# Local Binary Pattern(LBP)

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C N N on 13 Sep 2011
Commented: Image Analyst on 1 Jun 2022
I am trying to execute local binary pattern in MATLAB using the image processing toolbox. When i execute I can't get a LBP image and LBP histogram.
clear all;
close all;
clc;
figure,imshow(I)
%%Crop
I2 = imcrop(I);
figure, imshow(I2)
w=size(I2,1);
h=size(I2,2);
%%LBP
scale = 2.^[7 6 5; 0 -inf 4; 1 2 3];
for i=2:w-1
for j=2:h-1
J0=I2(i,j);
I3(i-1,j-1)=I2(i-1,j-1)>J0;
I3(i-1,j)=I2(i-1,j)>J0;
I3(i-1,j+1)=I2(i-1,j+1)>J0;
I3(i,j+1)=I2(i,j+1)>J0;
I3(i+1,j+1)=I2(i+1,j+1)>J0;
I3(i+1,j)=I2(i+1,j)>J0;
I3(i+1,j-1)=I2(i+1,j-1)>J0;
I3(i,j-1)=I2(i,j-1)>J0;
LBP(i,j)=I3(i-1,j-1)*2^7+I3(i-1,j)*2^6+I3(i-1,j+1)*2^5+I3(i,j+1)*2^4+I3(i+1,j+1)*2^3+I3(i+1,j)*2^2+I3(i+1,j-1)*2^1+I3(i,j-1)*2^0;
end
end
figure,imshow(LBP)
figure,imhist(LBP)
what is the issue.i am supposed to get numbers from 0 to 255 but i am not getting it. how to correct it?
sajjad nasiri on 28 May 2019

madhulika jain on 22 Sep 2011
I think you shpuld be getting LBP in the range 0 to 255. I ran this code and i got the values in this range only
C N N on 22 Sep 2011
The problem with this code that after you get LBP, you need to convert it to uint8 before displaying it.

Izzo on 3 Dec 2011
hey i think i got your problem, because i already did LBP. First, make sure that u have converted the image into grayscale form. Second, you have to replace the test image with the new LBP image, therefore: instead of this code
LBP(i,j)=I3(i-1,j-1)*2^7+I3(i-1,j)*2^6+I3(i-1,j+1)*2^5+I3(i,j+1)*2^4+I3(i+1,j+1)*2^3+I3(i+1,j)*2^2+I3(i+1,j-1)*2^1+I3(i,j-1)*2^0;
try to change into:
I2(i-1,j-1)=I3(i-1,j-1)*2^7+I3(i-1,j)*2^6+I3(i-1,j+1)*2^5+I3(i,j+1)*2^4+I3(i+1,j+1)*2^3+I3(i+1,j)*2^2+I3(i+1,j-1)*2^1+I3(i,j-1)*2^0;
because i have problem like that too, then my supervisor solve into this for me :P
elfrida on 16 Aug 2013
sir, can you help me to make the program if i want to get LBP value from 13x13 image size?

Image Analyst on 22 Sep 2011
Change the last few lines of your code to this:
subplot(2,2,3);
imshow(LBP, []);
subplot(2,2,4);
[pixelCounts, GLs] = imhist(uint8(LBP));
bar(GLs, pixelCounts);
Walter Roberson on 14 Feb 2021
But I think it does not show me what I want
We do not have your image to test with, and we do not know what kinds of output you expect to see.

remo on 1 Dec 2011
hi CNN,
Sorry for being a Noob. I just cant understand why you use 'scale = 2.^[7 6 5; 0 -inf 4; 1 2 3]; ' in this program. can you explain?
Image Analyst on 29 Mar 2019
Unhide the comments above and you'll see a desciption of how those are computed.

Fa Fa on 13 Jun 2012
Hi All, has somebody the Result, i runed this code but i didnt get the result. Thanks

anusha on 24 Sep 2014
Edited: anusha on 24 Sep 2014
u can try this code
clear all; close all; I=imread('cameraman.tif'); I=rgb2gray(I); I1=imcrop(I); [w h]=size(I1);
for i=2:w-1 for j=2:h-1 val=I1(i,j); scale=2.^[0 1 2;7 -inf 3;6 5 4]; mat=[I1(i-1,j-1) I1(i-1,j) I1(i-1,j+1);I1(i,j-1) I1(i,j) I1(i,j+1);I1(i+1,j-1) I1(i+1,j) I1(i+1,j+1)]; mat=mat>=val; fin=mat.*scale; I1(i,j)=sum(sum(fin)); end end imshow(I1,[]);

Nikolay S. on 16 Mar 2015
You can try one of the following LBP implementations: http://www.mathworks.com/matlabcentral/fileexchange/49787-shift-based-lbp http://www.mathworks.com/matlabcentral/fileexchange/36484-local-binary-patterns I'm using both for a while, updating them fomr time to time.

Ktk on 29 Oct 2017
the code is not giving me any values. can anyone help? thanks!
Image Analyst on 1 Jun 2022
@SATISH SONWANE, isn't that the usual way? You can change it if you want. Why would you want to?

Rahul Pandey on 5 Feb 2020
will u please explain the code which u have written for lbp?@sajjad nasari

Vrushali Patil on 2 Apr 2020
Image Analyst on 2 Apr 2020
Who said you had to crop the image? Just go ahead and do it on your image without cropping it if you don't want to crop it. What's the problem?

Image Analyst on 2 Apr 2020
See my attached LBP demo.

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