I am using gunicorn with multiple workers for my machine learning project. But the problem is when I send a train request only the worker getting the training request gets updated with the latest model after training is done. Here it is worth to mention that, to make the inference faster I have programmed to load the model once after each training. This is why, the only worker which is used for current training operation loads the latest model and the other workers still keeps the previously loaded model. Right now the model file (binary format) is loaded once after each training in a global dictionary variable where key is the model name and the value is the model file. Obviously, this problem won't occur if I program it to load the model every time from disk for each prediction, but I cannot do it, as it will make the prediction slower.
I studied further on global variables and further investigation shows that, in a multi-processing environment, all the workers (processes) create their own copies of global variables. Apart from the binary model file, I also have some other global variables (in dictionary type) need to be synced across all processes. So, how to handle this situation?
TL;DR: I need some approach which can help me to store variable which will be common across all the processes (workers). Any way to do this? With multiprocessing.Manager, dill etc.?
Update 1: I have multiple machine learning algorithms in my project and they have their own model files, which are being loaded to memory in a dictionary where the key is the model name and the value is the corresponding model object. I need to share all of them (in other words, I need to share the dictionary). But some of the models are not pickle serializable like - FastText. So, when I try to use a proxy variable (in my case a dictionary to hold models) with multiprocessing.Manager I get error for those non-pickle-serializable object while assigning the loaded model file to this dictionary. Like: can't pickle fasttext_pybind.fasttext objects. More information on multiprocessing.Manager can be found here: Proxy Objects
Following is the summary what I have done:
import multiprocessing
import fasttext
mgr = multiprocessing.Manager()
model_dict = mgr.dict()
model_file = fasttext.load_model("path/to/model/file/which/is/in/.bin/format")
model_dict["fasttext"] = model_file # This line throws this error
Error:
can't pickle fasttext_pybind.fasttext objects
I printed the model_file which I am trying to assign, it is:
<fasttext.FastText._FastText object at 0x7f86e2b682e8>
Update 2:
According to this answer I modified my code a little bit:
import fasttext
from multiprocessing.managers import SyncManager
def Manager():
m = SyncManager()
m.start()
return m
# As the model file has a type of "<fasttext.FastText._FastText object at 0x7f86e2b682e8>" so, using "fasttext.FastText._FastText" as the class of it
SyncManager.register("fast", fasttext.FastText._FastText)
# Now this is the Manager as a replacement of the old one.
mgr = Manager()
ft = mgr.fast() # This line gives error.
This gives me EOFError.
Update 3: I tried using dill both with multiprocessing and multiprocess. The summary of changes are as the following:
import multiprocessing
import multiprocess
import dill
# Any one of the following two lines
mgr = multiprocessing.Manager() # Or,
mgr = multiprocess.Manager()
model_dict = mgr.dict()
... ... ...
... ... ...
model_file = dill.dumps(model_file) # This line throws the error
model_dict["fasttext"] = model_file
... ... ...
... ... ...
# During loading
model_file = dill.loads(model_dict["fasttext"])
But still getting the error: can't pickle fasttext_pybind.fasttext objects.
Update 4:
This time I am using another library called jsonpickle. It seems to be that serialization and de-serialization occurs properly (as it is not reporting any issue while running). But surprisingly enough, after de-serialization whenever I am making a prediction, it faces segmentation fault. More details and the steps to reproduce it can be found here: Segmentation fault (core dumped)
Update 5: Tried cloudpickle, srsly, but couldn't make the program working.
For the sake of completeness I am providing the solution that worked for me. All the approaches I have tried to serialize FastText went in vain. Finally, as #MedetTleukabiluly mentioned in the comment, I managed to share the message of loading the model from the disk with other workers with redis-pubsub. Obviously, it is not actually sharing the model from the same memory space, rather, just sharing the message to other workers to inform them they should load the model from the disk (as a new training just happened). Following is the general solution:
# redis_pubsub.py
import logging
import os
import fasttext
import socket
import threading
import time
"""The whole purpose of GLOBAL_NAMESPACE is to keep the whole pubsub mechanism separate.
As this might be a case another service also publishing in the same channel.
