K-NN Classsification on images
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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?
采纳的回答
Arthur Goldsipe
2017-3-16
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 个评论
Arthur Goldsipe
2017-3-16
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.
更多回答(1 个)
Image Analyst
2017-3-17
I know you've already accepted an answer, but you might want to check out my knn demo.
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