What makes this regular expression very slow? [duplicate] - python

Specification of the problem:
I'm searching through really great amount of lines of a log file and I'm distributing those lines to groups in order to regular expressions(RegExses) I have stored using the re.match() function. Unfortunately some of my RegExses are too complicated and Python sometimes gets himself to backtracking hell. Due to this I need to protect it with some kind of timeout.
Problems:
re.match, I'm using, is Python's function and as I found out somewhere here on StackOverflow (I'm really sorry, I can not find the link now :-( ). It is very difficult to interrupt thread with running Python's library. For this reason threads are out of the game.
Because evaluating of re.match function takes relatively short time and I want to analyse with this function great amount of lines, I need some timeout function that wont't take too long to execute (this makes threads even less suitable, it takes really long time to initialise new thread) and can be set to less than one second.
For those reasons, answers here - Timeout on a function call
and here - Timeout function if it takes too long to finish with decorator (alarm - 1sec and more) are off the table.
I've spent this morning searching for solution to this question but I did not find any satisfactory answer.

Solution:
I've just modified a script posted here: Timeout function if it takes too long to finish.
And here is the code:
from functools import wraps
import errno
import os
import signal
class TimeoutError(Exception):
pass
def timeout(seconds=10, error_message=os.strerror(errno.ETIME)):
def decorator(func):
def _handle_timeout(signum, frame):
raise TimeoutError(error_message)
def wrapper(*args, **kwargs):
signal.signal(signal.SIGALRM, _handle_timeout)
signal.setitimer(signal.ITIMER_REAL,seconds) #used timer instead of alarm
try:
result = func(*args, **kwargs)
finally:
signal.alarm(0)
return result
return wraps(func)(wrapper)
return decorator
And then you can use it like this:
from timeout import timeout
from time import time
#timeout(0.01)
def loop():
while True:
pass
try:
begin = time.time()
loop()
except TimeoutError, e:
print "Time elapsed: {:.3f}s".format(time.time() - begin)
Which prints
Time elapsed: 0.010s

Related

Training a model based on time rather than epochs [duplicate]

