When i apply a matrix to a buffergeometry
I want to get the updated position attributes fast , i am dealing with 1000000+ vertex .
I have tried Matrix4.applyToBufferAttribute() , but the buffer attribute is still the same
What is the most proper way to perform this ?
I have tried Matrix4.applyToBufferAttribute() , but the buffer attribute is still the same
Then it seems you are doing something wrong in your application. Matrix4.applyToBufferAttribute() does apply the matrix to the given attribute. The method is used multiple times in the core of three.js for example in BufferGeometry.applyMatrix():
https://github.com/mrdoob/three.js/blob/9f7f38b543c8a51d5614b72c04d657a4cfad68da/src/core/BufferGeometry.js#L141-L142
Ensure to set BufferAttribute.needsUpdate to true after the method invocation. And yes, it's the intended way to apply a 4x4 transformation matrix to a buffer attribute.
Related
I'm trying to finetune the pretrained Transformer-XL model transfo-xl-wt103 for a language modeling task. Therfore, I use the model class TransfoXLLMHeadModel.
To iterate over my dataset I use the LMOrderedIterator from the file tokenization_transfo_xl.py which yields a tensor with the data and its target for each batch (and the sequence length).
Let's assume the following data with batch_size = 1 and bptt = 8:
data = tensor([[1,2,3,4,5,6,7,8]])
target = tensor([[2,3,4,5,6,7,8,9]])
mems # from the previous output
My question is: I currently pass this data into the model like this:
output = model(input_ids=data, labels=target, mems=mems)
Is this correct?
I am wondering because the documentation says for the labels parameter:
labels (:obj:torch.LongTensor of shape :obj:(batch_size, sequence_length), optional, defaults to :obj:None):
Labels for language modeling.
Note that the labels are shifted inside the model, i.e. you can set lm_labels = input_ids
So what is it about the parameter lm_labels? I only see labels defined in the forward method.
And when the labels "are shifted" inside the model, does this mean I have to pass data twice (additionally instead of targets) because its shifted inside? But how does the model then know the next token to predict?
I also read through this bug and the fix in this pull request but I don't quite understand how to treat the model now (before vs. after fix)
Thanks in advance for some help!
Edit: Link to issue on Github
That does sound like a typo from another model's convention. You do have to pass data twice, once to input_ids and once to labels (in your case, [1, ... , 8] for both). The model will then attempt to predict [2, ... , 8] from [1, ... , 7]). I am not sure adding something at the beginning of the target tensor would work as that would probably cause size mismatches later down the line.
Passing twice is the default way to do this in transformers; before the aforementioned PR, TransfoXL did not shift labels internally and you had to shift the labels yourself. The PR changed it to be consistent with the library and the documentation, where you have to pass the same data twice.
How can I access the perspective camera's projection matrix directly and change one or more of the 16 values?
I tried the code bellow with and without .updateProjectionMatrix() and it doesn't work, probably it is overiden by an internal function:
cameraPersp.projectionMatrix.elements.set =
(a,b,c,d,
e,f,g,h,
i,j,k,l,
m,n,o,p);
cameraPersp.updateProjectionMatrix();
Also, I have no idea if it can be multiplied, added etc using .set (lack of documentation) -it doesn't raise an error though.
So, to set a given matrix to custom float values, directly:
cameraPersp.projectionMatrix.set
(a,b,c,d,
e,f,g,h,
i,j,k,l,
m,n,o,p);
To change a single matrix value directly:
cameraPersp.projectionMatrix.elements[n] = yourfloatvalue;
In both cases, updateProjectionMatrix() should NOT be called.
In an old version of my code, I used to do a hardcopy() with a given resolution, ie:
frame = hardcopy(figHandle, ['-d' renderer], ['-r' num2str(round(pixelsperinch))]);
For reference, hardcopy saves a figure window to file.
Then I would typically perform:
ZZ = rgb2gray(frame) < 255/2;
se = strel('disk',diskSize);
ZZ2 = imdilate(ZZ,se); %perform dilation.
Surface = bwarea(ZZ2); %get estimated surface (in pixels)
This worked until I switched to Matlab 2017, in which the hardcopy() function is deprecated and we are left with the print() function instead.
I am unable to extract the data from figure handler at a specific resolution using print. I've tried many things, including:
frame = print(figHandle, '-opengl', strcat('-r',num2str(round(pixelsperinch))));
But it doesn't work. How can I overcome this?
EDIT
I don't want to 'save' nor create a figure file, my aim is to extract the data from the figure in order to mesure a surface after a dilation process. I just want to keep this information and since 'im processing a LOT of different trajectories (total is approx. 1e7 trajectories), i don't want to save each file to disk (this is costly, time execution speaking). I'm running this code on a remote server (without a graphic card).
