How I can find the indices of 4 consecutive elements in the same row?

I have a big binary matrix, and I try to find the indices of certain elements. Let's say for example I have this matrix

SA = [ 0 0 0 0 0 1 1 0 0 0 1 0 1 0 1 1 0 0 1 1; 

       1 0 0 0 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 1;

       0 1 0 0 0 0 1 0 1 1 1 0 1 0 1 0 0 1 1 1];

each 4 consecutive elements is considered together, so how I can find the indices of 1 0 1 0? I just need the indices for the first element, and assume no repetition for the same consecutive 4 elements!

Answer 1

[KSSV]

clc; clear all 

SA = [ 0 0 0 0 0 1 1 0 0 0 1 0 1 0 1 1 0 0 1 1; 
     1 0 0 0 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 1;
     0 1 0 0 0 0 1 0 1 1 1 0 1 0 1 0 0 1 1 1];   % Your matrix

[m,n] = size(SA) ;                                 % Dimensions of your matrix
B = [1 0 1 0] ;                                    % Matrix to compare 
myidx = [] ;                              % Initialize your indices needed                                         
% Loop for each row and column 
for i = 1:m 
    for j = 1:n-length(B)
        if SA(i,j) == B(1)
            if SA(i,j+1)==B(2) && SA(i,j+2) == B(3) && SA(i,j+3) == B(4) 
                myidx = [myidx ; [i,j]] ;
            end
        end
     end
 end

[Osama Hussein]

Thank you, The code works, I just need to choose the indices which begins at 1 or 5 or 9 ... since each 4 consecutive elements are together. I think I can do this, Thank you very much :)

[Stephen23]

See Image Analyst's answer for a much faster and neater solution.

Answer 2

[Image Analyst]

I offer a much simpler solution:

for row = 1 : size(SA, 1)   
  columns{row} = strfind(SA(row,:), [1,0,1,0])
end

[Osama Hussein]

Thank you, but I think it gives me wrong answer or I don't understand the what columns is. I try to change sequence but still there is something unclear, could you clarify it, please?

[Stephen23]

Columns is simply a cell array of the column indices. You could even do it on one line using cellfun:

 col = cellfun(@(v)strfind(v,[1,0,1,0]),num2cell(SA,2),'Uni',0);

The answer you accepted has two nested loops, thirteen lines of code, and multiple temporary variables. MATLAB code does not need to be so complicated to perform trivial tasks like this!

[Image Analyst]

Thanks Stephen. Osama, look at the output of it:

columns = 
    [11]    [15]    [1x2 double]

So columns{1} tells where 1 0 1 0 shows up in row #1. You can see that that happens at column #11 in row #1. So that one is correct.

columns{2} tells where 1 0 1 0 shows up in row #2. You can see that that happens at column #15 in row #2. So that one is also correct.

columns{3} is a 2 element array which is [11, 13]. It tells where 1 0 1 0 shows up in row #3. You can see that that happens both at column #11 and at column #13 in row #3. So that one is also correct.

So why do you think it may be giving you the wrong answer? What columns do you think the pattern should show up in? What do you think the right answer should be?