"""
GLOBAL_NAMESPACE = "SERVICE_0"
def get_ip():
s = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
try:
# doesn't even have to be reachable
s.connect(('10.255.255.255', 1))
IP = s.getsockname()[0]
except Exception:
IP = '127.0.0.1'
finally:
s.close()
return IP
class RedisPubSub:
def __init__(self):
self.redis_client = get_redis_client() #TODO: A SAMPLE METHOD WHICH CAN RETURN YOUR REDIS CLIENT (you have to implement)
# Unique ID is used, to identify which worker from which server is the publisher. Just to avoid updating
# getting a message which message is indeed sent by itself.
self.unique_id = "IP_" + get_ip() + "__" + str(GLOBAL_NAMESPACE) + "__" + "PID_" + str(os.getpid())
def listen_to_channel_and_update_models(self, channel):
try:
pubsub = self.redis_client.pubsub()
pubsub.subscribe(channel)
except Exception as exception:
logging.error(f"REDIS_ERROR: Model Update Listening: {exception}")
while True:
try:
message = pubsub.get_message()
# Successful operation gives 1 and unsuccessful gives 0
# ..we are not interested to receive these flags
if message and message["data"] != 1 and message["data"] != 0:
message = message["data"].decode("utf-8")
message = str(message)
splitted_msg = message.split("__SEPERATOR__")
# Not only making sure the message is coming from another worker
# but also we have to make sure the message sender and receiver (i.e, both of the workers) are under the same namespace
if (splitted_msg[0] != self.unique_id) and (splitted_msg[0].split('__')[1] == GLOBAL_NAMESPACE):
algo_name = splitted_msg[1]
model_path = splitted_msg[2]
# Fasttext
if "fasttext" in algo_name:
try:
#TODO: YOU WILL GET THE LOADED NEW FILE IN model_file. USE IT TO UPDATE THE OLD ONE.
model_file = fasttext.load_model(model_path + '.bin')
except Exception as exception:
logging.error(exception)
else:
logging.info(f"{algo_name} model is updated for process with unique_id: {self.unique_id} by process with unique_id: {splitted_msg[0]}")
time.sleep(1) # sleeping for 1 second to avoid hammering the CPU too much
except Exception as exception:
time.sleep(1)
logging.error(f"PUBSUB_ERROR: Model or component update: {exception}")
def publish_to_channel(self, channel, algo_name, model_path):
def _publish_to_channel():
try:
message = self.unique_id + '__SEPERATOR__' + str(algo_name) + '__SEPERATOR__' + str(model_path)
time.sleep(3)
self.redis_client.publish(channel, message)
except Exception as exception:
logging.error(f"PUBSUB_ERROR: Model or component publishing: {exception}")
# As the delay before pubsub can pause the next activities which are independent, hence, doing this publishing in another thread.
thread = threading.Thread(target = _publish_to_channel)
thread.start()
Also you have to start the listener:
from redis_pubsub import RedisPubSub
pubsub = RedisPubSub()
# start the listener:
thread = threading.Thread(target = pubsub.listen_to_channel_and_update_models, args = ("sync-ml-models", ))
thread.start()
From fasttext training module, when you finish the training, publish this message to other workers, such that the other workers get a chance to re-load the model from the disk:
# fasttext_api.py
from redis_pubsub import RedisPubSub
pubsub = RedisPubSub()
pubsub.publish_to_channel(channel = "sync-ml-models", # a sample name for the channel
algo_name = f"fasttext",
model_path = "path/to/fasttext/model")
When I use a standard Queue to send samples to a process, everything works fine. However, since my needs are simple, I tried to use a SimpleQueue and for some reason the 'empty' method doesn't work. Here's the details:
Error comes from the consumer process (when sample_queue is Queue, everything works, when sample_queue is SimpleQueue, things break):
def frame_update(i):
while not self.sample_queue.empty():
sample = self.sample_queue.get()
for line in lines:
While executing sample_queue.empty() -- SimpleQueue.empty(), from Python 3.6 on windows (queues.py) we get:
def empty(self):
return not self._poll()
Where self._poll() has been set in init by:
def __init__(self, *, ctx):
self._reader, self._writer = connection.Pipe(duplex=False)
self._rlock = ctx.Lock()
self._poll = self._reader.poll
if sys.platform == 'win32':
self._wlock = None
else:
self._wlock = ctx.Lock()
So I follow the self._reader which is set from connection.Pipe (connection.py):
...