In Python, for a toy example:
for x in range(0, 3):
# Call function A(x)
I want to continue the for loop if function A takes more than five seconds by skipping it so I won't get stuck or waste time.
By doing some search, I realized a subprocess or thread may help, but I have no idea how to implement it here.
I think creating a new process may be overkill. If you're on Mac or a Unix-based system, you should be able to use signal.SIGALRM to forcibly time out functions that take too long. This will work on functions that are idling for network or other issues that you absolutely can't handle by modifying your function. I have an example of using it in this answer:
Option for SSH to timeout after a short time? ClientAlive & ConnectTimeout don't seem to do what I need them to do
Editing my answer in here, though I'm not sure I'm supposed to do that:
import signal
class TimeoutException(Exception): # Custom exception class
pass
def timeout_handler(signum, frame): # Custom signal handler
raise TimeoutException
# Change the behavior of SIGALRM
signal.signal(signal.SIGALRM, timeout_handler)
for i in range(3):
# Start the timer. Once 5 seconds are over, a SIGALRM signal is sent.
signal.alarm(5)
# This try/except loop ensures that
# you'll catch TimeoutException when it's sent.
try:
A(i) # Whatever your function that might hang
except TimeoutException:
continue # continue the for loop if function A takes more than 5 second
else:
# Reset the alarm
signal.alarm(0)
This basically sets a timer for 5 seconds, then tries to execute your code. If it fails to complete before time runs out, a SIGALRM is sent, which we catch and turn into a TimeoutException. That forces you to the except block, where your program can continue.
Maybe someone find this decorator useful, based on TheSoundDefense answer:
import time
import signal
class TimeoutException(Exception): # Custom exception class
pass
def break_after(seconds=2):
def timeout_handler(signum, frame): # Custom signal handler
raise TimeoutException
def function(function):
def wrapper(*args, **kwargs):
signal.signal(signal.SIGALRM, timeout_handler)
signal.alarm(seconds)
try:
res = function(*args, **kwargs)
signal.alarm(0) # Clear alarm
return res
except TimeoutException:
print u'Oops, timeout: %s sec reached.' % seconds, function.__name__, args, kwargs
return
return wrapper
return function
Test:
#break_after(3)
def test(a, b, c):
return time.sleep(10)
>>> test(1,2,3)
Oops, timeout: 3 sec reached. test (1, 2, 3) {}
If you can break your work up and check every so often, that's almost always the best solution. But sometimes that's not possible—e.g., maybe you're reading a file off an slow file share that every once in a while just hangs for 30 seconds. To deal with that internally, you'd have to restructure your whole program around an async I/O loop.
If you don't need to be cross-platform, you can use signals on *nix (including Mac and Linux), APCs on Windows, etc. But if you need to be cross-platform, that doesn't work.
So, if you really need to do it concurrently, you can, and sometimes you have to. In that case, you probably want to use a process for this, not a thread. You can't really kill a thread safely, but you can kill a process, and it can be as safe as you want it to be. Also, if the thread is taking 5+ seconds because it's CPU-bound, you don't want to fight with it over the GIL.
There are two basic options here.
First, you can put the code in another script and run it with subprocess:
subprocess.check_call([sys.executable, 'other_script.py', arg, other_arg],
timeout=5)
Since this is going through normal child-process channels, the only communication you can use is some argv strings, a success/failure return value (actually a small integer, but that's not much better), and optionally a hunk of text going in and a chunk of text coming out.
Alternatively, you can use multiprocessing to spawn a thread-like child process:
p = multiprocessing.Process(func, args)
p.start()
p.join(5)
if p.is_alive():
p.terminate()
As you can see, this is a little more complicated, but it's better in a few ways:
You can pass arbitrary Python objects (at least anything that can be pickled) rather than just strings.
Instead of having to put the target code in a completely independent script, you can leave it as a function in the same script.
It's more flexible—e.g., if you later need to, say, pass progress updates, it's very easy to add a queue in either or both directions.
The big problem with any kind of parallelism is sharing mutable data—e.g., having a background task update a global dictionary as part of its work (which your comments say you're trying to do). With threads, you can sort of get away with it, but race conditions can lead to corrupted data, so you have to be very careful with locking. With child processes, you can't get away with it at all. (Yes, you can use shared memory, as Sharing state between processes explains, but this is limited to simple types like numbers, fixed arrays, and types you know how to define as C structures, and it just gets you back to the same problems as threads.)
Ideally, you arrange things so you don't need to share any data while the process is running—you pass in a dict as a parameter and get a dict back as a result. This is usually pretty easy to arrange when you have a previously-synchronous function that you want to put in the background.
But what if, say, a partial result is better than no result? In that case, the simplest solution is to pass the results over a queue. You can do this with an explicit queue, as explained in Exchanging objects between processes, but there's an easier way.
If you can break the monolithic process into separate tasks, one for each value (or group of values) you wanted to stick in the dictionary, you can schedule them on a Pool—or, even better, a concurrent.futures.Executor. (If you're on Python 2.x or 3.1, see the backport futures on PyPI.)
Let's say your slow function looked like this:
def spam():
global d
for meat in get_all_meats():
count = get_meat_count(meat)
d.setdefault(meat, 0) += count
Instead, you'd do this:
def spam_one(meat):
count = get_meat_count(meat)
return meat, count
with concurrent.futures.ProcessPoolExecutor(max_workers=1) as executor:
results = executor.map(spam_one, get_canned_meats(), timeout=5)
for (meat, count) in results:
d.setdefault(meat, 0) += count
As many results as you get within 5 seconds get added to the dict; if that isn't all of them, the rest are abandoned, and a TimeoutError is raised (which you can handle however you want—log it, do some quick fallback code, whatever).
And if the tasks really are independent (as they are in my stupid little example, but of course they may not be in your real code, at least not without a major redesign), you can parallelize the work for free just by removing that max_workers=1. Then, if you run it on an 8-core machine, it'll kick off 8 workers and given them each 1/8th of the work to do, and things will get done faster. (Usually not 8x as fast, but often 3-6x as fast, which is still pretty nice.)
This seems like a better idea (sorry, I am not sure of the Python names of thing yet):
import signal
def signal_handler(signum, frame):
raise Exception("Timeout!")
signal.signal(signal.SIGALRM, signal_handler)
signal.alarm(3) # Three seconds
try:
for x in range(0, 3):
# Call function A(x)
except Exception, msg:
print "Timeout!"
signal.alarm(0) # Reset
The comments are correct in that you should check inside. Here is a potential solution. Note that an asynchronous function (by using a thread for example) is different from this solution. This is synchronous which means it will still run in series.
import time
for x in range(0,3):
someFunction()
def someFunction():
start = time.time()
while (time.time() - start < 5):
# do your normal function
return;

