revisit get_package_dir

[sbillinge]

Problem

no problem, but I think we are handling this task differently in other packages so let's revisit and see if there is a standard way to do it. In general, I would rather not have it in the __init__.py file. I am not sure why, but I never really like importing things from there as a I don't fully understand the behavior.

Proposed solution

Right now it has to be there because files(__name) returns the package name when it is called from the init for some reason, and then this is returned as a context manager. I think in other packages in diffpy where we are trying to do the same thing, we do something like find the path of current file as a Pathlib object, then recurse around using .parent and so on to get the path we are after and then pass that to a function that opens the file. This just seems more explicit and less magical to me.

[cadenmyers13]

Please see this thread for more info #55

[ycexiao]

@sbillinge I didn't find the exact example of using Pathlib to get the package dir in other diffpy packages, but I'm wondering if the behavior below is what you want:

def get_package_dir(root_path=None):
    if root_path is None:
        resource = files(__name__)  # change into Path(__file__).parents[?],
    else:
        resource = root_path
    return as_file(resource)

So that the function can be moved to wherever in the diffpy.cmi. If it is the case, where do we want to put the function? packsmanager.py?

[sbillinge]

Yes, 100%correct on all