.pop('flush', False)
_print(*args, **kwargs)
if flush:
kwargs.get('file', sys.stdout).flush()
except NameError:
pass
Now we have imported our functionality from the standard library.
exit_after decorator
Next we need a function to terminate the main() from the child thread:
def quit_function(fn_name):
# print to stderr, unbuffered in Python 2.
print('{0} took too long'.format(fn_name), file=sys.stderr)
sys.stderr.flush() # Python 3 stderr is likely buffered.
thread.interrupt_main() # raises KeyboardInterrupt
And here is the decorator itself:
def exit_after(s):
'''
use as decorator to exit process if
function takes longer than s seconds
'''
def outer(fn):
def inner(*args, **kwargs):
timer = threading.Timer(s, quit_function, args=[fn.__name__])
timer.start()
try:
result = fn(*args, **kwargs)
finally:
timer.cancel()
return result
return inner
return outer
Usage
And here's the usage that directly answers your question about exiting after 5 seconds!:
#exit_after(5)
def countdown(n):
print('countdown started', flush=True)
for i in range(n, -1, -1):
print(i, end=', ', flush=True)
sleep(1)
print('countdown finished')
Demo:
>>> countdown(3)
countdown started
3, 2, 1, 0, countdown finished
>>> countdown(10)
countdown started
10, 9, 8, 7, 6, countdown took too long
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 11, in inner
File "<stdin>", line 6, in countdown
KeyboardInterrupt
The second function call will not finish, instead the process should exit with a traceback!
KeyboardInterrupt does not always stop a sleeping thread
Note that sleep will not always be interrupted by a keyboard interrupt, on Python 2 on Windows, e.g.:
#exit_after(1)
def sleep10():
sleep(10)
print('slept 10 seconds')
>>> sleep10()
sleep10 took too long # Note that it hangs here about 9 more seconds
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 11, in inner
File "<stdin>", line 3, in sleep10
KeyboardInterrupt
nor is it likely to interrupt code running in extensions unless it explicitly checks for PyErr_CheckSignals(), see Cython, Python and KeyboardInterrupt ignored
I would avoid sleeping a thread more than a second, in any case - that's an eon in processor time.
How do I call the function or what do I wrap it in so that if it takes longer than 5 seconds the script cancels it and does something else?
To catch it and do something else, you can catch the KeyboardInterrupt.
>>> try:
... countdown(10)
... except KeyboardInterrupt:
... print('do something else')
...
countdown started
10, 9, 8, 7, 6, countdown took too long
do something else
I have a different proposal which is a pure function (with the same API as the threading suggestion) and seems to work fine (based on suggestions on this thread)
def timeout(func, args=(), kwargs={}, timeout_duration=1, default=None):
import signal
class TimeoutError(Exception):
pass
def handler(signum, frame):
raise TimeoutError()
# set the timeout handler
signal.signal(signal.SIGALRM, handler)
signal.alarm(timeout_duration)
try:
result = func(*args, **kwargs)
except TimeoutError as exc:
result = default
finally:
signal.alarm(0)
return result
I ran across this thread when searching for a timeout call on unit tests. I didn't find anything simple in the answers or 3rd party packages so I wrote the decorator below you can drop right into code:
import multiprocessing.pool
import functools
def timeout(max_timeout):
"""Timeout decorator, parameter in seconds."""
def timeout_decorator(item):
"""Wrap the original function."""
#functools.wraps(item)
def func_wrapper(*args, **kwargs):
"""Closure for function."""
pool = multiprocessing.pool.ThreadPool(processes=1)
async_result = pool.apply_async(item, args, kwargs)
# raises a TimeoutError if execution exceeds max_timeout
return async_result.get(max_timeout)
return func_wrapper
return timeout_decorator
Then it's as simple as this to timeout a test or any function you like:
#timeout(5.0) # if execution takes longer than 5 seconds, raise a TimeoutError
def test_base_regression(self):
...
The stopit package, found on pypi, seems to handle timeouts well.
I like the #stopit.threading_timeoutable decorator, which adds a timeout parameter to the decorated function, which does what you expect, it stops the function.
Check it out on pypi: https://pypi.python.org/pypi/stopit
I am the author of wrapt_timeout_decorator.
Most of the solutions presented here work wunderfully under Linux on the first glance - because we have fork() and signals() - but on windows the things look a bit different.
And when it comes to subthreads on Linux, You cant use Signals anymore.
In order to spawn a process under Windows, it needs to be picklable - and many decorated functions or Class methods are not.
So you need to use a better pickler like dill and multiprocess (not pickle and multiprocessing) - thats why You cant use ProcessPoolExecutor (or only with limited functionality).
For the timeout itself - You need to define what timeout means - because on Windows it will take considerable (and not determinable) time to spawn the process. This can be tricky on short timeouts. Lets assume, spawning the process takes about 0.5 seconds (easily !!!). If You give a timeout of 0.2 seconds what should happen?
Should the function time out after 0.5 + 0.2 seconds (so let the method run for 0.2 seconds)?
Or should the called process time out after 0.2 seconds (in that case, the decorated function will ALWAYS timeout, because in that time it is not even spawned)?
Also nested decorators can be nasty and You cant use Signals in a subthread. If You want to create a truly universal, cross-platform decorator, all this needs to be taken into consideration (and tested).
Other issues are passing exceptions back to the caller, as well as logging issues (if used in the decorated function - logging to files in another process is NOT supported)
I tried to cover all edge cases, You might look into the package wrapt_timeout_decorator, or at least test Your own solutions inspired by the unittests used there.
#Alexis Eggermont - unfortunately I dont have enough points to comment - maybe someone else can notify You - I think I solved Your import issue.
There are a lot of suggestions, but none using concurrent.futures, which I think is the most legible way to handle this.
from concurrent.futures import ProcessPoolExecutor
# Warning: this does not terminate function if timeout
def timeout_five(fnc, *args, **kwargs):
with ProcessPoolExecutor() as p:
f = p.submit(fnc, *args, **kwargs)
return f.result(timeout=5)
Super simple to read and maintain.
We make a pool, submit a single process and then wait up to 5 seconds before raising a TimeoutError that you could catch and handle however you needed.
Native to python 3.2+ and backported to 2.7 (pip install futures).
Switching between threads and processes is as simple as replacing ProcessPoolExecutor with ThreadPoolExecutor.
If you want to terminate the Process on timeout I would suggest looking into Pebble.
Building on and and enhancing the answer by #piro , you can build a contextmanager. This allows for very readable code which will disable the alaram signal after a successful run (sets signal.alarm(0))
from contextlib import contextmanager
import signal
import time
#contextmanager
def timeout(duration):
def timeout_handler(signum, frame):
raise TimeoutError(f'block timedout after {duration} seconds')
signal.signal(signal.SIGALRM, timeout_handler)
signal.alarm(duration)
try:
yield
finally:
signal.alarm(0)
def sleeper(duration):
time.sleep(duration)
print('finished')
Example usage:
In [19]: with timeout(2):
...: sleeper(1)
...:
finished
In [20]: with timeout(2):
...: sleeper(3)
...:
---------------------------------------------------------------------------
Exception Traceback (most recent call last)
<ipython-input-20-66c78858116f> in <module>()
1 with timeout(2):
----> 2 sleeper(3)
3
<ipython-input-7-a75b966bf7ac> in sleeper(t)
1 def sleeper(t):
----> 2 time.sleep(t)
3 print('finished')
4
<ipython-input-18-533b9e684466> in timeout_handler(signum, frame)
2 def timeout(duration):
3 def timeout_handler(signum, frame):
----> 4 raise Exception(f'block timedout after {duration} seconds')
5 signal.signal(signal.SIGALRM, timeout_handler)
6 signal.alarm(duration)
Exception: block timedout after 2 seconds
Great, easy to use and reliable PyPi project timeout-decorator (https://pypi.org/project/timeout-decorator/)
installation:
pip install timeout-decorator
Usage:
import time
import timeout_decorator
#timeout_decorator.timeout(5)
def mytest():
print "Start"
for i in range(1,10):
time.sleep(1)
print "%d seconds have passed" % i
if __name__ == '__main__':
mytest()
timeout-decorator don't work on windows system as , windows didn't support signal well.
