[ITK-users] [ITK] PCA on 3D Data

Timothee Evain tevain at telecom-paristech.fr
Mon May 23 04:02:38 EDT 2016


Hi Robert,

That's quite difficult to help you without more insight on what you are expecting to get or to do with this shape model.
When you use ImagePCAShapeModelEstimator to generate your model, you should expect a mean image as a signed distance map (i.e. the mean shape is the set of pixels <0), and some "variation" images corresponding to the principal components. These images can multiplied to some magnitude and then added/subtracted to the mean image to get a particular instance of your model as a new distance map. To see the new shape, just threshold to get all pixels <0.
You can refer to: 
https://itk.org/Doxygen/html/classitk_1_1PCAShapeSignedDistanceFunction.html
which is the function used by itk in the geodesic level set segmentation with shape prior. 

HTH,

Tim

----- Mail original -----
De: "Robert" <scorpiuni at gmail.com>
À: insight-users at itk.org
Envoyé: Dimanche 22 Mai 2016 22:16:36
Objet: [ITK] [ITK-users] PCA on 3D Data

Hello,
I have a set of masked anatomical structures on a CT/MRT/... Scan.
It basically is segmented Volume Data, and I want a Shape model based on
those files.
So I used the ImagePCAShapeModelEstimator, which I input my data to. I've
read that you need to provide distance maps, so I used
SignedDanielssonDistanceMapImageFilter on my thresholded, binarized data.
Now when I write the mean image as well as the different components as .nrrd
volume, I get a "mean distance map" as mean image. This output as distance
map however is not the result that I need.
I then tried to generate some actual shape models using the previous mean
image with corresponding eigenvalues, but the result is not really helpful.
I guess I could get better results here when experimenting more with the
lower and upper threshold values of my final thresholding filter, but I
first wanted to ask if there is a better alternative way to create a mean
image and shape models, and to ask if my approach is even correct at all.
Thanks for your attention,
Robert



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