"Directory this source file is in (and a sibling)"

[John Machin]

Sion Arrowsmith wrote:
> I have a module which needs to know what directory it's in, and to
> refer to files in a sibling directory, something like App/src/foo.py
> wants to read App/data/conf.xml . But I have no idea in what context
> foo.py is going to be run -- it could be being run as a script, it
> could be being imported as a module by another script from anywhere in
> the directory structure, it's even possible someone will have called
> execfile on it. The following works for everything I've tried:
>
> thisdir = os.path.dirname(os.path.normpath(__file__))
> siblingdir = os.path.normpath(os.path.join(testdir, os.path.pardir, "sibling"))
>
> However, a colleague expressed disgust at this code, but not really
> being a Python programmer had no better suggestions. Is there a neater
> way of getting what I want?

If you plan to have your code executed out of a zip or an egg, you may
have problems because __file__ is not what you want, at least with eggs
and Python 2.[34]. There was a brief thread on this topic on the
distutils SIG mailing list within the last few days.

HTH,
John