How to get texture index from the local binary pattern histogram? I want to get a single value for an image and compare with multiple images.

I have tried 'extractLBPFeatures' function. However, it is providing 1x59 vector as output. I want to get a single value for an image as that can be easy to compare with the other images.

Can I use the standard deviation of the lbp histogram as a measure a texture?

Answer 1

[Vaibhav]

Hi Nabanita,

I understand that you are looking for a method to compare texture indices among images.

The squared error between LBP features serves as a metric to gauge texture similarity, with a smaller value suggesting a closer resemblance between images.

Suppose "lbp1" contains the LBP features for image 1, and "lbp2" contains the LBP features for image 2, the squared error can be computed as follows:

Img1vsImg2 = (lbp1 - lbp2).^2

The code snippet below can be used to visualize the difference using a bar graph:

Img1vsImg2 = (lbp1 - lbp2).^2

figure
bar(Img1vsImg2)
title('Squared Error of LBP Histograms')
xlabel('LBP Histogram Bins')
legend('Img1 vs Img2')

While considering a single extracted LBP value, options like taking the mean or median can also be explored. However, using squared error provides a clearer indication of texture differences among the images.

You can refer to the following MathWorks documentation to know more about "extractLBPFeatures" and squared error calculation:

https://www.mathworks.com/help/vision/ref/extractlbpfeatures.html?s_tid=doc_ta

Hope this helps!