If you use timeout-decorator in windows system you will get the following
AttributeError: module 'signal' has no attribute 'SIGALRM'
Some suggested to use use_signals=False but didn't worked for me.
Author #bitranox created the following package:
pip install https://github.com/bitranox/wrapt-timeout-decorator/archive/master.zip
Code Sample:
import time
from wrapt_timeout_decorator import *
#timeout(5)
def mytest(message):
print(message)
for i in range(1,10):
time.sleep(1)
print('{} seconds have passed'.format(i))
def main():
mytest('starting')
if __name__ == '__main__':
main()
Gives the following exception:
TimeoutError: Function mytest timed out after 5 seconds
Highlights
Raises TimeoutError uses exceptions to alert on timeout - can easily be modified
Cross Platform: Windows & Mac OS X
Compatibility: Python 3.6+ (I also tested on python 2.7 and it works with small syntax adjustments)
For full explanation and extension to parallel maps, see here https://flipdazed.github.io/blog/quant%20dev/parallel-functions-with-timeouts
Minimal Example
>>> #killer_call(timeout=4)
... def bar(x):
... import time
... time.sleep(x)
... return x
>>> bar(10)
Traceback (most recent call last):
...
__main__.TimeoutError: function 'bar' timed out after 4s
and as expected
>>> bar(2)
2
Full code
import multiprocessing as mp
import multiprocessing.queues as mpq
import functools
import dill
from typing import Tuple, Callable, Dict, Optional, Iterable, List, Any
class TimeoutError(Exception):
def __init__(self, func: Callable, timeout: int):
self.t = timeout
self.fname = func.__name__
def __str__(self):
return f"function '{self.fname}' timed out after {self.t}s"
def _lemmiwinks(func: Callable, args: Tuple, kwargs: Dict[str, Any], q: mp.Queue):
"""lemmiwinks crawls into the unknown"""
q.put(dill.loads(func)(*args, **kwargs))
def killer_call(func: Callable = None, timeout: int = 10) -> Callable:
"""
Single function call with a timeout
Args:
func: the function
timeout: The timeout in seconds
"""
if not isinstance(timeout, int):
raise ValueError(f'timeout needs to be an int. Got: {timeout}')
if func is None:
return functools.partial(killer_call, timeout=timeout)
#functools.wraps(killer_call)
def _inners(*args, **kwargs) -> Any:
q_worker = mp.Queue()
proc = mp.Process(target=_lemmiwinks, args=(dill.dumps(func), args, kwargs, q_worker))
proc.start()
try:
return q_worker.get(timeout=timeout)
except mpq.Empty:
raise TimeoutError(func, timeout)
finally:
try:
proc.terminate()
except:
pass
return _inners
if __name__ == '__main__':
#killer_call(timeout=4)
def bar(x):
import time
time.sleep(x)
return x
print(bar(2))
bar(10)
Notes
You will need to import inside the function because of the way dill works.
This will also mean these functions may not be not compatible with doctest if there are imports inside your target functions. You will get an issue with __import__ not found.
Just in case it is helpful for anyone, building on the answer by #piro, I've made a function decorator:
import time
import signal
from functools import wraps
def timeout(timeout_secs: int):
def wrapper(func):
#wraps(func)
def time_limited(*args, **kwargs):
# Register an handler for the timeout
def handler(signum, frame):
raise Exception(f"Timeout for function '{func.__name__}'")
# Register the signal function handler
signal.signal(signal.SIGALRM, handler)
# Define a timeout for your function
signal.alarm(timeout_secs)
result = None
try:
result = func(*args, **kwargs)
except Exception as exc:
raise exc
finally:
# disable the signal alarm
signal.alarm(0)
return result
return time_limited
return wrapper
Using the wrapper on a function with a 20 seconds timeout would look something like:
#timeout(20)
def my_slow_or_never_ending_function(name):
while True:
time.sleep(1)
print(f"Yet another second passed {name}...")
try:
results = my_slow_or_never_ending_function("Yooo!")
except Exception as e:
print(f"ERROR: {e}")
We can use signals for the same. I think the below example will be useful for you. It is very simple compared to threads.
import signal
def timeout(signum, frame):
raise myException
#this is an infinite loop, never ending under normal circumstances
def main():
print 'Starting Main ',
while 1:
print 'in main ',
#SIGALRM is only usable on a unix platform
signal.signal(signal.SIGALRM, timeout)
#change 5 to however many seconds you need
signal.alarm(5)
try:
main()
except myException:
print "whoops"
Another solution with asyncio :
If you want to cancel the background task and not just timeout on the running main code, then you need an explicit communication from main thread to ask the code of the task to cancel , like a threading.Event()
import asyncio
import functools
import multiprocessing
from concurrent.futures.thread import ThreadPoolExecutor
class SingletonTimeOut:
pool = None
#classmethod
def run(cls, to_run: functools.partial, timeout: float):
pool = cls.get_pool()
loop = cls.get_loop()
try:
task = loop.run_in_executor(pool, to_run)
return loop.run_until_complete(asyncio.wait_for(task, timeout=timeout))
except asyncio.TimeoutError as e:
error_type = type(e).__name__ #TODO
raise e
#classmethod
def get_pool(cls):
if cls.pool is None:
cls.pool = ThreadPoolExecutor(multiprocessing.cpu_count())
return cls.pool
#classmethod
def get_loop(cls):
try:
return asyncio.get_event_loop()
except RuntimeError:
asyncio.set_event_loop(asyncio.new_event_loop())
# print("NEW LOOP" + str(threading.current_thread().ident))
return asyncio.get_event_loop()
# ---------------
TIME_OUT = float('0.2') # seconds
def toto(input_items,nb_predictions):
return 1
to_run = functools.partial(toto,
input_items=1,
nb_predictions="a")
results = SingletonTimeOut.run(to_run, TIME_OUT)
#!/usr/bin/python2
import sys, subprocess, threading
proc = subprocess.Popen(sys.argv[2:])
timer = threading.Timer(float(sys.argv[1]), proc.terminate)
timer.start()
proc.wait()
timer.cancel()
exit(proc.returncode)
The func_timeout package by Tim Savannah has worked well for me.