c1 = PipeConnection(h1, writable=duplex)
c2 = PipeConnection(h2, readable=duplex)
Ok, great. The _reader is going to be a PipeConnection and pipe connection has this method:
def _poll(self, timeout):
if (self._got_empty_message or
_winapi.PeekNamedPipe(self._handle)[0] != 0):
return True
return bool(wait([self], timeout))
Alright -- So a couple of questions:
1) Shouldn't the init of SimpleQueue be assigning self.poll to self._reader._poll instead of self._reader.poll? Or am I missing something in the inheritance hierarchy?
2) The PipeConnection _poll routine takes a timeout parameter, so #1 shouldn't work...
*) -- Is there some other binding of PipeConnection _poll that I'm missing?
Am I missing something? I am using Python3.6, Windows, debugging in PyCharm and I follow all the paths and they're in the standard multiprocessing paths. I'd appreciate any help or advice. Thanks!
EDIT: After further review, I can see that PipeConnection is a subclass of _ConnectionBase which does indeed have a 'poll' method and it is bound with a default timeout parameter.
So the question is: When SimpleQueue is initializing and sets
self._poll = self._reader.poll
Why doesn't it go up the class hierarchy to grab that from _ConnectionBase?
After looking at why the Queue type works and why the SimpleQueue doesn't, I found that Queue sets the _poll method 'after_fork' as well as before. SimpleQueue doesn't. By changing the setstate method to add self._poll = self._reader.poll as follows (queues.py, line 338), SimpleQueue works
def __setstate__(self, state):
(self._reader, self._writer, self._rlock, self._wlock) = state
self._poll = self._reader.poll
Seems like a bug to me unless I'm really misunderstanding something. I'll submit a bug report and reference this post. Hope this helps someone!
http://bugs.python.org/issue30301
Is there a Pythonic way to have only one instance of a program running?
The only reasonable solution I've come up with is trying to run it as a server on some port, then second program trying to bind to same port - fails. But it's not really a great idea, maybe there's something more lightweight than this?
(Take into consideration that program is expected to fail sometimes, i.e. segfault - so things like "lock file" won't work)
The following code should do the job, it is cross-platform and runs on Python 2.4-3.2. I tested it on Windows, OS X and Linux.
from tendo import singleton
me = singleton.SingleInstance() # will sys.exit(-1) if other instance is running
The latest code version is available singleton.py. Please file bugs here.
You can install tend using one of the following methods:
easy_install tendo
pip install tendo
manually by getting it from http://pypi.python.org/pypi/tendo
Simple, cross-platform solution, found in another question by zgoda:
import fcntl
import os
import sys
def instance_already_running(label="default"):
"""
Detect if an an instance with the label is already running, globally
at the operating system level.
Using `os.open` ensures that the file pointer won't be closed
by Python's garbage collector after the function's scope is exited.
The lock will be released when the program exits, or could be
released if the file pointer were closed.
"""
lock_file_pointer = os.open(f"/tmp/instance_{label}.lock", os.O_WRONLY)
try:
fcntl.lockf(lock_file_pointer, fcntl.LOCK_EX | fcntl.LOCK_NB)
already_running = False
except IOError:
already_running = True
return already_running
A lot like S.Lott's suggestion, but with the code.
This code is Linux specific. It uses 'abstract' UNIX domain sockets, but it is simple and won't leave stale lock files around. I prefer it to the solution above because it doesn't require a specially reserved TCP port.
try:
import socket
s = socket.socket(socket.AF_UNIX, socket.SOCK_STREAM)
## Create an abstract socket, by prefixing it with null.
s.bind( '\0postconnect_gateway_notify_lock')
except socket.error as e:
error_code = e.args[0]
error_string = e.args[1]
print "Process already running (%d:%s ). Exiting" % ( error_code, error_string)
sys.exit (0)
The unique string postconnect_gateway_notify_lock can be changed to allow multiple programs that need a single instance enforced.
I don't know if it's pythonic enough, but in the Java world listening on a defined port is a pretty widely used solution, as it works on all major platforms and doesn't have any problems with crashing programs.
