How to get the file path of a module from a function executed but not declared in it?

Question

If I want the path of the current module, I\'ll use __file__.

Now let\'s say I want a function to return that. I can\'t do:

def get_path         


        

Answer 1

I found a way to do it with the inspect module. I'm ok with this solution, but if somebody find a way to do it without dumping the whole stacktrace, it would be cleaner and I would accept his answer gratefully:

def get_path():
    frame, filename, line_number, function_name, lines, index =\
        inspect.getouterframes(inspect.currentframe())[1]
    return filename