Installation:
pip install func_timeout
Usage:
import time
from func_timeout import func_timeout, FunctionTimedOut
def my_func(n):
time.sleep(n)
time_to_sleep = 10
# time out after 2 seconds using kwargs
func_timeout(2, my_func, kwargs={'n' : time_to_sleep})
# time out after 2 seconds using args
func_timeout(2, my_func, args=(time_to_sleep,))
I had a need for nestable timed interrupts (which SIGALARM can't do) that won't get blocked by time.sleep (which the thread-based approach can't do). I ended up copying and lightly modifying code from here: http://code.activestate.com/recipes/577600-queue-for-managing-multiple-sigalrm-alarms-concurr/
The code itself:
#!/usr/bin/python
# lightly modified version of http://code.activestate.com/recipes/577600-queue-for-managing-multiple-sigalrm-alarms-concurr/
"""alarm.py: Permits multiple SIGALRM events to be queued.
Uses a `heapq` to store the objects to be called when an alarm signal is
raised, so that the next alarm is always at the top of the heap.
"""
import heapq
import signal
from time import time
__version__ = '$Revision: 2539 $'.split()[1]
alarmlist = []
__new_alarm = lambda t, f, a, k: (t + time(), f, a, k)
__next_alarm = lambda: int(round(alarmlist[0][0] - time())) if alarmlist else None
__set_alarm = lambda: signal.alarm(max(__next_alarm(), 1))
class TimeoutError(Exception):
def __init__(self, message, id_=None):
self.message = message
self.id_ = id_
class Timeout:
''' id_ allows for nested timeouts. '''
def __init__(self, id_=None, seconds=1, error_message='Timeout'):
self.seconds = seconds
self.error_message = error_message
self.id_ = id_
def handle_timeout(self):
raise TimeoutError(self.error_message, self.id_)
def __enter__(self):
self.this_alarm = alarm(self.seconds, self.handle_timeout)
def __exit__(self, type, value, traceback):
try:
cancel(self.this_alarm)
except ValueError:
pass
def __clear_alarm():
"""Clear an existing alarm.
If the alarm signal was set to a callable other than our own, queue the
previous alarm settings.
"""
oldsec = signal.alarm(0)
oldfunc = signal.signal(signal.SIGALRM, __alarm_handler)
if oldsec > 0 and oldfunc != __alarm_handler:
heapq.heappush(alarmlist, (__new_alarm(oldsec, oldfunc, [], {})))
def __alarm_handler(*zargs):
"""Handle an alarm by calling any due heap entries and resetting the alarm.
Note that multiple heap entries might get called, especially if calling an
entry takes a lot of time.
"""
try:
nextt = __next_alarm()
while nextt is not None and nextt <= 0:
(tm, func, args, keys) = heapq.heappop(alarmlist)
func(*args, **keys)
nextt = __next_alarm()
finally:
if alarmlist: __set_alarm()
def alarm(sec, func, *args, **keys):
"""Set an alarm.
When the alarm is raised in `sec` seconds, the handler will call `func`,
passing `args` and `keys`. Return the heap entry (which is just a big
tuple), so that it can be cancelled by calling `cancel()`.
"""
__clear_alarm()
try:
newalarm = __new_alarm(sec, func, args, keys)
heapq.heappush(alarmlist, newalarm)
return newalarm
finally:
__set_alarm()
def cancel(alarm):
"""Cancel an alarm by passing the heap entry returned by `alarm()`.
It is an error to try to cancel an alarm which has already occurred.
"""
__clear_alarm()
try:
alarmlist.remove(alarm)
heapq.heapify(alarmlist)
finally:
if alarmlist: __set_alarm()
and a usage example:
import alarm
from time import sleep
try:
with alarm.Timeout(id_='a', seconds=5):
try:
with alarm.Timeout(id_='b', seconds=2):
sleep(3)
except alarm.TimeoutError as e:
print 'raised', e.id_
sleep(30)
except alarm.TimeoutError as e:
print 'raised', e.id_
else:
print 'nope.'
I have face the same problem but my situation is need work on sub thread, signal didn't work for me, so I wrote a python package: timeout-timer to solve this problem, support for use as context or decorator, use signal or sub thread module to trigger a timeout interrupt:
from timeout_timer import timeout, TimeoutInterrupt
class TimeoutInterruptNested(TimeoutInterrupt):
pass
def test_timeout_nested_loop_both_timeout(timer="thread"):
cnt = 0
try:
with timeout(5, timer=timer):
try:
with timeout(2, timer=timer, exception=TimeoutInterruptNested):
sleep(2)
except TimeoutInterruptNested:
cnt += 1
time.sleep(10)
except TimeoutInterrupt:
cnt += 1
assert cnt == 2
see more: https://github.com/dozysun/timeout-timer
Here is a simple example running one method with timeout and also retriev its value if successfull.
import multiprocessing
import time
ret = {"foo": False}
def worker(queue):
"""worker function"""
ret = queue.get()
time.sleep(1)
ret["foo"] = True
queue.put(ret)
if __name__ == "__main__":
queue = multiprocessing.Queue()
queue.put(ret)
p = multiprocessing.Process(target=worker, args=(queue,))
p.start()
p.join(timeout=10)
if p.exitcode is None:
print("The worker timed out.")
else:
print(f"The worker completed and returned: {queue.get()}")
Here is a slight improvement to the given thread-based solution.
The code below supports exceptions:
def runFunctionCatchExceptions(func, *args, **kwargs):
try:
result = func(*args, **kwargs)
except Exception, message:
return ["exception", message]
return ["RESULT", result]
def runFunctionWithTimeout(func, args=(), kwargs={}, timeout_duration=10, default=None):
import threading
class InterruptableThread(threading.Thread):
def __init__(self):
threading.Thread.__init__(self)
self.result = default
def run(self):
self.result = runFunctionCatchExceptions(func, *args, **kwargs)
it = InterruptableThread()
it.start()
it.join(timeout_duration)
if it.isAlive():
return default
if it.result[0] == "exception":
raise it.result[1]
return it.result[1]
Invoking it with a 5 second timeout:
result = timeout(remote_calculate, (myarg,), timeout_duration=5)
Here is a POSIX version that combines many of the previous answers to deliver following features:
Subprocesses blocking the execution.
Usage of the timeout function on class member functions.
Strict requirement on time-to-terminate.
Here is the code and some test cases:
import threading
import signal
import os
import time
class TerminateExecution(Exception):
"""
Exception to indicate that execution has exceeded the preset running time.
"""
def quit_function(pid):
# Killing all subprocesses
os.setpgrp()
os.killpg(0, signal.SIGTERM)
# Killing the main thread
os.kill(pid, signal.SIGTERM)
def handle_term(signum, frame):
raise TerminateExecution()
def invoke_with_timeout(timeout, fn, *args, **kwargs):
# Setting a sigterm handler and initiating a timer
old_handler = signal.signal(signal.SIGTERM, handle_term)
timer = threading.Timer(timeout, quit_function, args=[os.getpid()])
terminate = False
# Executing the function
timer.start()
try:
result = fn(*args, **kwargs)
except TerminateExecution:
terminate = True
finally:
# Restoring original handler and cancel timer
signal.signal(signal.SIGTERM, old_handler)
timer.cancel()
if terminate:
raise BaseException("xxx")
return result
### Test cases
def countdown(n):
print('countdown started', flush=True)
for i in range(n, -1, -1):
print(i, end=', ', flush=True)
time.sleep(1)
print('countdown finished')
return 1337
def really_long_function():
time.sleep(10)
def really_long_function2():
os.system("sleep 787")
# Checking that we can run a function as expected.
assert invoke_with_timeout(3, countdown, 1) == 1337
# Testing various scenarios
t1 = time.time()
try:
print(invoke_with_timeout(1, countdown, 3))
assert(False)
except BaseException:
assert(time.time() - t1 < 1.1)
print("All good", time.time() - t1)
t1 = time.time()
try:
print(invoke_with_timeout(1, really_long_function2))
assert(False)
except BaseException:
assert(time.time() - t1 < 1.1)
print("All good", time.time() - t1)
t1 = time.time()
try:
print(invoke_with_timeout(1, really_long_function))
assert(False)
except BaseException:
assert(time.time() - t1 < 1.1)
print("All good", time.time() - t1)
# Checking that classes are referenced and not
# copied (as would be the case with multiprocessing)
class X:
def __init__(self):
self.value = 0
def set(self, v):
self.value = v
x = X()
invoke_with_timeout(2, x.set, 9)
assert x.value == 9
I intend to kill the process if job not done , using thread and process both to achieve this.