Another advantage of listening to a port is that you could send a command to the running instance. For example when the users starts the program a second time, you could send the running instance a command to tell it to open another window (that's what Firefox does, for example. I don't know if they use TCP ports or named pipes or something like that, 'though).
Never written python before, but this is what I've just implemented in mycheckpoint, to prevent it being started twice or more by crond:
import os
import sys
import fcntl
fh=0
def run_once():
global fh
fh=open(os.path.realpath(__file__),'r')
try:
fcntl.flock(fh,fcntl.LOCK_EX|fcntl.LOCK_NB)
except:
os._exit(0)
run_once()
Found Slava-N's suggestion after posting this in another issue (http://stackoverflow.com/questions/2959474). This one is called as a function, locks the executing scripts file (not a pid file) and maintains the lock until the script ends (normal or error).
Use a pid file. You have some known location, "/path/to/pidfile" and at startup you do something like this (partially pseudocode because I'm pre-coffee and don't want to work all that hard):
import os, os.path
pidfilePath = """/path/to/pidfile"""
if os.path.exists(pidfilePath):
pidfile = open(pidfilePath,"r")
pidString = pidfile.read()
if <pidString is equal to os.getpid()>:
# something is real weird
Sys.exit(BADCODE)
else:
<use ps or pidof to see if the process with pid pidString is still running>
if <process with pid == 'pidString' is still running>:
Sys.exit(ALREADAYRUNNING)
else:
# the previous server must have crashed
<log server had crashed>
<reopen pidfilePath for writing>
pidfile.write(os.getpid())
else:
<open pidfilePath for writing>
pidfile.write(os.getpid())
So, in other words, you're checking if a pidfile exists; if not, write your pid to that file. If the pidfile does exist, then check to see if the pid is the pid of a running process; if so, then you've got another live process running, so just shut down. If not, then the previous process crashed, so log it, and then write your own pid to the file in place of the old one. Then continue.
The best solution for this on windows is to use mutexes as suggested by #zgoda.
import win32event
import win32api
from winerror import ERROR_ALREADY_EXISTS
mutex = win32event.CreateMutex(None, False, 'name')
last_error = win32api.GetLastError()
if last_error == ERROR_ALREADY_EXISTS:
print("App instance already running")
Some answers use fctnl (included also in #sorin tendo package) which is not available on windows and should you try to freeze your python app using a package like pyinstaller which does static imports, it throws an error.
Also, using the lock file method, creates a read-only problem with database files( experienced this with sqlite3).
Here is my eventual Windows-only solution. Put the following into a module, perhaps called 'onlyone.py', or whatever. Include that module directly into your __ main __ python script file.
import win32event, win32api, winerror, time, sys, os
main_path = os.path.abspath(sys.modules['__main__'].__file__).replace("\\", "/")
first = True
while True:
mutex = win32event.CreateMutex(None, False, main_path + "_{<paste YOUR GUID HERE>}")
if win32api.GetLastError() == 0:
break
win32api.CloseHandle(mutex)
if first:
print "Another instance of %s running, please wait for completion" % main_path
first = False
time.sleep(1)
Explanation
The code attempts to create a mutex with name derived from the full path to the script. We use forward-slashes to avoid potential confusion with the real file system.
Advantages
No configuration or 'magic' identifiers needed, use it in as many different scripts as needed.
No stale files left around, the mutex dies with you.
Prints a helpful message when waiting
This may work.
Attempt create a PID file to a known location. If you fail, someone has the file locked, you're done.
When you finish normally, close and remove the PID file, so someone else can overwrite it.
You can wrap your program in a shell script that removes the PID file even if your program crashes.
You can, also, use the PID file to kill the program if it hangs.
For anybody using wxPython for their application, you can use the function wx.SingleInstanceChecker documented here.
I personally use a subclass of wx.App which makes use of wx.SingleInstanceChecker and returns False from OnInit() if there is an existing instance of the app already executing like so:
import wx
class SingleApp(wx.App):
"""
class that extends wx.App and only permits a single running instance.
"""
def OnInit(self):
"""
wx.App init function that returns False if the app is already running.
