Problem to Threshold a Matrix

I need to threshold the surrounding pixels of the given matrix with respect to the centre pixel of the given matrix. If the surrounding values are greater than or equal to the center of the pixel they are given a 1 otherwise they are given a 0. Then I need to store all the values in the shown order to result in a vector which contains the binary value.

[James Tursa]

Have you tried coding this? What problems are you having? Not working, or too slow, or ???

[CharlesB]

matrix = [ 85 99 21; 54 54 86; 57 12 13];

%matrix(2,2) is the centre pixel

thres_mat = matrix > matrix(2,2); % which results in the binary matrix shown

my problem is to store those binary values in that order

Answer 1

[Image Analyst]

Let's call it what it is, okay? You're asking for the " local binary pattern".

For a FULL demo on the whole image, see the attached m-file. it creates this image

[CharlesB]

Simply amazing that's what I am working on the achieve :) thank you!! definitely the best answer.

[CharlesB]

Hi! I have used what you gave me to apply it to cells in a cell array where the cell array is an original image split into various cells 16x16 each. Now my problem is that the line

localBinaryPatternImage(rows, cols) = eightBitNumber;

which is found in the for loop only gives the values of the last 15x15 unit8 cell array. How can i concatenate all of the cell arrays to form the lbp image?

[Image Analyst]

You either need to extract the 16x16 image from the cell and do it on the small arrays, or just don't split the image up into cells. Not sure why you're doing that anyway.

[CharlesB]

I am doing it because I want to plot the histogram and lbp features for every cell then concatenate them rather than have 1 histogram and 1 lbp feature vector for the entire image.

[CharlesB]

%% Grayscale Baboon Image I2 = imread('baboon.png'); [rows,columns,dim] = size (I2); sz2 = [rows, columns];

chunk_size2 = [16 16]; % your desired size of the chunks image is broken into sc2 = sz2 ./ chunk_size2; % number of chunks in each dimension; must be integer % split to chunk_size(1) by chunk_size(2) chunks X2 = mat2cell(I2, chunk_size2(1) * ones(sc2(1),1), chunk_size2(2) *ones(sc2(2),1));

[r, c] = size(X2); z = cell2mat(X2(1));

%Extracting LBP features for each cell and concatinating them into a %histogram

% localBinaryPatternImage = zeros(size(I2), 'uint8'); %for celliter3 = 1:numel(X2)

result = []; for row = 1 : r for col = 1 : c

Z = cell2mat(X2(row, col)); [row_cell, col_cell] = size(Z);

             for rows = 2 : row_cell - 1 
                 for cols = 2 : col_cell - 1 

            centerPixel = Z(rows,cols);
            pixel7= Z(rows-1, cols-1) > centerPixel;  
            pixel6= Z(rows-1, cols) > centerPixel;   
            pixel5= Z(rows-1, cols+1) > centerPixel;  
            pixel4= Z(rows, cols+1) > centerPixel;     
            pixel3= Z(rows+1, cols+1) > centerPixel;    
            pixel2= Z(rows+1, cols) > centerPixel;      
            pixel1= Z(rows+1, cols-1) > centerPixel;     
            pixel0= Z(rows, cols-1) > centerPixel;    
            eightBitNumber = uint8(...
                pixel7 * 2^7 + pixel6 * 2^6 + ...
                pixel5 * 2^5 + pixel4 * 2^4 + ...
                pixel3 * 2^3 + pixel2 * 2^2 + ...
                pixel1 * 2 + pixel0);
            % Or you can use the built-in function bwpack(), which is somewhat simpler but a lot slower.
            % eightBitNumber = uint8(bwpack([pixel0; pixel1; pixel2; pixel3; pixel4; pixel5; pixel6; pixel7]));
            localBinaryPatternImage(rows, cols) = eightBitNumber;

                 end 
              end 
        end 
    end

[Image Analyst]

I don't understand why you want to do that. And anyway, you don't have one LBP feature for the entire image. Every pixel has its own local binary pattern, so you have millions of patterns.

Answer 2

[James Tursa]

Using your small example:

>> x = 2;
>> y = 2;
>> matrix = [ 85 99 21; 54 54 86; 57 12 13]
matrix =
    85    99    21
    54    54    86
    57    12    13
>> t = matrix >= matrix(y,x)
t =
     1     1     0
     1     1     1
     1     0     0
>> b = [t(y,x-1) t(y+1,x-1:x+1) t(y,x+1) t(y-1,x+1:-1:x-1)]
b =
     1     1     0     0     1     0     1     1
>> d = sum(b.*2.^(7:-1:0))
d =
   203

[CharlesB]

thank you :)