Plotter's mode=1 strange behaviour

[sergei9838]

Hi, Marco,
This is what I have discovered in the development version of vedo.

mode=1 option in Plotter is supposed to be rotating the actor. Rotate the cube in the renderer somehow and save the camera position:

from vedo import *
C = Cube()
cam1 = show(C, mode=1).camera
close()

Get the camera position:

cam1.GetPosition()
# (0.0, 0.0, 3.3460652149512318)

The camera is in the default position, as expected for mode=1. The view of the cube is as expected too:

show(C).close()

BUT: the vertices of the cube are at the initial position!

C.vertices
# array([[-0.5, -0.5, -0.5],
#       [-0.5, -0.5,  0.5],
#       [-0.5,  0.5, -0.5],
...

So what figure does the plotter show then?!
Moreover, let's clone the cube and display both:

C1 = C.clone()
show(C, C1.c("yellow"), axes=1).close()

So C1 is the correct copy of the initial cube. And not of the rotated C shown in the plotter.
This does not happen in vedo 2023.4.6: C.points() show rotated positions.
I am puzzled )
Cheers,
Sergei

[marcomusy]

Hi, that happens because you are changing the representation of the object on the gpu but not the underlying object. You can still apply the transform from the actor with:

from vedo import *
C = Cube()
cam1 = show(C, mode=1).camera
close()

C.apply_transform_from_actor()
print(C.vertices)

[sergei9838]

Thanks Marco, but now I am even more confused. If you call show(C) the second time, it will display the rotated cube from the first call. It means this GPU representation is stored somewhere in C. I guess in C.actor since show(C.actor) produces identical view to show(C). But then it means that calling show(C) I am actually viewing C.actor? How then I can be sure calling show on a mesh that I am actually viewing the mesh and not its actor from some previous calls?

This was not the way it worked in the earlier versions of vedo, as I mentioned. It would be more logical to apply_transform_from_actor() on the window's quit keeping the object and its actor in sync. This is my understanding of mode=1, otherwise it's identical to mode=0 plus additional confusion.

PS. Thus C1=C.clone() does not copy its actor since C1.apply_transform_from_actor() returns the identical transform. Is it intended?

Have a nice weekend, Marco, and thank you for your efforts and time!
Sergei

[marcomusy]

This is my understanding of mode=1, otherwise it's identical to mode=0 plus additional confusion.

this (an others) are very good points.
There is no ambiguity with the actor because now vedo does not expose any api to modify the actor-related vtkTransform object.
so all actors always have Identity transform.
Except in .user_mode(1) and .user_mode(3): for this modes we should indeed check if actors on screen were modified.

I'm flagging this as a bug to fix, thanks for forcing me to think better :)

[marcomusy]

You can try with the latest dev?

[sergei9838]

Oh, just noticed that MousePan() mode also creates discrepancy between the mesh and its actor

[marcomusy]

Uhm.. this seems to work fine:

from vedo import *

cube = Cube()
mode = interactor_modes.MousePan()

plt = Plotter(axes=1)
plt.user_mode(mode)
plt.show(cube)

print(cube.actor.GetMatrix())

I always get:

vtkMatrix4x4 (0x600001be0800)
  Debug: Off
  Modified Time: 194
  Reference Count: 2
  Registered Events: (none)
  Elements:
    1 0 0 0
    0 1 0 0
    0 0 1 0
    0 0 0 1

[sergei9838]

Probably the issue is not with MousePan but with viewing Image.

dog = Image(dataurl+"images/dog.jpg")
show(dog.shift((100,100,0))).close()

produces error:

[vedo.plotter] INFO: object 'Image' was manually moved. Updated to its current position.
...
    a.retrieve_object().apply_transform_from_actor()
AttributeError: 'Image' object has no attribute 'apply_transform_from_actor'

[marcomusy]

try again on the latest dev..

[sergei9838]

Yes, no complaints any more!
One small suggestion: scale for Image does not have origin= option . Could you implement it? It is important for making a stencil from image standing in from of the object like I showed in this discussion.
Right now I am doing it like this:

S = LinearTransform()
S.scale(s=0.9, origin=camera_position)
dog.apply_transform(S)

Not much of overhead, but, I think, it would be more unified to have the origin option.

[marcomusy]

Makes sense! Would you mind opening a PR, you just need add origin=True in visual.Actor3DHelper.scale().

[sergei9838]

Yes, this works just fine - thank you!!!

def scale(self, s, origin=True):
    """Multiply object size by `s` factor with the homothetic center in `origin`"""
    LT = vedo.LinearTransform().scale(s, origin)
    return self.apply_transform(LT)