Effacer les filtres
Effacer les filtres

K-NN Classsification on images

3 vues (au cours des 30 derniers jours)
prahlad h
prahlad h le 16 Mar 2017
Commenté : Walter Roberson le 28 Déc 2017
i1=dicomread('1.dcm');
i2=dicomread('2.dcm');
i3=dicomread('3.dcm');
i4=dicomread('4.dcm');
Sample=[i1;
i2];
Training=[i3;
i4];
Group=['1';
'2'];
k=2;
Class = knnclassify(Sample, Training, Group, k);
disp(Class);
I get an error - The length of GROUP must equal the number of rows in TRAINING.
The size of both the images are same.
It works when I'm using co-ordinates instead of i1, i2 and so on.
I know I should be using fitcknn, but it should work as well. Any inputs please?

Réponse acceptée

Arthur Goldsipe
Arthur Goldsipe le 16 Mar 2017
Hi,
Your input arguments Training and Group must have the same number of rows. I see that Group has only 2 rows, but Training probably has many more rows that than. Training consists of tow matrices stacked on top of each other. These matrices represent images as M-by-N matrices. So if all your images (i1, i2, i3, and i4) are all of size M-by-N, then Sample and Training have 2M rows. If your intention is to treat each image as a single entity, then you could construct Sample and Training as follows:
Sample = [i1(:) i2(:)]';
Training = [i3(:) i4(:)]';
That said, I don't think this classification approach is very useful if you only have one training example from each class.
-Arthur
  3 commentaires
Arthur Goldsipe
Arthur Goldsipe le 16 Mar 2017
i1(:) reshapes the 512x512 matrix into a (512x1) column vector. The square brackets concatenate . And ' is the transpose operator. So Sample ends up being a 2x512 matrix.
In this case, what I mean by a "single entity" is a "single row" in the matrix, because that's how knnclassify identifies what numbers belong to the same observation.
prahlad h
prahlad h le 17 Mar 2017
Sorry for the noob question but what do you mean by a column vector? When you're reshaping 512x512 matrix into a 512x1 column vector, is there any loss in the image?
Also, a single row, does that mean that all the pixels are not compared when knnclassify is used?
It's working fine but I just wanted to know for my understanding!

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Image Analyst
Image Analyst le 17 Mar 2017
I know you've already accepted an answer, but you might want to check out my knn demo.
  1 commentaire
prahlad h
prahlad h le 17 Mar 2017
Sure. I'll look at it. It would help me with understanding it a little better. I've been through your other demos on segmentation and thresholding. They are amazing!

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