from concurrent.futures import ThreadPoolExecutor
from time import sleep
import multiprocessing
# test case 1
def worker_1(a,b,c):
for _ in range(2):
print('very time consuming sleep')
sleep(1)
return a+b+c
# test case 2
def worker_2(in_name):
for _ in range(10):
print('very time consuming sleep')
sleep(1)
return 'hello '+in_name
Actual class as a contextmanager
class FuncTimer():
def __init__(self,fn,args,runtime):
self.fn = fn
self.args = args
self.queue = multiprocessing.Queue()
self.runtime = runtime
self.process = multiprocessing.Process(target=self.thread_caller)
def thread_caller(self):
with ThreadPoolExecutor() as executor:
future = executor.submit(self.fn, *self.args)
self.queue.put(future.result())
def __enter__(self):
return self
def start_run(self):
self.process.start()
self.process.join(timeout=self.runtime)
if self.process.exitcode is None:
self.process.kill()
if self.process.exitcode is None:
out_res = None
print('killed premature')
else:
out_res = self.queue.get()
return out_res
def __exit__(self, exc_type, exc_value, exc_traceback):
self.process.kill()
How to use it
print('testing case 1')
with FuncTimer(fn=worker_1,args=(1,2,3),runtime = 5) as fp:
res = fp.start_run()
print(res)
print('testing case 2')
with FuncTimer(fn=worker_2,args=('ram',),runtime = 5) as fp:
res = fp.start_run()
print(res)
This is the standard language-neutral approach
import logging
logger = logging.getLogger(__file__)
class SomeClass(object):
max_retry = 2
def get(self, key):
try:
return self.__get_value(key)
except Exception:
logger.error("Exception occured even after retrying...")
raise
def __get_value(self, key, retry_num=0):
try:
return self.connection().get(key)
except Exception:
logger.error("Exception occured")
if retry_num < self.max_retry:
retry_num += 1
logger.warning("Retrying!!! Retry count - %s", retry_num)
self.__get_value(key, retry_num)
else:
raise
Is there any better pythonic way to do this?
Cleaner approach would be not to change state of the class since it's retry just for the function call (current implementation wouldn't work as expected when the method is called multiple times). I'd prefer a retry decorator (as loop with break when succeed) used as:
...
#retry(n_times=2, catch=Exception)
def get (self, key):
...
I'm calling a function in Python which I know may stall and force me to restart the script.
How do I call the function or what do I wrap it in so that if it takes longer than 5 seconds the script cancels it and does something else?
You may use the signal package if you are running on UNIX:
In [1]: import signal
# Register an handler for the timeout
In [2]: def handler(signum, frame):
...: print("Forever is over!")
...: raise Exception("end of time")
...:
# This function *may* run for an indetermined time...
In [3]: def loop_forever():
...: import time
...: while 1:
...: print("sec")
...: time.sleep(1)
...:
...:
# Register the signal function handler
In [4]: signal.signal(signal.SIGALRM, handler)
Out[4]: 0
# Define a timeout for your function
In [5]: signal.alarm(10)
Out[5]: 0
In [6]: try:
...: loop_forever()
...: except Exception, exc:
...: print(exc)
....:
sec
sec
sec
sec
sec
sec
sec
sec
Forever is over!
end of time
# Cancel the timer if the function returned before timeout
# (ok, mine won't but yours maybe will :)
In [7]: signal.alarm(0)
Out[7]: 0
10 seconds after the call signal.alarm(10), the handler is called. This raises an exception that you can intercept from the regular Python code.
This module doesn't play well with threads (but then, who does?)
Note that since we raise an exception when timeout happens, it may end up caught and ignored inside the function, for example of one such function:
def loop_forever():
while 1:
print('sec')
try:
time.sleep(10)
except:
continue
You can use multiprocessing.Process to do exactly that.
Code
import multiprocessing
import time
# bar
def bar():
for i in range(100):
print "Tick"
time.sleep(1)
if __name__ == '__main__':
# Start bar as a process
p = multiprocessing.Process(target=bar)
p.start()
# Wait for 10 seconds or until process finishes
p.join(10)
# If thread is still active
if p.is_alive():
print "running... let's kill it..."
# Terminate - may not work if process is stuck for good
p.terminate()
# OR Kill - will work for sure, no chance for process to finish nicely however
# p.kill()
p.join()
How do I call the function or what do I wrap it in so that if it takes longer than 5 seconds the script cancels it?
I posted a gist that solves this question/problem with a decorator and a threading.Timer. Here it is with a breakdown.
Imports and setups for compatibility
It was tested with Python 2 and 3. It should also work under Unix/Linux and Windows.
First the imports. These attempt to keep the code consistent regardless of the Python version:
from __future__ import print_function
import sys
import threading
from time import sleep
try:
import thread
except ImportError:
import _thread as thread
Use version independent code:
try:
range, _print = xrange, print
def print(*args, **kwargs):
flush = kwargs.pop('flush', False)
_print(*args, **kwargs)
if flush:
kwargs.get('file', sys.stdout).flush()
except NameError:
pass
Now we have imported our functionality from the standard library.
exit_after decorator
Next we need a function to terminate the main() from the child thread:
def quit_function(fn_name):
# print to stderr, unbuffered in Python 2.
print('{0} took too long'.format(fn_name), file=sys.stderr)
sys.stderr.flush() # Python 3 stderr is likely buffered.
thread.interrupt_main() # raises KeyboardInterrupt
And here is the decorator itself:
def exit_after(s):
'''
use as decorator to exit process if
function takes longer than s seconds
'''
def outer(fn):
def inner(*args, **kwargs):
timer = threading.Timer(s, quit_function, args=[fn.__name__])
timer.start()
try:
result = fn(*args, **kwargs)
finally:
timer.cancel()
return result
return inner
return outer
Usage
And here's the usage that directly answers your question about exiting after 5 seconds!:
#exit_after(5)
def countdown(n):
print('countdown started', flush=True)
for i in range(n, -1, -1):
print(i, end=', ', flush=True)
sleep(1)
print('countdown finished')
Demo:
>>> countdown(3)
countdown started
3, 2, 1, 0, countdown finished
>>> countdown(10)
countdown started
10, 9, 8, 7, 6, countdown took too long
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 11, in inner
File "<stdin>", line 6, in countdown
KeyboardInterrupt
The second function call will not finish, instead the process should exit with a traceback!
