K-NN Classsification on images
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prahlad h
on 16 Mar 2017
Commented: Walter Roberson
on 28 Dec 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?
Accepted Answer
Arthur Goldsipe
on 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 Comments
Arthur Goldsipe
on 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.
More Answers (1)
Image Analyst
on 17 Mar 2017
I know you've already accepted an answer, but you might want to check out my knn demo.
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