Setting timeout limit on Windows

I have a function that should not take too long to run. I would like to set a timeout limit on it. I can find one proposed solutions on the internet. See the following SO post. Timeout on a function call
The solution uses signals, which is not available on Windows. There is a similar use of signals for making a ticker, which has a windows port as explained on this SO post: python: windows equivalent of SIGALRM this is not an answer to the timeout directly, but could be adapted to work for timeouts. It is written for python 2.7 though.
Since the answers are roughly 10 years old, my question is: Is there any more modern python (e.g. python 3.7) way to create a context manager/decorator/similar wrapper do make a "normal function" into a timeout-limited function on a windows system?
Here's a way to convert #Praveenkumar's answer into an easy-to-use decorator, which it seems you now would like to know:
import time
import concurrent.futures as futures
def timeout(timelimit):
def decorator(func):
def decorated(*args, **kwargs):
with futures.ThreadPoolExecutor(max_workers=1) as executor:
future = executor.submit(func, *args, **kwargs)
try:
result = future.result(timelimit)
except futures.TimeoutError:
print('Timedout!')
raise TimeoutError from None
else:
print(result)
executor._threads.clear()
futures.thread._threads_queues.clear()
return result
return decorated
return decorator
#timeout(3)
def test(n):
print(f'Sleeping for {n} seconds')
time.sleep(n)
# Real code here.
return 'Done'
test(2) # OK
test(5) # -> Causes timeout.

how to set a time bound for Regular expression match [duplicate]

Specification of the problem:
I'm searching through really great amount of lines of a log file and I'm distributing those lines to groups in order to regular expressions(RegExses) I have stored using the re.match() function. Unfortunately some of my RegExses are too complicated and Python sometimes gets himself to backtracking hell. Due to this I need to protect it with some kind of timeout.
Problems:
re.match, I'm using, is Python's function and as I found out somewhere here on StackOverflow (I'm really sorry, I can not find the link now :-( ). It is very difficult to interrupt thread with running Python's library. For this reason threads are out of the game.
Because evaluating of re.match function takes relatively short time and I want to analyse with this function great amount of lines, I need some timeout function that wont't take too long to execute (this makes threads even less suitable, it takes really long time to initialise new thread) and can be set to less than one second.
For those reasons, answers here - Timeout on a function call
and here - Timeout function if it takes too long to finish with decorator (alarm - 1sec and more) are off the table.
I've spent this morning searching for solution to this question but I did not find any satisfactory answer.
Solution:
I've just modified a script posted here: Timeout function if it takes too long to finish.
And here is the code:
from functools import wraps
import errno
import os
import signal
class TimeoutError(Exception):
pass
def timeout(seconds=10, error_message=os.strerror(errno.ETIME)):
def decorator(func):
def _handle_timeout(signum, frame):
raise TimeoutError(error_message)
def wrapper(*args, **kwargs):
signal.signal(signal.SIGALRM, _handle_timeout)
signal.setitimer(signal.ITIMER_REAL,seconds) #used timer instead of alarm
try:
result = func(*args, **kwargs)
finally:
signal.alarm(0)
return result
return wraps(func)(wrapper)
return decorator
And then you can use it like this:
from timeout import timeout
from time import time
#timeout(0.01)
def loop():
while True:
pass
try:
begin = time.time()
loop()
except TimeoutError, e:
print "Time elapsed: {:.3f}s".format(time.time() - begin)
Which prints
Time elapsed: 0.010s