KeyboardInterrupt does not always stop a sleeping thread
Note that sleep will not always be interrupted by a keyboard interrupt, on Python 2 on Windows, e.g.:
#exit_after(1)
def sleep10():
sleep(10)
print('slept 10 seconds')
>>> sleep10()
sleep10 took too long # Note that it hangs here about 9 more seconds
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 11, in inner
File "<stdin>", line 3, in sleep10
KeyboardInterrupt
nor is it likely to interrupt code running in extensions unless it explicitly checks for PyErr_CheckSignals(), see Cython, Python and KeyboardInterrupt ignored
I would avoid sleeping a thread more than a second, in any case - that's an eon in processor time.
How do I call the function or what do I wrap it in so that if it takes longer than 5 seconds the script cancels it and does something else?
To catch it and do something else, you can catch the KeyboardInterrupt.
>>> try:
... countdown(10)
... except KeyboardInterrupt:
... print('do something else')
...
countdown started
10, 9, 8, 7, 6, countdown took too long
do something else
I have a different proposal which is a pure function (with the same API as the threading suggestion) and seems to work fine (based on suggestions on this thread)
def timeout(func, args=(), kwargs={}, timeout_duration=1, default=None):
import signal
class TimeoutError(Exception):
pass
def handler(signum, frame):
raise TimeoutError()
# set the timeout handler
signal.signal(signal.SIGALRM, handler)
signal.alarm(timeout_duration)
try:
result = func(*args, **kwargs)
except TimeoutError as exc:
result = default
finally:
signal.alarm(0)
return result
I ran across this thread when searching for a timeout call on unit tests. I didn't find anything simple in the answers or 3rd party packages so I wrote the decorator below you can drop right into code:
import multiprocessing.pool
import functools
def timeout(max_timeout):
"""Timeout decorator, parameter in seconds."""
def timeout_decorator(item):
"""Wrap the original function."""
#functools.wraps(item)
def func_wrapper(*args, **kwargs):
"""Closure for function."""
pool = multiprocessing.pool.ThreadPool(processes=1)
async_result = pool.apply_async(item, args, kwargs)
# raises a TimeoutError if execution exceeds max_timeout
return async_result.get(max_timeout)
return func_wrapper
return timeout_decorator
Then it's as simple as this to timeout a test or any function you like:
#timeout(5.0) # if execution takes longer than 5 seconds, raise a TimeoutError
def test_base_regression(self):
...
The stopit package, found on pypi, seems to handle timeouts well.
I like the #stopit.threading_timeoutable decorator, which adds a timeout parameter to the decorated function, which does what you expect, it stops the function.
Check it out on pypi: https://pypi.python.org/pypi/stopit
I am the author of wrapt_timeout_decorator.
Most of the solutions presented here work wunderfully under Linux on the first glance - because we have fork() and signals() - but on windows the things look a bit different.
And when it comes to subthreads on Linux, You cant use Signals anymore.
In order to spawn a process under Windows, it needs to be picklable - and many decorated functions or Class methods are not.
So you need to use a better pickler like dill and multiprocess (not pickle and multiprocessing) - thats why You cant use ProcessPoolExecutor (or only with limited functionality).
For the timeout itself - You need to define what timeout means - because on Windows it will take considerable (and not determinable) time to spawn the process. This can be tricky on short timeouts. Lets assume, spawning the process takes about 0.5 seconds (easily !!!). If You give a timeout of 0.2 seconds what should happen?
Should the function time out after 0.5 + 0.2 seconds (so let the method run for 0.2 seconds)?
Or should the called process time out after 0.2 seconds (in that case, the decorated function will ALWAYS timeout, because in that time it is not even spawned)?
Also nested decorators can be nasty and You cant use Signals in a subthread. If You want to create a truly universal, cross-platform decorator, all this needs to be taken into consideration (and tested).
Other issues are passing exceptions back to the caller, as well as logging issues (if used in the decorated function - logging to files in another process is NOT supported)
I tried to cover all edge cases, You might look into the package wrapt_timeout_decorator, or at least test Your own solutions inspired by the unittests used there.
#Alexis Eggermont - unfortunately I dont have enough points to comment - maybe someone else can notify You - I think I solved Your import issue.
There are a lot of suggestions, but none using concurrent.futures, which I think is the most legible way to handle this.
from concurrent.futures import ProcessPoolExecutor
# Warning: this does not terminate function if timeout
def timeout_five(fnc, *args, **kwargs):
with ProcessPoolExecutor() as p:
f = p.submit(fnc, *args, **kwargs)
return f.result(timeout=5)
Super simple to read and maintain.
We make a pool, submit a single process and then wait up to 5 seconds before raising a TimeoutError that you could catch and handle however you needed.
Native to python 3.2+ and backported to 2.7 (pip install futures).
Switching between threads and processes is as simple as replacing ProcessPoolExecutor with ThreadPoolExecutor.
If you want to terminate the Process on timeout I would suggest looking into Pebble.
Building on and and enhancing the answer by #piro , you can build a contextmanager. This allows for very readable code which will disable the alaram signal after a successful run (sets signal.alarm(0))
from contextlib import contextmanager
import signal
import time
#contextmanager
def timeout(duration):
def timeout_handler(signum, frame):
raise TimeoutError(f'block timedout after {duration} seconds')
signal.signal(signal.SIGALRM, timeout_handler)
signal.alarm(duration)
try:
yield
finally:
signal.alarm(0)
def sleeper(duration):
time.sleep(duration)
print('finished')
Example usage:
In [19]: with timeout(2):
...: sleeper(1)
...:
finished
In [20]: with timeout(2):
...: sleeper(3)
...:
---------------------------------------------------------------------------
Exception Traceback (most recent call last)
<ipython-input-20-66c78858116f> in <module>()
1 with timeout(2):
----> 2 sleeper(3)
3
<ipython-input-7-a75b966bf7ac> in sleeper(t)
1 def sleeper(t):
----> 2 time.sleep(t)
3 print('finished')
4
<ipython-input-18-533b9e684466> in timeout_handler(signum, frame)
2 def timeout(duration):
3 def timeout_handler(signum, frame):
----> 4 raise Exception(f'block timedout after {duration} seconds')
5 signal.signal(signal.SIGALRM, timeout_handler)
6 signal.alarm(duration)
Exception: block timedout after 2 seconds
Great, easy to use and reliable PyPi project timeout-decorator (https://pypi.org/project/timeout-decorator/)
installation:
pip install timeout-decorator
Usage:
import time
import timeout_decorator
#timeout_decorator.timeout(5)
def mytest():
print "Start"
for i in range(1,10):
time.sleep(1)
print "%d seconds have passed" % i
if __name__ == '__main__':
mytest()
timeout-decorator don't work on windows system as , windows didn't support signal well.
If you use timeout-decorator in windows system you will get the following
AttributeError: module 'signal' has no attribute 'SIGALRM'
Some suggested to use use_signals=False but didn't worked for me.
