How do I get the path of the current executed file in Python?

Question

This may seem like a newbie question, but it is not. Some common approaches don\'t work in all cases:

sys.argv[0]

This means using path = os.path.abs

Answer 1

I was running into a similar problem, and I think this might solve the problem:

def module_path(local_function):
   ''' returns the module path without the use of __file__.  Requires a function defined
   locally in the module.
   from http://stackoverflow.com/questions/729583/getting-file-path-of-imported-module'''
   return os.path.abspath(inspect.getsourcefile(local_function))

It works for regular scripts and in idle. All I can say is try it out for others!

My typical usage:

from toolbox import module_path
def main():
   pass # Do stuff

global __modpath__
__modpath__ = module_path(main)

Now I use __modpath__ instead of __file__.