Question

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)

Update: the solutions offered are much more complex and less reliant than just having a port occupied with a non-existent server, so I'd have to go with that one.

Perhaps your life would be easier if you tracked down and fixed the segfault. Not that it's an easy thing to do.
It's not in my library, it's in python's libxml bindings and extremely shy - fires only once a couple days.
Python's standard library supports flock(), which is The Right Thing for modern UNIX programs. Opening a port uses a spot in a much more constrained namespace, whereas pidfiles are more complex as you need to check running processes to invalidate them safely; flock has neither problem.

score 22 · Answer 1

The above 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

singleton Python: single instance of program Stack Overflow

10 Answers