Author #bitranox created the following package:
pip install https://github.com/bitranox/wrapt-timeout-decorator/archive/master.zip
Code Sample:
import time
from wrapt_timeout_decorator import *
#timeout(5)
def mytest(message):
print(message)
for i in range(1,10):
time.sleep(1)
print('{} seconds have passed'.format(i))
def main():
mytest('starting')
if __name__ == '__main__':
main()
Gives the following exception:
TimeoutError: Function mytest timed out after 5 seconds
Highlights
Raises TimeoutError uses exceptions to alert on timeout - can easily be modified
Cross Platform: Windows & Mac OS X
Compatibility: Python 3.6+ (I also tested on python 2.7 and it works with small syntax adjustments)
For full explanation and extension to parallel maps, see here https://flipdazed.github.io/blog/quant%20dev/parallel-functions-with-timeouts
Minimal Example
>>> #killer_call(timeout=4)
... def bar(x):
... import time
... time.sleep(x)
... return x
>>> bar(10)
Traceback (most recent call last):
...
__main__.TimeoutError: function 'bar' timed out after 4s
and as expected
>>> bar(2)
2
Full code
import multiprocessing as mp
import multiprocessing.queues as mpq
import functools
import dill
from typing import Tuple, Callable, Dict, Optional, Iterable, List, Any
class TimeoutError(Exception):
def __init__(self, func: Callable, timeout: int):
self.t = timeout
self.fname = func.__name__
def __str__(self):
return f"function '{self.fname}' timed out after {self.t}s"
def _lemmiwinks(func: Callable, args: Tuple, kwargs: Dict[str, Any], q: mp.Queue):
"""lemmiwinks crawls into the unknown"""
q.put(dill.loads(func)(*args, **kwargs))
def killer_call(func: Callable = None, timeout: int = 10) -> Callable:
"""
Single function call with a timeout
Args:
func: the function
timeout: The timeout in seconds
"""
if not isinstance(timeout, int):
raise ValueError(f'timeout needs to be an int. Got: {timeout}')
if func is None:
return functools.partial(killer_call, timeout=timeout)
#functools.wraps(killer_call)
def _inners(*args, **kwargs) -> Any:
q_worker = mp.Queue()
proc = mp.Process(target=_lemmiwinks, args=(dill.dumps(func), args, kwargs, q_worker))
proc.start()
try:
return q_worker.get(timeout=timeout)
except mpq.Empty:
raise TimeoutError(func, timeout)
finally:
try:
proc.terminate()
except:
pass
return _inners
if __name__ == '__main__':
#killer_call(timeout=4)
def bar(x):
import time
time.sleep(x)
return x
print(bar(2))
bar(10)
Notes
You will need to import inside the function because of the way dill works.
This will also mean these functions may not be not compatible with doctest if there are imports inside your target functions. You will get an issue with __import__ not found.
Just in case it is helpful for anyone, building on the answer by #piro, I've made a function decorator:
import time
import signal
from functools import wraps
def timeout(timeout_secs: int):
def wrapper(func):
#wraps(func)
def time_limited(*args, **kwargs):
# Register an handler for the timeout
def handler(signum, frame):
raise Exception(f"Timeout for function '{func.__name__}'")
# Register the signal function handler
signal.signal(signal.SIGALRM, handler)
# Define a timeout for your function
signal.alarm(timeout_secs)
result = None
try:
result = func(*args, **kwargs)
except Exception as exc:
raise exc
finally:
# disable the signal alarm
signal.alarm(0)
return result
return time_limited
return wrapper
Using the wrapper on a function with a 20 seconds timeout would look something like:
#timeout(20)
def my_slow_or_never_ending_function(name):
while True:
time.sleep(1)
print(f"Yet another second passed {name}...")
try:
results = my_slow_or_never_ending_function("Yooo!")
except Exception as e:
print(f"ERROR: {e}")
We can use signals for the same. I think the below example will be useful for you. It is very simple compared to threads.
import signal
def timeout(signum, frame):
raise myException
#this is an infinite loop, never ending under normal circumstances
def main():
print 'Starting Main ',
while 1:
print 'in main ',
#SIGALRM is only usable on a unix platform
signal.signal(signal.SIGALRM, timeout)
#change 5 to however many seconds you need
signal.alarm(5)
try:
main()
except myException:
print "whoops"
Another solution with asyncio :
If you want to cancel the background task and not just timeout on the running main code, then you need an explicit communication from main thread to ask the code of the task to cancel , like a threading.Event()
import asyncio
import functools
import multiprocessing
from concurrent.futures.thread import ThreadPoolExecutor
class SingletonTimeOut:
pool = None
#classmethod
def run(cls, to_run: functools.partial, timeout: float):
pool = cls.get_pool()
loop = cls.get_loop()
try:
task = loop.run_in_executor(pool, to_run)
return loop.run_until_complete(asyncio.wait_for(task, timeout=timeout))
except asyncio.TimeoutError as e:
error_type = type(e).__name__ #TODO
raise e
#classmethod
def get_pool(cls):
if cls.pool is None:
cls.pool = ThreadPoolExecutor(multiprocessing.cpu_count())
return cls.pool
#classmethod
def get_loop(cls):
try:
return asyncio.get_event_loop()
except RuntimeError:
asyncio.set_event_loop(asyncio.new_event_loop())
# print("NEW LOOP" + str(threading.current_thread().ident))
return asyncio.get_event_loop()
# ---------------
TIME_OUT = float('0.2') # seconds
def toto(input_items,nb_predictions):
return 1
to_run = functools.partial(toto,
input_items=1,
nb_predictions="a")
results = SingletonTimeOut.run(to_run, TIME_OUT)
#!/usr/bin/python2
import sys, subprocess, threading
proc = subprocess.Popen(sys.argv[2:])
timer = threading.Timer(float(sys.argv[1]), proc.terminate)
timer.start()
proc.wait()
timer.cancel()
exit(proc.returncode)
The func_timeout package by Tim Savannah has worked well for me.
Installation:
pip install func_timeout
Usage:
import time
from func_timeout import func_timeout, FunctionTimedOut
def my_func(n):
time.sleep(n)
time_to_sleep = 10
# time out after 2 seconds using kwargs
func_timeout(2, my_func, kwargs={'n' : time_to_sleep})
# time out after 2 seconds using args
func_timeout(2, my_func, args=(time_to_sleep,))
I had a need for nestable timed interrupts (which SIGALARM can't do) that won't get blocked by time.sleep (which the thread-based approach can't do). I ended up copying and lightly modifying code from here: http://code.activestate.com/recipes/577600-queue-for-managing-multiple-sigalrm-alarms-concurr/
The code itself:
#!/usr/bin/python
# lightly modified version of http://code.activestate.com/recipes/577600-queue-for-managing-multiple-sigalrm-alarms-concurr/
"""alarm.py: Permits multiple SIGALRM events to be queued.
Uses a `heapq` to store the objects to be called when an alarm signal is
raised, so that the next alarm is always at the top of the heap.
"""
import heapq
import signal
from time import time
__version__ = '$Revision: 2539 $'.split()[1]
alarmlist = []
__new_alarm = lambda t, f, a, k: (t + time(), f, a, k)
__next_alarm = lambda: int(round(alarmlist[0][0] - time())) if alarmlist else None
__set_alarm = lambda: signal.alarm(max(__next_alarm(), 1))
class TimeoutError(Exception):
def __init__(self, message, id_=None):
self.message = message
self.id_ = id_
class Timeout:
''' id_ allows for nested timeouts. '''
def __init__(self, id_=None, seconds=1, error_message='Timeout'):
self.seconds = seconds
self.error_message = error_message
self.id_ = id_
def handle_timeout(self):
raise TimeoutError(self.error_message, self.id_)
def __enter__(self):
self.this_alarm = alarm(self.seconds, self.handle_timeout)
def __exit__(self, type, value, traceback):
try:
cancel(self.this_alarm)
except ValueError:
pass
def __clear_alarm():
"""Clear an existing alarm.
