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

I have a package that looks like the following

  package / __init__.py Module     in the module   
 
  def function (file_name): open (file_name) As with F: # do stuff  
 
 Somewhere else in a arbitrary directory I have a python file that looks like something 
 
  package package import Please. Function ("some_file .txt")  
 
 But upon turning it on, it is giving me the  FileNotFoundError: [Errno 2] There is no such file or directory: "some_file. Txt ". 
 
 The problem is that the complete path of  some_file.txt  may look like  c: \ user \ user \ document \ some_file.txt , but < Code> package.function  Path complete path is something like  c: \ user \ user \ document \ package \ some_file.txt . Is there a way I can create this from which to open the file by calling  package.function  from some file outside the  package  directory? Can I join? 
 
 Sorry if my terminology is really ambiguous, then I am actually unfamiliar with the  os  stuff. 
 
  Edit:  The exact file setup looks like this: 
 
  directory / foo.py package / __init__.py module.py a Other directories / bar.txt  
 
 and  foo  
 
 
  
 
  < P> 
 
  Class = "post-text" itemprop = "text"> 
 I think you are missing points. 
 
 It is not important that where the source code is, the relative path (and the pure file name  is  relative path) are always interpreted relative to that directory in the process of program Walks (ie directory, when you type  dragon c: \ path \ to \ my \ python \ code \ code.py )