I have a Python module which uses some resources in a subdirectory of the module directory. After searching around on stack overflow and finding related answers, I managed to d
Building on lunaryorn's answer, I keep a function at the top of my modules in which I have to build multiple paths. This saves me repeated typing of join
s.
def package_path(*paths, package_directory=os.path.dirname(os.path.abspath(__file__))):
return os.path.join(package_directory, *paths)
To build the path, call it like this:
font_file = package_path('fonts', 'myfont.ttf')
Or if you just need the package directory:
package_directory = package_path()