How to get filename of the __main__ module in Python?

import __main__print __main__.__file__

Perhaps this will do the trick:

import sysfrom os import pathprint path.abspath(sys.modules['__main__'].__file__)

Note that, for safety, you should check whether the __main__ module has a __file__ attribute. If it's dynamically created, or is just being run in the interactive python console, it won't have a __file__:

python>>> import sys>>> print sys.modules['__main__']<module '__main__' (built-in)>>>> print sys.modules['__main__'].__file__AttributeError: 'module' object has no attribute '__file__'

A simple hasattr() check will do the trick to guard against scenario 2 if that's a possibility in your app.

The python code below provides additional functionality, including that it works seamlessly with py2exe executables.

I use similar code to like this to find paths relative to the running script, aka __main__. as an added benefit, it works cross-platform including Windows.

import impimport osimport sysdef main_is_frozen():   return (hasattr(sys, "frozen") or # new py2exe           hasattr(sys, "importers") # old py2exe           or imp.is_frozen("__main__")) # tools/freezedef get_main_dir():   if main_is_frozen():       # print 'Running from path', os.path.dirname(sys.executable)       return os.path.dirname(sys.executable)   return os.path.dirname(sys.argv[0])# find path to where we are runningpath_to_script=get_main_dir()# OPTIONAL:# add the sibling 'lib' dir to our module search pathlib_path = os.path.join(get_main_dir(), os.path.pardir, 'lib')sys.path.insert(0, lib_path)# OPTIONAL: # use info to find relative data files in 'data' subdirdatafile1 = os.path.join(get_main_dir(), 'data', 'file1')

Hopefully the above example code can provide additional insight into how to determine the path to the running script...