Python 3.4 - Opening a file in a module with a different directory -

i have package looks following

package/     __init__.py     module.py 

in module.py have this

def function(file_name):     open(file_name) f:         # stuff 

somewhere else in arbitrary directory have python file looks this

import package  package.function("some_file.txt") 

but upon running it, it's giving me filenotfounderror: [errno 2] no such file or directory: "some_file.txt".

the problem absolute path of some_file.txt may c:\users\user\documents\some_file.txt, in package.function path absolute path c:\users\user\documents\package\some_file.txt. there way can make calling package.function file outside package directory automatically includes absolute path of file want open?

sorry if terminology ambiguous, i'm unfamiliar os stuff.

edit: exact file setup have looks this:

directory/     foo.py     package/         __init__.py         module.py     another_directory/         bar.txt 

and foo.py looks this

import package  package.function("another_directory/bar.txt") 

i think you're missing point.

it's not important source code lies, relative paths (and pure file names are relative paths) interpreted relative directory program process runs in (ie. directory when type python c:\path\to\my\python\code\code.py )