Inserting non zero element in diagonal position of a nonzero matrix

I m trying to insert Insert set of non zero elements in diagonal position of a nonzero matrix

I now have.

A1= [2 3 4 6 7 8

1 3 4 5 7 8

1 2 4 5 6 8]

A2=[1 5

2 6

3 7]

My desired matrix =

A = [ 1 2 3 4 5 6 7 8

1 2 3 4 5 6 7 8

1 2 3 4 5 6 7 8]

A is divided into 2blocks, each of size 3x4. The first approaching diagonal elements are extracted and given as A2.

Help me insert A2 into A1 so as to get my desired matrix A.

[James Tursa]

It is not clear what is really needed because your numbers don't seem to match up. That is, simply inserting A2 somehow into A1 can't give you the A result you show. E.g., there are only two 8's in A1 and A2, but there are three 8's in A. How? What is the actual rule for creating A?

[Madhumitha Jayaram]

A1= [2 3 4 6 7 8

1 3 4 5 7 8

1 2 4 5 6 8]

I apologise for the typo, this is my A1 matrix.

initially, A is the matrix from which A1 and A2 are extracted.

For each block of size 3x4,

1st element of row one, 2nd element of row 2 and third element of row 3 are extracted at the encoding side and the matrix is separated as A1 and A2. At the decoding, I need them back in my desired matrix A

Answer 1

[DGM]

This isn't that much different than Benjamin's answer now that I look at it, but eh. I guess it's my own spin.

This should work for any A1 array with size integer-divisible by 3

A1 = [2 3 4 6 7 8;
      1 3 4 5 7 8;
      1 2 4 5 6 8];
A2 = [1 5;
      2 6;
      3 7]*11;

% use larger versions of the test arrays to demonstrate generality
A1 = repmat(A1,2,2);
A2 = repmat(A2,2,2);

% assumes that size(A1) is integer-divisible by 3 in both dimensions
brow = size(A1,1)/3;
bcol = size(A1,2)/3;
C1 = mat2cell(A1,ones(1,brow)*3,ones(1,bcol)*3);
C2 = mat2cell(A2,ones(1,brow)*3,ones(1,bcol));

diagmask = logical([1 0 0; 0 1 0; 0 0 1; 0 0 0]);
A = cell(size(C1));
for br = 1:brow
    for bc = 1:bcol
        newblock = zeros(4,3);
        newblock(diagmask) = C2{br,bc};
        newblock(~diagmask) = C1{br,bc}.';
        A{br,bc} = newblock.';
    end
end
A = cell2mat(A)

A = 6×16

11 2 3 4 55 6 7 8 11 2 3 4 55 6 7 8 1 22 3 4 5 66 7 8 1 22 3 4 5 66 7 8 1 2 33 4 5 6 77 8 1 2 33 4 5 6 77 8 11 2 3 4 55 6 7 8 11 2 3 4 55 6 7 8 1 22 3 4 5 66 7 8 1 22 3 4 5 66 7 8 1 2 33 4 5 6 77 8 1 2 33 4 5 6 77 8

[Madhumitha Jayaram]

Thank you so much. This works.

Answer 2

[_]

If I understand what you are trying to do, here is one way. I got some discrepancies between this code's output A and the A you say should result (in the 2nd row, 7th column and in the 3rd row, 8th column), but I believe it was due to typos in your A. It's also possible I misunderstood the objective. I've added 10 to your A2 in order to better see the elements that have been added in.

A1 = [ ...
    2 3 4 6 7 8; ...
    1 3 4 5 6 8; ...
    1 2 4 5 6 7];
A2 = [ ...
    1 5; ...
    2 6; ...
    3 7]+10;
display(A1);

A1 = 3×6

2 3 4 6 7 8 1 3 4 5 6 8 1 2 4 5 6 7

display(A2);

A2 = 3×2

11 15 12 16 13 17

[m,n] = size(A1);
n = n/2;
A = repmat({zeros(m,n+1)},1,2);
for i = [1 2]
    for j = 1:m
        A{i}(j,[1:j-1 j+1:end]) = A1(j,(i-1)*n+(1:n));
        A{i}(j,j) = A2(j,i);
    end
end
A = cat(2,A{:});
display(A);

A = 3×8

11 2 3 4 15 6 7 8 1 12 3 4 5 16 6 8 1 2 13 4 5 6 17 7

[Madhumitha Jayaram]

Thank you so much

Answer 3

[Image Analyst]

Your desired matrix doesn't seem to match your verbal description, as @James Tursa pointed out. But anyway, try this. It does what you say and almost gives your desired matrix (which I suspect may be wrong, or else A1 is wrong):

A1= [2 3 4 6 7 8
    1 3 4 5 6 8
    1 2 4 5 6 7];
A2=[1 5
    2 6
    3 7];
% According to verbal instructions:
A = [A2(1,1), A1(1,1:3), A2(1,2), A1(1,4:end);
    A1(2,1), A2(2,1), A1(2, 2:4), A2(2,2), A1(2,5:end);
    A1(3,1:2), A2(3,1), A1(3, 3:5), A2(3,2), A1(3,6:end)]

A = 3×8

1 2 3 4 5 6 7 8 1 2 3 4 5 6 6 8 1 2 3 4 5 6 7 7

% It almost matches the desired matrix except for locations (2, 7) and (3, 8).
% My desired matrix =
Adesired = [ 1 2 3 4 5 6 7 8
    1 2 3 4 5 6 7 8
    1 2 3 4 5 6 7 8]

Adesired = 3×8

1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8