If the alarm signal was set to a callable other than our own, queue the
previous alarm settings.
"""
oldsec = signal.alarm(0)
oldfunc = signal.signal(signal.SIGALRM, __alarm_handler)
if oldsec > 0 and oldfunc != __alarm_handler:
heapq.heappush(alarmlist, (__new_alarm(oldsec, oldfunc, [], {})))
def __alarm_handler(*zargs):
"""Handle an alarm by calling any due heap entries and resetting the alarm.
Note that multiple heap entries might get called, especially if calling an
entry takes a lot of time.
"""
try:
nextt = __next_alarm()
while nextt is not None and nextt <= 0:
(tm, func, args, keys) = heapq.heappop(alarmlist)
func(*args, **keys)
nextt = __next_alarm()
finally:
if alarmlist: __set_alarm()
def alarm(sec, func, *args, **keys):
"""Set an alarm.
When the alarm is raised in `sec` seconds, the handler will call `func`,
passing `args` and `keys`. Return the heap entry (which is just a big
tuple), so that it can be cancelled by calling `cancel()`.
"""
__clear_alarm()
try:
newalarm = __new_alarm(sec, func, args, keys)
heapq.heappush(alarmlist, newalarm)
return newalarm
finally:
__set_alarm()
def cancel(alarm):
"""Cancel an alarm by passing the heap entry returned by `alarm()`.
It is an error to try to cancel an alarm which has already occurred.
"""
__clear_alarm()
try:
alarmlist.remove(alarm)
heapq.heapify(alarmlist)
finally:
if alarmlist: __set_alarm()
and a usage example:
import alarm
from time import sleep
try:
with alarm.Timeout(id_='a', seconds=5):
try:
with alarm.Timeout(id_='b', seconds=2):
sleep(3)
except alarm.TimeoutError as e:
print 'raised', e.id_
sleep(30)
except alarm.TimeoutError as e:
print 'raised', e.id_
else:
print 'nope.'
I have face the same problem but my situation is need work on sub thread, signal didn't work for me, so I wrote a python package: timeout-timer to solve this problem, support for use as context or decorator, use signal or sub thread module to trigger a timeout interrupt:
from timeout_timer import timeout, TimeoutInterrupt
class TimeoutInterruptNested(TimeoutInterrupt):
pass
def test_timeout_nested_loop_both_timeout(timer="thread"):
cnt = 0
try:
with timeout(5, timer=timer):
try:
with timeout(2, timer=timer, exception=TimeoutInterruptNested):
sleep(2)
except TimeoutInterruptNested:
cnt += 1
time.sleep(10)
except TimeoutInterrupt:
cnt += 1
assert cnt == 2
see more: https://github.com/dozysun/timeout-timer
Here is a simple example running one method with timeout and also retriev its value if successfull.
import multiprocessing
import time
ret = {"foo": False}
def worker(queue):
"""worker function"""
ret = queue.get()
time.sleep(1)
ret["foo"] = True
queue.put(ret)
if __name__ == "__main__":
queue = multiprocessing.Queue()
queue.put(ret)
p = multiprocessing.Process(target=worker, args=(queue,))
p.start()
p.join(timeout=10)
if p.exitcode is None:
print("The worker timed out.")
else:
print(f"The worker completed and returned: {queue.get()}")
Here is a slight improvement to the given thread-based solution.
The code below supports exceptions:
def runFunctionCatchExceptions(func, *args, **kwargs):
try:
result = func(*args, **kwargs)
except Exception, message:
return ["exception", message]
return ["RESULT", result]
def runFunctionWithTimeout(func, args=(), kwargs={}, timeout_duration=10, default=None):
import threading
class InterruptableThread(threading.Thread):
def __init__(self):
threading.Thread.__init__(self)
self.result = default
def run(self):
self.result = runFunctionCatchExceptions(func, *args, **kwargs)
it = InterruptableThread()
it.start()
it.join(timeout_duration)
if it.isAlive():
return default
if it.result[0] == "exception":
raise it.result[1]
return it.result[1]
Invoking it with a 5 second timeout:
result = timeout(remote_calculate, (myarg,), timeout_duration=5)
Here is a POSIX version that combines many of the previous answers to deliver following features:
Subprocesses blocking the execution.
Usage of the timeout function on class member functions.
Strict requirement on time-to-terminate.
Here is the code and some test cases:
import threading
import signal
import os
import time
class TerminateExecution(Exception):
"""
Exception to indicate that execution has exceeded the preset running time.
"""
def quit_function(pid):
# Killing all subprocesses
os.setpgrp()
os.killpg(0, signal.SIGTERM)
# Killing the main thread
os.kill(pid, signal.SIGTERM)
def handle_term(signum, frame):
raise TerminateExecution()
def invoke_with_timeout(timeout, fn, *args, **kwargs):
# Setting a sigterm handler and initiating a timer
old_handler = signal.signal(signal.SIGTERM, handle_term)
timer = threading.Timer(timeout, quit_function, args=[os.getpid()])
terminate = False
# Executing the function
timer.start()
try:
result = fn(*args, **kwargs)
except TerminateExecution:
terminate = True
finally:
# Restoring original handler and cancel timer
signal.signal(signal.SIGTERM, old_handler)
timer.cancel()
if terminate:
raise BaseException("xxx")
return result
### Test cases
def countdown(n):
print('countdown started', flush=True)
for i in range(n, -1, -1):
print(i, end=', ', flush=True)
time.sleep(1)
print('countdown finished')
return 1337
def really_long_function():
time.sleep(10)
def really_long_function2():
os.system("sleep 787")
# Checking that we can run a function as expected.
assert invoke_with_timeout(3, countdown, 1) == 1337
# Testing various scenarios
t1 = time.time()
try:
print(invoke_with_timeout(1, countdown, 3))
assert(False)
except BaseException:
assert(time.time() - t1 < 1.1)
print("All good", time.time() - t1)
t1 = time.time()
try:
print(invoke_with_timeout(1, really_long_function2))
assert(False)
except BaseException:
assert(time.time() - t1 < 1.1)
print("All good", time.time() - t1)
t1 = time.time()
try:
print(invoke_with_timeout(1, really_long_function))
assert(False)
except BaseException:
assert(time.time() - t1 < 1.1)
print("All good", time.time() - t1)
# Checking that classes are referenced and not
# copied (as would be the case with multiprocessing)
class X:
def __init__(self):
self.value = 0
def set(self, v):
self.value = v
x = X()
invoke_with_timeout(2, x.set, 9)
assert x.value == 9
I intend to kill the process if job not done , using thread and process both to achieve this.