The issue I'm struggling with is: "One or more output arguments not assigned during call to "varargout"."
getframe() does not allow for setting a specific resolution (it uses current resolution instead as far as I know)
EDIT2
Ok, figured out how to do, you need to pass the '-RGBImage' argument like this:
frame = print(figHandle, ['-' renderer], ['-r' num2str(round(pixelsperinch))], '-RGBImage');
it also accept custom resolution and renderer as specified in the documentation.
I think you must specify formattype too (-dtiff in my case). I've tried this in Matlab 2016b with no problem:
print(figHandle,'-dtiff', '-opengl', '-r600', 'nameofmyfig');
EDIT:
If you need the CData just find the handle of the corresponding axes and get its CData
f = findobj('Tag','mytag')
Then depending on your matlab version use:
mycdata = get(f,'CData');
or directly
mycdta = f.CData;
EDIT 2:
You can set the tag of your image programatically and then do what I said previously:
a = imshow('peppers.png');
set(a,'Tag','mytag');
I am using Paraview 5.0.1. If any solution requires updating, I can try.
I want to programmatically obtain field plots (and corresponding PlotOverLine) of displacements and stresses in rotated coordinate systems.
What are appropriate/convenient/possible ways of doing this?
So far, I have created one Calculator filter for each component of displacements and stresses.
For instance, I used Calculators in 2D with results
(displacement.iHat)*cos(0.7853981625)+(displacement.jHat)*sin(0.7853981625)
(stress_3-stress_0)*sin(45.0*3.14159265/180)*cos(45.0*3.14159265/180)+stress_1*((cos(45.0*3.14159265/180))^2-(sin(45.0*3.14159265/180))^2)
It works fine, but it is quite cumbersome, in several aspects:
Creating them (one filter per component).
Plotting several of them in a single XY plot
Exporting them (one export per component).
Is there a simple way to do this?
PS: The Transform filter does not accomplish this. It rotates the view, not the fields.
Two solutions:
Ugly, inneficient solution
Use Transform and check "Transform All Input vectors"
Add a calculator and add a dummy array
Use transform the other way around, without checking "Transform All Input vectors"
Correct solution :
Compute the transformation yourself in a programmable filter
input = self.GetUnstructuredGridInput();
output = self.GetUnstructuredGridOutput();
output.ShallowCopy(input)
data = input.GetPointData().GetArray("YourArray")
vec = vtk.vtkDoubleArray();
vec.SetNumberOfComponents(3);
vec.SetName("TransformedVectors");
numPoints = input.GetNumberOfPoints()
for i in xrange(0, numPoints):
tuple = data.GetTuple(i)
transform(tuple) # implement the transform in python
vec.InsertNextTuple(tuple)
output.GetPointData().AddArray(vec)
I am migrating a THREE.js app from r60 to r70. Amongst other changes I notice that r60 constructs of the following form no longer work in r70.
mesh.position.set(0,0,0);
myVector3 = new THREE.Vector3 (100,200,300);
mesh.position = myVector3;
This applies to meshes, pointLights, presumably to all Object3D's but I havent tested further.
In the above example the mesh.position x,y,z values remain unchanged at (0,0,0). For illustration see this JSFiddle and compare lines 70 and 73.
//...The next line DOES NOT update Sphere_mesh.position.x
Sphere_mesh.position = NewPos_vector3;//...
//...The next line DOES update Sphere_mesh.position.x
Sphere_mesh.position.x = NewPos_vector3.x
In a debugger no console warning is given during execution that the assignment has not worked. In the very brief THREE.js migration notes for (r68 --> r69) I see something about an Object3D's position no longer being immutable but I don't know what that means.
Anyway, my question
Is there a standard THREE construct or function I can use to copy x,y,z values from a Vector3 object to the x,y,z values of a mesh.position rather than the effective, but verbose, form such as
mesh.position.x = myVector3.x;
mesh.position.y = myVector3.y;
mesh.position.z = myVector3.z; ?
e.g. something such as
mesh.position = F_Get_Object3DPosition_from_Vector3(myVector3); ?
I know it would be easy to write my own function but a standard THREE function would be more likely to evolve smoothly with future versions of THREE.
position beeing immutable means that the position property can not be changed.
so
mesh.position = anything;
won't work (but you already discovered this)
what you can do is not change the position, but you have to change position values.
in your case, the easiest way is
mesh.position.copy (myVector3);
I think you meant myVector3 not myVector3() in the last line... Anyway, I though that would work too but the thing is, you are applying a Vector to something that supposed to be a point/Vertex. Even if that worked in my opinion it wasn't the right way to do it. How about using a simple array:
mesh.position.set(0,0,0);
new_position = [100,200,300]
mesh.position.fromArray(new_position,0)
in which 0 is the starting index. So you can have multiple position sets in one array