Break the function after certain time

In Python, for a toy example:
for x in range(0, 3):
# Call function A(x)
I want to continue the for loop if function A takes more than five seconds by skipping it so I won't get stuck or waste time.
By doing some search, I realized a subprocess or thread may help, but I have no idea how to implement it here.
I think creating a new process may be overkill. If you're on Mac or a Unix-based system, you should be able to use signal.SIGALRM to forcibly time out functions that take too long. This will work on functions that are idling for network or other issues that you absolutely can't handle by modifying your function. I have an example of using it in this answer:
Option for SSH to timeout after a short time? ClientAlive & ConnectTimeout don't seem to do what I need them to do
Editing my answer in here, though I'm not sure I'm supposed to do that:
import signal
class TimeoutException(Exception): # Custom exception class
pass
def timeout_handler(signum, frame): # Custom signal handler
raise TimeoutException
# Change the behavior of SIGALRM
signal.signal(signal.SIGALRM, timeout_handler)
for i in range(3):
# Start the timer. Once 5 seconds are over, a SIGALRM signal is sent.
signal.alarm(5)
# This try/except loop ensures that
# you'll catch TimeoutException when it's sent.
try:
A(i) # Whatever your function that might hang
except TimeoutException:
continue # continue the for loop if function A takes more than 5 second
else:
# Reset the alarm
signal.alarm(0)
This basically sets a timer for 5 seconds, then tries to execute your code. If it fails to complete before time runs out, a SIGALRM is sent, which we catch and turn into a TimeoutException. That forces you to the except block, where your program can continue.
Maybe someone find this decorator useful, based on TheSoundDefense answer:
import time
import signal
class TimeoutException(Exception): # Custom exception class
pass
def break_after(seconds=2):
def timeout_handler(signum, frame): # Custom signal handler
raise TimeoutException
def function(function):
def wrapper(*args, **kwargs):
signal.signal(signal.SIGALRM, timeout_handler)
signal.alarm(seconds)
try:
res = function(*args, **kwargs)
signal.alarm(0) # Clear alarm
return res
except TimeoutException:
print u'Oops, timeout: %s sec reached.' % seconds, function.__name__, args, kwargs
return
return wrapper
return function
Test:
#break_after(3)
def test(a, b, c):
return time.sleep(10)
>>> test(1,2,3)
Oops, timeout: 3 sec reached. test (1, 2, 3) {}
If you can break your work up and check every so often, that's almost always the best solution. But sometimes that's not possible—e.g., maybe you're reading a file off an slow file share that every once in a while just hangs for 30 seconds. To deal with that internally, you'd have to restructure your whole program around an async I/O loop.
If you don't need to be cross-platform, you can use signals on *nix (including Mac and Linux), APCs on Windows, etc. But if you need to be cross-platform, that doesn't work.
So, if you really need to do it concurrently, you can, and sometimes you have to. In that case, you probably want to use a process for this, not a thread. You can't really kill a thread safely, but you can kill a process, and it can be as safe as you want it to be. Also, if the thread is taking 5+ seconds because it's CPU-bound, you don't want to fight with it over the GIL.
There are two basic options here.
First, you can put the code in another script and run it with subprocess:
subprocess.check_call([sys.executable, 'other_script.py', arg, other_arg],
timeout=5)
Since this is going through normal child-process channels, the only communication you can use is some argv strings, a success/failure return value (actually a small integer, but that's not much better), and optionally a hunk of text going in and a chunk of text coming out.
Alternatively, you can use multiprocessing to spawn a thread-like child process:
