Description
pymat.vis._model.Vis which gets returned e.g. from stainless.vis looks like an internal implementation class
Steps to Reproduce
In [17]: from pymat import stainless
In [18]: stainless.vis
Out[18]: Vis(source='ambientcg', material_id='Metal012', tier='1k', finishes={'brushed': {'source': 'ambientcg', 'id': 'Metal012'}, 'polished': {'source': 'ambientcg', 'id': 'Metal049A'}, 'dirty': {'source': 'ambientcg', 'id': 'Metal049B'}}, roughness=0.3, metallic=1.0, base_color=(0.75, 0.75, 0.77, 1.0), ior=None, transmission=0.0, clearcoat=None, emissive=None)
In [19]: type(stainless.vis)
Out[19]: pymat.vis._model.Vis
In my code I now need to say
pbr: pymat.vis._model.Vis = stainless.vis
which looks like using an internal model not meant to be exposed.
Do I miss something?
Expected Behavior
I would expect Vis to be a clearly public class (now private model in the module path)
Actual Behavior
see above
Environment
any
Additional Context
No response
Possible Solution
No response
Changelog Category
Fixed
Description
pymat.vis._model.Viswhich gets returned e.g. fromstainless.vislooks like an internal implementation classSteps to Reproduce
In my code I now need to say
which looks like using an internal model not meant to be exposed.
Do I miss something?
Expected Behavior
I would expect
Visto be a clearly public class (now private model in the module path)Actual Behavior
see above
Environment
any
Additional Context
No response
Possible Solution
No response
Changelog Category
Fixed