On Sat, Dec 20, 2008 at 08:16:03AM -0800, C. Titus Brown wrote: -> -> On Fri, Dec 19, 2008 at 05:58:14PM -0800, Istvan Albert wrote: -> -> On Dec 19, 6:03 pm, Christopher Lee <[email protected]> wrote: -> -> -> -> > - I'd like to keep the interface for accessing this as simple as -> -> > possible, e.g. to access the namespace object, the user need do -> -> > nothing more than type -> -> > from pygr import Data -> -> > I guess you were suggesting a two step process to access the namespace -> -> > object: -> -> > import pygr -> -> > data = pygr.Data() -> -> -> -> Maybe it could be made even be simpler than the way it currently -> -> works. Meaning that the default pygr.Data instance is constructed upon -> -> importing pygr. One may use that in most cases or make their own with -> -> some aptly named and customizable constructor. For example the default -> -> Data would in fact correspond to an initialization like this: -> -> -> -> Data = pygr.DataPath( rcfile='.pygrdata', namespaces=["Bio", "Test"], -> -> layers=["my", "here"] ) () -> -> This makes sense to me, too. It would also help in testing pygr.Data, -> because you wouldn't be at the mercy of the import framework details for -> reloading/mocking/stubbing. -> -> Unfortunately, you can't do -> -> from pygr.Data.x.y.z import a -> -> if pygr.Data is not a module, though :(. There may be a way around -> this, but I haven't run across it; does anyone know?
Err, now that I think about it pygr already does something like this... --t -- C. Titus Brown, [email protected] --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "pygr-dev" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/pygr-dev?hl=en -~----------~----~----~----~------~----~------~--~---