p = multiprocessing.Process(func, args)
p.start()
p.join(5)
if p.is_alive():
p.terminate()
As you can see, this is a little more complicated, but it's better in a few ways:
You can pass arbitrary Python objects (at least anything that can be pickled) rather than just strings.
Instead of having to put the target code in a completely independent script, you can leave it as a function in the same script.
It's more flexible—e.g., if you later need to, say, pass progress updates, it's very easy to add a queue in either or both directions.
The big problem with any kind of parallelism is sharing mutable data—e.g., having a background task update a global dictionary as part of its work (which your comments say you're trying to do). With threads, you can sort of get away with it, but race conditions can lead to corrupted data, so you have to be very careful with locking. With child processes, you can't get away with it at all. (Yes, you can use shared memory, as Sharing state between processes explains, but this is limited to simple types like numbers, fixed arrays, and types you know how to define as C structures, and it just gets you back to the same problems as threads.)
Ideally, you arrange things so you don't need to share any data while the process is running—you pass in a dict as a parameter and get a dict back as a result. This is usually pretty easy to arrange when you have a previously-synchronous function that you want to put in the background.
But what if, say, a partial result is better than no result? In that case, the simplest solution is to pass the results over a queue. You can do this with an explicit queue, as explained in Exchanging objects between processes, but there's an easier way.
If you can break the monolithic process into separate tasks, one for each value (or group of values) you wanted to stick in the dictionary, you can schedule them on a Pool—or, even better, a concurrent.futures.Executor. (If you're on Python 2.x or 3.1, see the backport futures on PyPI.)
Let's say your slow function looked like this:
def spam():
global d
for meat in get_all_meats():
count = get_meat_count(meat)
d.setdefault(meat, 0) += count
Instead, you'd do this:
def spam_one(meat):
count = get_meat_count(meat)
return meat, count
with concurrent.futures.ProcessPoolExecutor(max_workers=1) as executor:
results = executor.map(spam_one, get_canned_meats(), timeout=5)
for (meat, count) in results:
d.setdefault(meat, 0) += count
As many results as you get within 5 seconds get added to the dict; if that isn't all of them, the rest are abandoned, and a TimeoutError is raised (which you can handle however you want—log it, do some quick fallback code, whatever).
And if the tasks really are independent (as they are in my stupid little example, but of course they may not be in your real code, at least not without a major redesign), you can parallelize the work for free just by removing that max_workers=1. Then, if you run it on an 8-core machine, it'll kick off 8 workers and given them each 1/8th of the work to do, and things will get done faster. (Usually not 8x as fast, but often 3-6x as fast, which is still pretty nice.)
This seems like a better idea (sorry, I am not sure of the Python names of thing yet):
import signal
def signal_handler(signum, frame):
raise Exception("Timeout!")
signal.signal(signal.SIGALRM, signal_handler)
signal.alarm(3) # Three seconds
try:
for x in range(0, 3):
# Call function A(x)
except Exception, msg:
print "Timeout!"
signal.alarm(0) # Reset
The comments are correct in that you should check inside. Here is a potential solution. Note that an asynchronous function (by using a thread for example) is different from this solution. This is synchronous which means it will still run in series.
import time
for x in range(0,3):
someFunction()
def someFunction():
start = time.time()
while (time.time() - start < 5):
# do your normal function
return;