"""
self.name = "SingleApp-%s".format(wx.GetUserId())
self.instance = wx.SingleInstanceChecker(self.name)
if self.instance.IsAnotherRunning():
wx.MessageBox(
"An instance of the application is already running",
"Error",
wx.OK | wx.ICON_WARNING
)
return False
return True
This is a simple drop-in replacement for wx.App that prohibits multiple instances. To use it simply replace wx.App with SingleApp in your code like so:
app = SingleApp(redirect=False)
frame = wx.Frame(None, wx.ID_ANY, "Hello World")
frame.Show(True)
app.MainLoop()
Using a lock-file is a quite common approach on unix. If it crashes, you have to clean up manually. You could stor the PID in the file, and on startup check if there is a process with this PID, overriding the lock-file if not. (However, you also need a lock around the read-file-check-pid-rewrite-file). You will find what you need for getting and checking pid in the os-package. The common way of checking if there exists a process with a given pid, is to send it a non-fatal signal.
Other alternatives could be combining this with flock or posix semaphores.
Opening a network socket, as saua proposed, would probably be the easiest and most portable.
I'm posting this as an answer because I'm a new user and Stack Overflow won't let me vote yet.
Sorin Sbarnea's solution works for me under OS X, Linux and Windows, and I am grateful for it.
However, tempfile.gettempdir() behaves one way under OS X and Windows and another under other some/many/all(?) *nixes (ignoring the fact that OS X is also Unix!). The difference is important to this code.
OS X and Windows have user-specific temp directories, so a tempfile created by one user isn't visible to another user. By contrast, under many versions of *nix (I tested Ubuntu 9, RHEL 5, OpenSolaris 2008 and FreeBSD 8), the temp dir is /tmp for all users.
That means that when the lockfile is created on a multi-user machine, it's created in /tmp and only the user who creates the lockfile the first time will be able to run the application.
A possible solution is to embed the current username in the name of the lock file.
It's worth noting that the OP's solution of grabbing a port will also misbehave on a multi-user machine.
Building upon Roberto Rosario's answer, I come up with the following function:
SOCKET = None
def run_single_instance(uniq_name):
try:
import socket
global SOCKET
SOCKET = socket.socket(socket.AF_UNIX, socket.SOCK_STREAM)
## Create an abstract socket, by prefixing it with null.
# this relies on a feature only in linux, when current process quits, the
# socket will be deleted.
SOCKET.bind('\0' + uniq_name)
return True
except socket.error as e:
return False
We need to define global SOCKET vaiable since it will only be garbage collected when the whole process quits. If we declare a local variable in the function, it will go out of scope after the function exits, thus the socket be deleted.
All the credit should go to Roberto Rosario, since I only clarify and elaborate upon his code. And this code will work only on Linux, as the following quoted text from https://troydhanson.github.io/network/Unix_domain_sockets.html explains:
Linux has a special feature: if the pathname for a UNIX domain socket
begins with a null byte \0, its name is not mapped into the
filesystem. Thus it won’t collide with other names in the filesystem.
Also, when a server closes its UNIX domain listening socket in the
abstract namespace, its file is deleted; with regular UNIX domain
sockets, the file persists after the server closes it.
