Reputation: 8945
How to set a limit on the maximum amount of time a function can run for?
For example, using time.sleep
as a placeholder function, how to limit the amount of time time.sleep
can run to a maximum of 5 mins (300 seconds)?
import time
try:
# As noted above `time.sleep` is a placeholder for a function
# which takes 10 minutes to complete.
time.sleep(600)
except:
print('took too long')
That is, how can time.sleep(600)
above be limited and interrupted after 300 seconds?
Upvotes: 3
Views: 7133
Reputation: 363546
On POSIX you have a simple and clean solution available in signal
module.
import signal
import time
class Timeout(Exception):
pass
def handler(sig, frame):
raise Timeout
signal.signal(signal.SIGALRM, handler) # register interest in SIGALRM events
signal.alarm(2) # timeout in 2 seconds
try:
time.sleep(60)
except Timeout:
print('took too long')
Caveats:
For other readers where the caveats above are a deal breaker, you will need a more heavyweight approach. The best option is usually to run the code in a separate process (or possibly a thread), and terminate that process if it takes too long. See multiprocessing
module for example.
Upvotes: 4
Reputation:
One of currently probably preferred options to accomplish what you want is usage of Pythons
multiprocessing ( especially its proc.join(timeoutTime) method )
module ( see tutorial )
Just copy/paste the code example below and run it. Does it what you are after?
def beBusyFor(noOfSeconds):
import time
print(" beBusyFor() message: going to rest for", noOfSeconds, "seconds")
time.sleep(noOfSeconds)
print(" beBusyFor() message: was resting", noOfSeconds, "seconds, now AWAKE")
import multiprocessing
noOfSecondsBusy = 5; timeoutTime = 3
print("--- noOfSecondsBusy = 5; timeoutTime = 3 ---")
proc = multiprocessing.Process(target=beBusyFor, args=(noOfSecondsBusy, ))
print("Start beBusyFor()")
proc.start()
print("beBusyFor() is running")
proc.join(timeoutTime)
if proc.is_alive():
print(timeoutTime, "seconds passed, beBusyFor() still running, terminate()" )
proc.terminate()
else:
print("OK, beBusyFor() has finished its work in time.")
#:if
print()
noOfSecondsBusy = 2; timeoutTime = 3
print("--- noOfSecondsBusy = 2; timeoutTime = 3 ---")
proc = multiprocessing.Process(target=beBusyFor, args=(noOfSecondsBusy, ))
print("Start beBusyFor()")
proc.start()
print("beBusyFor() started")
proc.join(timeoutTime)
if proc.is_alive():
print(timeoutTime, "seconds passed, beBusyFor() still running, terminate()" )
proc.terminate()
else:
print("OK, beBusyFor() has finished its work in time.")
#:if
it outputs:
--- noOfSecondsBusy = 5; timeoutTime = 3 ---
Start beBusyFor()
beBusyFor() is running
beBusyFor() message: going to rest for 5 seconds
3 seconds passed, beBusyFor() still running, terminate()
--- noOfSecondsBusy = 2; timeoutTime = 3 ---
Start beBusyFor()
beBusyFor() started
beBusyFor() message: going to rest for 2 seconds
beBusyFor() message: was resting 2 seconds, now AWAKE
OK, beBusyFor() has finished its work in time.
Another known to me option is using a
decorator function and the signal module
Checkout the web page with origin of the code I have provided here (only one small adjustment was necessary to make it run on Python 3.6):
import signal
class TimeoutError(Exception):
def __init__(self, value = "Timed Out"):
self.value = value
def __str__(self):
return repr(self.value)
def timeout(seconds_before_timeout):
def decorate(f):
def handler(signum, frame):
raise TimeoutError()
def new_f(*args, **kwargs):
old = signal.signal(signal.SIGALRM, handler)
signal.alarm(seconds_before_timeout)
try:
result = f(*args, **kwargs)
finally:
signal.signal(signal.SIGALRM, old)
signal.alarm(0)
return result
# new_f.func_name = f.func_name
new_f.__name__ = f.__name__
return new_f
return decorate
# Try it out:
import time
@timeout(5)
def mytest():
print( "mytest() message: Started" )
for i in range(1,10):
time.sleep(1)
print( "mytest() message: %d seconds have passed" % i )
try:
mytest()
except TimeoutError as e:
print("stopped executing mytest() because it", e)
print("continuing script execution past call of mytest()")
The code above outputs:
mytest() message: Started
mytest() message: 1 seconds have passed
mytest() message: 2 seconds have passed
mytest() message: 3 seconds have passed
mytest() message: 4 seconds have passed
stopped executing mytest() because it 'Timed Out'
continuing script execution past call of mytest()
Upvotes: 2