from concurrent.futures import ThreadPoolExecutor
from time import sleep
import multiprocessing
# test case 1
def worker_1(a,b,c):
for _ in range(2):
print('very time consuming sleep')
sleep(1)
return a+b+c
# test case 2
def worker_2(in_name):
for _ in range(10):
print('very time consuming sleep')
sleep(1)
return 'hello '+in_name
Actual class as a contextmanager
class FuncTimer():
def __init__(self,fn,args,runtime):
self.fn = fn
self.args = args
self.queue = multiprocessing.Queue()
self.runtime = runtime
self.process = multiprocessing.Process(target=self.thread_caller)
def thread_caller(self):
with ThreadPoolExecutor() as executor:
future = executor.submit(self.fn, *self.args)
self.queue.put(future.result())
def __enter__(self):
return self
def start_run(self):
self.process.start()
self.process.join(timeout=self.runtime)
if self.process.exitcode is None:
self.process.kill()
if self.process.exitcode is None:
out_res = None
print('killed premature')
else:
out_res = self.queue.get()
return out_res
def __exit__(self, exc_type, exc_value, exc_traceback):
self.process.kill()
How to use it
print('testing case 1')
with FuncTimer(fn=worker_1,args=(1,2,3),runtime = 5) as fp:
res = fp.start_run()
print(res)
print('testing case 2')
with FuncTimer(fn=worker_2,args=('ram',),runtime = 5) as fp:
res = fp.start_run()
print(res)
I'm trying to put a timeout on a function send.
I have found some elements in these posts :
https://stackoverflow.com/a/494273/3824723
https://stackoverflow.com/a/2282656/3824723
https://stackoverflow.com/a/11731208/3824723
The first one seems to apply to every function and not to a precise one, this is why I chose a decorator implementation like the second one.
I tried to mix it up and I have this :
from functools import wraps
import os
import signal
class TimeoutError(Exception):
pass
def timeout_func(error_message="Timeout in send pipe!"):
def decorator(func):
def _handle_timeout(signum, frame):
if args[0].action=="warn":
print "WARNING : ",error_message
elif args[0].action=="kill":
raise TimeoutError(error_message)
def wrapper(*args, **kwargs):
print args
signal.signal(signal.SIGALRM, _handle_timeout,args[0].action)
signal.alarm(args[0].seconds)
print str(args)
try:
result = func(*args, **kwargs)
finally:
signal.alarm(0)
return result
return wraps(func)(wrapper)
return decorator
class Link(object):
def __init__(self,timeout=1,action="warn"):
self.timeout=timeout
self.action=action
#timeout_func
def send(self,value):
print "working : ", value
It gives me this :
In [6]: l=Link()
In [7]: l.send(1)
--------------------------------------------------------------------------- TypeError Traceback (most recent call
last) in ()
----> 1 l.send(1)
TypeError: decorator() takes exactly 1 argument (2 given)
My issue is that I would like to pass the timeout value second to the decorator through the Link's self. I don't fully understand the whole decorator mechanism here, and can't figure out what is wrong.
Can someone explain me how this decorator works, and what should I modify to fix it? Or if you think of a simpler/more explicit solution to implement it ?
So I've been debugging my issue, and I found a working solution:
from functools import wraps
import signal
class TimeoutError(Exception):
pass
def timeout_func(f):
def _handle_timeout(signum, frame):
raise TimeoutError("timeout error")
def wrapper(*args):
signal.signal(signal.SIGALRM, _handle_timeout)
signal.setitimer(signal.ITIMER_REAL,args[0].timeout) #args[0] is self of Link class here
try:
result = f(*args,**kwargs)
finally:
signal.alarm(0)
return result
return wrapper
class Link(object):
def __init__(self,timeout=0.1,action="warn"):
self.timeout=timeout
self.action=action
def send(self,value): # I use this func to handle the exceptions
try:
self.send_timeout(value) # call the timeout function
except TimeoutError as e: # handle Timeout exception
if self.action=="warn":
print "WARNING : Timeout error in pipe send!"
elif self.action=="kill":
print "Killing Link : ", e
raise
except (Exception,KeyboardInterrupt) as e:
print "Exception in link : ", e
raise
#timeout_func # apply timeout decorator to this function
def send_timeout(self,value):
# DO STUFF HERE
To call it :
l=Link()
l.send("any value")
I use signal.setitimer(signal.ITIMER_REAL,args[0].timeout) because it allows setting a timeout < 1 second, wich was not the case with signal.signal(), wich only accept integer as timer.
Background and question
I'm using Django 1.5.1 and django-celery 3.0.17. I want to write a custom decorator to ensure that only one instance of the function runs at a time, similar to this but without all the repeated try/finally.
How can I write a decorator so that Celery methods like apply and delay can be called with arguments?
Others have created such decorators with apparent success. What am I missing?
Implementations
I've tried writing the decorator both a function and as a class, but with either implementation, when attempting to decorate a function with both my decorator and #celery.task, the arguments are not passed to the decorated function, resulting in the error message:
TypeError: foo() takes exactly 1 argument (0 given), where foo is the name of the decorated function.
Functional implementation
# util.py
from functools import wraps
from django.core.cache import get_cache
cache = get_cache('filesystem')
def cache_lock(lock_id, timeout=cache.get('TIMEOUT', 720)):
def _decorator(func):
try:
timeout_secs = timeout.total_seconds()
except AttributeError:
# Timeout is None (forever) or number of seconds.
timeout_secs = timeout
acquire_lock = lambda: cache.add(lock_id, 'true', timeout_secs) if timeout_secs else cache.add(lock_id, 'true')
release_lock = lambda: cache.delete(lock_id)
#wraps(func)
def _apply_lock(*args, **kwargs):
if acquire_lock():
try:
return func(*args, **kwargs)
finally:
release_lock()
else:
return False
return _apply_lock
return _decorator
Class-based implementation
# util.py
from functools import wraps
from django.core.cache import get_cache
cache = get_cache('filesystem')
class cache_lock(object):
def __init__(self, lock_id, timeout=cache.get('TIMEOUT', 720)):
self.lock_id = lock_id
self.timeout = timeout
def __call__(self, func):
try:
timeout_secs = self.timeout.total_seconds()
except AttributeError:
# Timeout is None (forever) or number of seconds.
timeout_secs = self.timeout
acquire_lock = lambda: cache.add(self.lock_id, 'true', timeout_secs) if timeout_secs else cache.add(self.lock_id, 'true')
release_lock = lambda: cache.delete(self.lock_id)
#wraps(func)
def _apply_lock(*args, **kwargs):
if acquire_lock():
try:
return func(*args, **kwargs)
finally:
release_lock()
else:
return False
return _apply_lock
Test case
For both implementations, the first test method succeeds and the second fails.
# tests.py
from datetime import timedelta
from celery import task # using celery.task.task does not help
from django.test import TestCase
from django.test.utils import override_settings
from .util import cache_lock
class UtilTests(TestCase):
def test_cache_lock_without_celery(self):
#cache_lock('recursive', timedelta(seconds=1))
def call_count(i):
self.assertFalse(call_count(i + 1))
return i + 1
self.assertEqual(call_count(0), 1) # succeeds
celery_settings = {
'CELERY_ALWAYS_EAGER': True,
'CELERY_EAGER_PROPAGATES_EXCEPTIONS': True,
'DEBUG': True,
}
#override_settings(**celery_settings)
def test_cache_lock_with_celery(self):
#task(name='test_cache_lock_with_celery')
#cache_lock('recursive', timedelta(seconds=600))
def call_count(i):
self.assertFalse(call_count.apply(i + 1).result)
return i + 1
self.assertEqual(call_count.apply(0).result, 1) # fails!