locate element using find function in 3D matrix

hello,

I am working in the code bellow. The main purpose of question is to determine the location of varibale B(:,:,i). respecting the 3 indeex of the matrix.

%------ the declation of matrix and choose the 9 largest value of the matrix

t = input('Enter indeex 3 of your matrix - ')
matrix2 = rand(6,6,t)
for i = 1:t
    T2(:,:,i) = unique(matrix2(:,:,i))   % Determine the unique of variable matrix. tunique size 6*6*1 , 6*6*2 and 6*6*3.......6*6*t
    B(:,:,i) = T2(27:36,:,i)             % I choose only the last 9 nums from t unique
end


% I need to know the position of each B(:,:,i) in  the main matrix2 and
% their indeex
for d = 1 : t
    for j = 1:9 
    [a(j) b(j)] = find(matrix2(:,:,d) == B(j,:,d)); % I used the function find to locate the varivale o in the matrix
    end
end

I really appreciate any help

Answer 1

[DGM]

I am not really sure what the end goal here is, but to just get the row/column indexes of the subset of values selected

t = input('Enter size of your matrix along dim 3 - ')
matrix2 = rand(6,6,t);

num=9; % define this as a paramter.  call it whatever is appropriate

B=zeros([num t]); % preallocate B
for i = 1:t
    % unique() returns a vector, not a 2D array.
    % if we don't need to keep T2 for anything, don't.
    % we can't guarantee the length of T2, so flip it
    % that way we're not indexing from the free end
    T2 = flipud(unique(matrix2(:,:,i)));
    B(:,i) = T2(1:num);
end

% i'm going to throw the results in a cell array
% since there's no guarantee that the results will have a common size
locations=cell(t,num);
for d = 1:t
    for j = 1:num
        [row col] = find(matrix2(:,:,d) == B(j,d)); 
        locations{d,j}=[row col];
    end
end

% look at the results
celldisp(locations);

This will give the results in a cell array, with each cell containing the row and column indexes corresponding to each value in the vector B. For an array with t=1

locations{1} =
     1     3
locations{2} =
     2     4
locations{3} =
     6     2
locations{4} =
     3     1
locations{5} =
     4     2
locations{6} =
     2     3
locations{7} =
     5     6
locations{8} =
     4     5
locations{9} =
     3     3

Why did I choose to use a cell array? Since you are using unique() to filter the values in matrix2, I can only assume that you expect the values in matrix2 to be potentially nonunique and that is important for your tasks. If that is the case, then the results returned by find() will be of variable length. This can be demonstrated by using randi() instead of rand():

matrix2 = randi(9,6,6,t);

In this case, we are guaranteeing nonuniqueness for the purpose of demonstration, so the output will look like this for the same size array:

locations{1} =
     6     1
     1     2
     5     2
     4     3
     1     5
     3     5
locations{2} =
     5     4
     4     6
locations{3} =
     1     3
     2     3
     1     4
     6     5
locations{4} =
     4     2
     6     2
     3     4
     4     4
     6     4
     2     6
     5     6
locations{5} =
     3     1
     3     2
locations{6} =
     6     3
locations{7} =
     5     3
     2     5
     4     5
     5     5
     3     6
locations{8} =
     1     1
     2     1
     4     1
     5     1
     3     3
locations{9} =
     2     2
     2     4
     1     6
     6     6

If on the other hand, you were only using unique() because you meant to be using sort() instead, then things could be simplified to use a numeric output array.