Late answer, but for windows you can use:
from win32event import CreateMutex
from win32api import CloseHandle, GetLastError
from winerror import ERROR_ALREADY_EXISTS
import sys
class singleinstance:
""" Limits application to single instance """
def __init__(self):
self.mutexname = "testmutex_{D0E858DF-985E-4907-B7FB-8D732C3FC3B9}"
self.mutex = CreateMutex(None, False, self.mutexname)
self.lasterror = GetLastError()
def alreadyrunning(self):
return (self.lasterror == ERROR_ALREADY_EXISTS)
def __del__(self):
if self.mutex:
CloseHandle(self.mutex)
Usage
# do this at beginnig of your application
myapp = singleinstance()
# check is another instance of same program running
if myapp.alreadyrunning():
print ("Another instance of this program is already running")
sys.exit(1)
Here is a cross platform example that I've tested on Windows Server 2016 and Ubuntu 20.04 using Python 3.7.9:
import os
class SingleInstanceChecker:
def __init__(self, id):
if isWin():
ensure_win32api()
self.mutexname = id
self.lock = win32event.CreateMutex(None, False, self.mutexname)
self.running = (win32api.GetLastError() == winerror.ERROR_ALREADY_EXISTS)
else:
ensure_fcntl()
self.lock = open(f"/tmp/isnstance_{id}.lock", 'wb')
try:
fcntl.lockf(self.lock, fcntl.LOCK_EX | fcntl.LOCK_NB)
self.running = False
except IOError:
self.running = True
def already_running(self):
return self.running
def __del__(self):
if self.lock:
try:
if isWin():
win32api.CloseHandle(self.lock)
else:
os.close(self.lock)
except Exception as ex:
pass
# ---------------------------------------
# Utility Functions
# Dynamically load win32api on demand
# Install with: pip install pywin32
win32api=winerror=win32event=None
def ensure_win32api():
global win32api,winerror,win32event
if win32api is None:
import win32api
import winerror
import win32event
# Dynamically load fcntl on demand
# Install with: pip install fcntl
fcntl=None
def ensure_fcntl():
global fcntl
if fcntl is None:
import fcntl
def isWin():
return (os.name == 'nt')
# ---------------------------------------
Here is it in use:
import time, sys
def main(argv):
_timeout = 10
print("main() called. sleeping for %s seconds" % _timeout)
time.sleep(_timeout)
print("DONE")
if __name__ == '__main__':
SCR_NAME = "my_script"
sic = SingleInstanceChecker(SCR_NAME)
if sic.already_running():
print("An instance of {} is already running.".format(SCR_NAME))
sys.exit(1)
else:
main(sys.argv[1:])
I use single_process on my gentoo;
pip install single_process
example:
from single_process import single_process
#single_process
def main():
print 1
if __name__ == "__main__":
main()
refer: https://pypi.python.org/pypi/single_process/
I keep suspecting there ought to be a good POSIXy solution using process groups, without having to hit the file system, but I can't quite nail it down. Something like:
On startup, your process sends a 'kill -0' to all processes in a particular group. If any such processes exist, it exits. Then it joins the group. No other processes use that group.
However, this has a race condition - multiple processes could all do this at precisely the same time and all end up joining the group and running simultaneously. By the time you've added some sort of mutex to make it watertight, you no longer need the process groups.
This might be acceptable if your process only gets started by cron, once every minute or every hour, but it makes me a bit nervous that it would go wrong precisely on the day when you don't want it to.
I guess this isn't a very good solution after all, unless someone can improve on it?
I ran into this exact problem last week, and although I did find some good solutions, I decided to make a very simple and clean python package and uploaded it to PyPI. It differs from tendo in that it can lock any string resource name. Although you could certainly lock __file__ to achieve the same effect.
Install with: pip install quicklock
Using it is extremely simple:
[nate#Nates-MacBook-Pro-3 ~/live] python
Python 2.7.6 (default, Sep 9 2014, 15:04:36)
[GCC 4.2.1 Compatible Apple LLVM 6.0 (clang-600.0.39)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> from quicklock import singleton
>>> # Let's create a lock so that only one instance of a script will run
...
>>> singleton('hello world')
>>>
>>> # Let's try to do that again, this should fail
...
>>> singleton('hello world')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/Users/nate/live/gallery/env/lib/python2.7/site-packages/quicklock/quicklock.py", line 47, in singleton
raise RuntimeError('Resource <{}> is currently locked by <Process {}: "{}">'.format(resource, other_process.pid, other_process.name()))
RuntimeError: Resource <hello world> is currently locked by <Process 24801: "python">
>>>
>>> # But if we quit this process, we release the lock automatically
...
>>> ^D
[nate#Nates-MacBook-Pro-3 ~/live] python
Python 2.7.6 (default, Sep 9 2014, 15:04:36)
[GCC 4.2.1 Compatible Apple LLVM 6.0 (clang-600.0.39)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> from quicklock import singleton
>>> singleton('hello world')
>>>
>>> # No exception was thrown, we own 'hello world'!
