How can i manipulate a matrix with this characteristics?

Hi, i have a matrix wtih repeated elements like:

[1 4 1 4 2 3 3 5 5 6 6 5]

i want break that matrix. And then in each other matrix, running in a cycle, have only the index referent to each number (1, 2, 3, ...), something like:

1        2      3      4     5        6
[1 4]  [2 3]  [2 3   [1 4]  [3 5    [5 6
               3 5]          5 6     6 5]
                             6 5]

I tried creating a 3 dimension matrix (i,j,k) and each 3rd dimension having above matrix. That hypothesis gives me the repeated solutions inside. I tried to get them out but I might not be able to.

I hope someone can help me out.

PS: Some question that may have, just do it pls.

Regards, Vitor Ribeiro.

Answer 1

[Joseph Cheng]

You can use cells for this. I do not see the pattern you have for [1 4 1 4 2 3 3 5 5 6 6 5] to the broken up matrix. but to set the broken up matrix with index references, cells should work.

MATRIX{1} = [1 4];
MATRIX{2} = [2 3];
MATRIX{3} = [2 3;3 5];
etc.

Which should be simple enough to accomplish in a for loop with whatever initial matrix breakup pattern is.

[Vitor Ribeiro]

I have implemented this code, following your ideia.

    z=1;
    for a=1:size(bus,1)
        z=1;
        for i=1:size(branch_,1)
            for j=1:2
                if branch_(i,j)==bus(a)
                    brx{a,z}=branch_(i,1:2);
                    z=z+1;
                end
            end
        end
    end
Now i need to ensure that in the matrix brx i don't have repeated contents in each row.

[1,4]  [1,4]  []  []  []  []
[8,2]  [8,2]  []  []  []  []
[3,6]  [3,6]  []  []  []  []
[1,4]  [1,4]  [4,5]  [9,4]  [4,5]  [9,4]
[4,5]  [4,5]  [5,6]  [5,6]  []  []
[3,6]  [5,6]  [5,6]  [3,6]  [6,7]  [6,7]
[6,7]  [6,7]  [7,8]  [7,8]  []  []
[8,2]  [7,8]  [7,8]  [8,2]  [8,9]  [8,9]
[9,4]  [8,9]  [8,9]  [9,4]  []  []
Considering that the real aspect of inside matrix brx is represented above. I tried to do a comparison with a conditional structure if. Just like:

if brx{i,j}==brx{i,j+1}
       brx{i,j+1}={};
  end

But in fact, a comparison between this 2 elements results in a [1 1] matrix, cause it compares each one of the 2 element of inside cells of the matrix brx.

If you know a way to eliminate repeated cells of brx matrix I appreciate you answer me.

I tried below code, and obviously did not result. I have no experience working with cells. Hope someone can help me out.

for i=1:size(brx,1)
      for j=1:size(brx,2)-1
          for k=1:size(brx,1)
              if brx{i,j}==brx{i,k}
                  brx{i,j+1}={};
              end
          end
      end
  end

Best reagards, Vitor Ribeiro.

Answer 2

[Vitor Ribeiro]

I have implemented this code, following your ideia.

    z=1;
    for a=1:size(bus,1)
        z=1;
        for i=1:size(branch_,1)
            for j=1:2

                if branch_(i,j)==bus(a)

                    brx{a,z}=branch_(i,1:2);
                    z=z+1;
                end
            end
        end
    end

Now i need to ensure that in the matrix brx i don't have repeated contents in each row.

[1,4]  [1,4]  []  []  []  []
[8,2]  [8,2]  []  []  []  []
[3,6]  [3,6]  []  []  []  []
[1,4]  [1,4]  [4,5]  [9,4]  [4,5]  [9,4]
[4,5]  [4,5]  [5,6]  [5,6]  []  []
[3,6]  [5,6]  [5,6]  [3,6]  [6,7]  [6,7]
[6,7]  [6,7]  [7,8]  [7,8]  []  []
[8,2]  [7,8]  [7,8]  [8,2]  [8,9]  [8,9]
[9,4]  [8,9]  [8,9]  [9,4]  []  []

Considering that the real aspect of inside matrix brx is represented above. I tried to do a comparison with a conditional structure if. Just like:

if brx{i,j}==brx{i,j+1}

       brx{i,j+1}={};
  end

But in fact, a comparison between this 2 elements results in a [1 1] matrix, cause it compares each one of the 2 element of inside cells of the matrix brx.

If you know a way to eliminate repeated cells of brx matrix I appreciate you answer me.

I tried below code, and obviously did not result. I have no experience working with cells. Hope someone can help me out.

for i=1:size(brx,1)

      for j=1:size(brx,2)-1

          for k=1:size(brx,1)

              if brx{i,j}==brx{i,k}

                  brx{i,j+1}={};

              end
          end

      end
  end

Best reagards, Vitor Ribeiro.