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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.

python singleton locking
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Perhaps your life would be easier if you tracked down and fixed the segfault. Not that it's an easy thing to do. –David Locke Dec 19 '08 at 15:53
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It's not in my library, it's in python's libxml bindings and extremely shy - fires only once a couple days. –Slava N Dec 20 '08 at 17:51
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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. –Charles Duffy Dec 20 '08 at 19:02

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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
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