right way to run some code with timeout in Python

I looked online and found some SO discussing and ActiveState recipes for running some code with a timeout. It looks there are some common approaches:
Use thread that run the code, and join it with timeout. If timeout elapsed - kill the thread. This is not directly supported in Python (used private _Thread__stop function) so it is bad practice
Use signal.SIGALRM - but this approach not working on Windows!
Use subprocess with timeout - but this is too heavy - what if I want to start interruptible task often, I don't want fire process for each!
So, what is the right way? I'm not asking about workarounds (eg use Twisted and async IO), but actual way to solve actual problem - I have some function and I want to run it only with some timeout. If timeout elapsed, I want control back. And I want it to work on Linux and Windows.
A completely general solution to this really, honestly does not exist. You have to use the right solution for a given domain.
If you want timeouts for code you fully control, you have to write it to cooperate. Such code has to be able to break up into little chunks in some way, as in an event-driven system. You can also do this by threading if you can ensure nothing will hold a lock too long, but handling locks right is actually pretty hard.
If you want timeouts because you're afraid code is out of control (for example, if you're afraid the user will ask your calculator to compute 9**(9**9)), you need to run it in another process. This is the only easy way to sufficiently isolate it. Running it in your event system or even a different thread will not be enough. It is also possible to break things up into little chunks similar to the other solution, but requires very careful handling and usually isn't worth it; in any event, that doesn't allow you to do the same exact thing as just running the Python code.
What you might be looking for is the multiprocessing module. If subprocess is too heavy, then this may not suit your needs either.
import time
import multiprocessing
def do_this_other_thing_that_may_take_too_long(duration):
time.sleep(duration)
return 'done after sleeping {0} seconds.'.format(duration)
pool = multiprocessing.Pool(1)
print 'starting....'
res = pool.apply_async(do_this_other_thing_that_may_take_too_long, [8])
for timeout in range(1, 10):
try:
print '{0}: {1}'.format(duration, res.get(timeout))
except multiprocessing.TimeoutError:
print '{0}: timed out'.format(duration)
print 'end'
If it's network related you could try:
import socket
socket.setdefaulttimeout(number)
I found this with eventlet library:
http://eventlet.net/doc/modules/timeout.html
from eventlet.timeout import Timeout
timeout = Timeout(seconds, exception)
try:
... # execution here is limited by timeout
finally:
timeout.cancel()
For "normal" Python code, that doesn't linger prolongued times in C extensions or I/O waits, you can achieve your goal by setting a trace function with sys.settrace() that aborts the running code when the timeout is reached.
Whether that is sufficient or not depends on how co-operating or malicious the code you run is. If it's well-behaved, a tracing function is sufficient.
An other way is to use faulthandler:
import time
import faulthandler
faulthandler.enable()
try:
faulthandler.dump_tracebacks_later(3)
time.sleep(10)
finally:
faulthandler.cancel_dump_tracebacks_later()
N.B: The faulthandler module is part of stdlib in python3.3.
If you're running code that you expect to die after a set time, then you should write it properly so that there aren't any negative effects on shutdown, no matter if its a thread or a subprocess. A command pattern with undo would be useful here.
So, it really depends on what the thread is doing when you kill it. If its just crunching numbers who cares if you kill it. If its interacting with the filesystem and you kill it , then maybe you should really rethink your strategy.
What is supported in Python when it comes to threads? Daemon threads and joins. Why does python let the main thread exit if you've joined a daemon while its still active? Because its understood that someone using daemon threads will (hopefully) write the code in a way that it wont matter when that thread dies. Giving a timeout to a join and then letting main die, and thus taking any daemon threads with it, is perfectly acceptable in this context.
I've solved that in that way:
For me is worked great (in windows and not heavy at all) I'am hope it was useful for someone)
import threading
import time
class LongFunctionInside(object):
lock_state = threading.Lock()
working = False
def long_function(self, timeout):
self.working = True
timeout_work = threading.Thread(name="thread_name", target=self.work_time, args=(timeout,))
timeout_work.setDaemon(True)
timeout_work.start()
while True: # endless/long work
time.sleep(0.1) # in this rate the CPU is almost not used
if not self.working: # if state is working == true still working
break
self.set_state(True)
def work_time(self, sleep_time): # thread function that just sleeping specified time,
# in wake up it asking if function still working if it does set the secured variable work to false
time.sleep(sleep_time)
if self.working:
self.set_state(False)
def set_state(self, state): # secured state change
while True:
self.lock_state.acquire()
try:
self.working = state
break
finally:
self.lock_state.release()
lw = LongFunctionInside()
lw.long_function(10)
The main idea is to create a thread that will just sleep in parallel to "long work" and in wake up (after timeout) change the secured variable state, the long function checking the secured variable during its work.
I'm pretty new in Python programming, so if that solution has a fundamental errors, like resources, timing, deadlocks problems , please response)).
solving with the 'with' construct and merging solution from -
Timeout function if it takes too long to finish
this thread which work better.
import threading, time
class Exception_TIMEOUT(Exception):
pass
class linwintimeout:
def __init__(self, f, seconds=1.0, error_message='Timeout'):
self.seconds = seconds
self.thread = threading.Thread(target=f)
self.thread.daemon = True
self.error_message = error_message
def handle_timeout(self):
raise Exception_TIMEOUT(self.error_message)
def __enter__(self):
try:
self.thread.start()
self.thread.join(self.seconds)
except Exception, te:
raise te
def __exit__(self, type, value, traceback):
if self.thread.is_alive():
return self.handle_timeout()
def function():
while True:
print "keep printing ...", time.sleep(1)
try:
with linwintimeout(function, seconds=5.0, error_message='exceeded timeout of %s seconds' % 5.0):
pass
except Exception_TIMEOUT, e:
print " attention !! execeeded timeout, giving up ... %s " % e

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