I read a tutorial on OpenIMAJ & got the following code from that tutorial. According to the code, I get the center of clusters for an image. But from that point onward I don't know how to use these values in order to compare 2 images. This is the code. I have added the comments as it was in the documentation, so you can get an idea.
public static void main(String[] args) throws IOException
{
final String input_1Str ="/compareimage/clip6.jpg";
MBFImage input = ImageUtilities.readMBF(objectRecognition.class.getResourceAsStream(input_1Str));
input = ColourSpace.convert(input, ColourSpace.CIE_Lab); //apply a colour-space transform to the image
FloatKMeans cluster = FloatKMeans.createExact(2); //construct the K-Means algorithm. The parameter (2) is the number of clusters or classes we wish the algorithm to generate
FloatKMeans.createKDTreeEnsemble(2);
float[][] imageData = input.getPixelVectorNative(new float[input.getWidth() * input.getHeight()][3]); //The FloatKMeans algorithm takes its input as an array of floating point vectors (float[][]). We can flatten the pixels of an image into the required form using the getPixelVectorNative() method
FloatCentroidsResult result = cluster.cluster(imageData);
float[][] centroids = result.centroids; //The K-Means algorithm can then be run to group all the pixels into the requested number of classes
for (float[] fs : centroids) {
System.out.println(Arrays.toString(fs)); //Each class or cluster produced by the K-Means algorithm has an index, starting from 0. Each class is represented by its centroid. Running it prints the (L, a, b) coordinates of each of the classes
}
After running the above code for an image I got the following result.
Image 1:
[95.99463, 3.6134863, 2.641686]
[32.080914, 14.114657, -48.14841]
After running the above code on another image gives me the following
Image 2:
[32.11119, 13.966739, -47.994236]
[95.64963, 3.9856236, 2.9136417]
By using these values how can I compare the 2 images. I don't have any ideas on how to compare the images. Please help me. Thanks in advance.