Take a look: https://pypi.python.org/pypi/quicklock
linux example
This method is based on the creation of a temporary file automatically deleted after you close the application.
the program launch we verify the existence of the file;
if the file exists ( there is a pending execution) , the program is closed ; otherwise it creates the file and continues the execution of the program.
from tempfile import *
import time
import os
import sys
f = NamedTemporaryFile( prefix='lock01_', delete=True) if not [f for f in os.listdir('/tmp') if f.find('lock01_')!=-1] else sys.exit()
YOUR CODE COMES HERE
On a Linux system one could also ask
pgrep -a for the number of instances, the script
is found in the process list (option -a reveals the
full command line string). E.g.
import os
import sys
import subprocess
procOut = subprocess.check_output( "/bin/pgrep -u $UID -a python", shell=True,
executable="/bin/bash", universal_newlines=True)
if procOut.count( os.path.basename(__file__)) > 1 :
sys.exit( ("found another instance of >{}<, quitting."
).format( os.path.basename(__file__)))
Remove -u $UID if the restriction should apply to all users.
Disclaimer: a) it is assumed that the script's (base)name is unique, b) there might be race conditions.
Here's a good example for django with contextmanager and memcached:
https://docs.celeryproject.org/en/latest/tutorials/task-cookbook.html
Can be used to protect simultaneous operation on different hosts.
Can be used to manage multiple tasks.
Can also be changed for simple python scripts.
My modification of the above code is here:
import time
from contextlib import contextmanager
from django.core.cache import cache
#contextmanager
def memcache_lock(lock_key, lock_value, lock_expire):
timeout_at = time.monotonic() + lock_expire - 3
# cache.add fails if the key already exists
status = cache.add(lock_key, lock_value, lock_expire)
try:
yield status
finally:
# memcache delete is very slow, but we have to use it to take
# advantage of using add() for atomic locking
if time.monotonic() < timeout_at and status:
# don't release the lock if we exceeded the timeout
# to lessen the chance of releasing an expired lock owned by someone else
# also don't release the lock if we didn't acquire it
cache.delete(lock_key)
LOCK_EXPIRE = 60 * 10 # Lock expires in 10 minutes
def main():
lock_name, lock_value = "lock_1", "locked"
with memcache_lock(lock_name, lock_value, LOCK_EXPIRE) as acquired:
if acquired:
# single instance code here:
pass
if __name__ == "__main__":
main()
Here is a cross-platform implementation, creating a temporary lock file using a context manager.
Can be used to manage multiple tasks.
import os
from contextlib import contextmanager
from time import sleep
class ExceptionTaskInProgress(Exception):
pass
# Context manager for suppressing exceptions
class SuppressException:
def __init__(self):
pass
def __enter__(self):
return self
def __exit__(self, *exc):
return True
# Context manager for task
class TaskSingleInstance:
def __init__(self, task_name, lock_path):
self.task_name = task_name
self.lock_path = lock_path
self.lock_filename = os.path.join(self.lock_path, self.task_name + ".lock")
if os.path.exists(self.lock_filename):
raise ExceptionTaskInProgress("Resource already in use")
def __enter__(self):
self.fl = open(self.lock_filename, "w")
return self
def __exit__(self, exc_type, exc_val, exc_tb):
self.fl.close()
os.unlink(self.lock_filename)
# Here the task is silently interrupted
# if it is already running on another instance.
def main1():
task_name = "task1"
tmp_filename_path = "."
with SuppressException():
with TaskSingleInstance(task_name, tmp_filename_path):
print("The task `{}` has started.".format(task_name))
# The single task instance code is here.
sleep(5)
print("The task `{}` has completed.".format(task_name))
# Here the task is interrupted with a message
# if it is already running in another instance.
def main2():
task_name = "task1"
tmp_filename_path = "."
try:
with TaskSingleInstance(task_name, tmp_filename_path):
print("The task `{}` has started.".format(task_name))
# The single task instance code is here.
sleep(5)
print("Task `{}` completed.".format(task_name))
except ExceptionTaskInProgress as ex:
print("The task `{}` is already running.".format(task_name))
if __name__ == "__main__":
main1()
main2()
import sys,os
# start program
try: # (1)
os.unlink('lock') # (2)
fd=os.open("lock", os.O_CREAT|os.O_EXCL) # (3)
except:
try: fd=os.open("lock", os.O_CREAT|os.O_EXCL) # (4)
except:
print "Another Program running !.." # (5)
sys.exit()
# your program ...
# ...
# exit program
try: os.close(fd) # (6)
except: pass
try: os.unlink('lock')
except: pass
sys.exit()