How to transform a matrix in Matlab?

Matrix A is as follows:

A = [0  240  245  250
     25  1  2  1
     63  3  2  1];

I want matrix A to be transformed to B (like follows):

    B = [0  240  245  250  240  245  250  240  245  250
      25  1  0  1  0  1  0  0  0  0
      63  0  0  1  0  1  0  1  0  0];

there are three different variables in matrix A, so, 204 to 250 (first row) in matrix B is repeated 3 times (e.g. if there were 5 variables, then 240 to 250 should be repeated 5 times). Then, value of ID = 25 has 1, so 1 is added to B(2,2). Again, A(2,3) = 2, then B(2,6) should by =1 and A(2,4)=1, then A(2,10) should be equal by =1. And same for ID#63

[Walter Roberson]

I cannot figure out why it is the last 250 for ID 2 that is getting the 1 and not the first 250. Everything else seems to work out for me. I think the output should be

B = [0  240  245  250  240  245  250  240  245  250
    25  1  0  1  0  1  0  0  0  0
    63  0  0  1  0  1  0  1  0  0];

[John BG]

Walter

reading a similar question asked by Mohammad earlier on may help to understand

http://uk.mathworks.com/matlabcentral/answers/269885-how-to-transform-a-matrix-to-a-binary-type-in-matlab

A is used as a table.

There is already an answer to similar question, stacking the binary lines rather than lining them up horizontally.

[Moe]

Hi Walter,

You are right, the output should be same as you wrote in your comment. I edited in main question. Do you have any idea how to get this result?

[Priti Gujar]

Use the readall function to import all the data. Check that the preprocessing function was applied to each file by plotting the Y variable as a function of Time.

Answer 1

[John BG]

Hi Mohammad

MATLAB has the command linsolve to solve linear equation systems of the type A*x=b

Your question has one A:

A = [0   240   245   250;  25     1     2     1;  63     3     2     1]

and ten b:

B= [0  240  245  250  240  245  250  240  245  250;
    25  1  0  0  0  1  0  0  0  1;
    63  0  0  1  0  1  0  1  0  0]

One way to solve them is with a for loop:

C=zeros(4,10)
for k=1:1:10
   C(:,k)=linsolve(A,B(:,k))
end

answer:

C =
  Columns 1 through 6
          1.00          0.00         -0.04         -0.04         -0.04          0.00
         -0.00         -0.55          0.77          1.30          0.75         -0.02
             0             0             0             0             0             0
          0.00          1.49          0.24         -0.25          0.24          1.00
  Columns 7 through 10
         -0.04         -0.04         -0.04          0.00
          0.78          1.27          0.77         -0.52
             0             0             0             0
          0.25         -0.26          0.24          1.50

test it's the correct answer

A*C
ans =
  Columns 1 through 6
          0.00        240.00        245.00        250.00        240.00        245.00
         25.00          1.00         -0.00         -0.00         -0.00          1.00
         63.00         -0.00             0          1.00             0          1.00
  Columns 7 through 10
        250.00        240.00        245.00        250.00
             0         -0.00         -0.00          1.00
             0          1.00             0         -0.00

note the type (class) has changed to double. To bring it back to, for instance, range [0 255] use uint8.

does this answer help? if so click on the thumbs-up icon link on the top of this page, thanks in advance

John

[e-mail]

[Walter Roberson]

Mohammad Hesam comments

This is not "linsolve" problem.

[John BG]

understood, had to read the previous question to realize that A is a table and the kind of the variable tagging sought.

B has to be built by the answer, not used by it to find a transform matrix.

Thanks for mentioning Mohammad's comment.

Answer 2

[Andrei Bobrov]

[m,n] = size(A);
[ii,k] = ndgrid(1:n-1,1:m-1);
jj = A(2:end,2:end)';
b0 = accumarray([ii(:),jj(:),k(:)],1,[n-1,n-1,m-1]);
B = [nan,repmat(A(1,2:end),1,n-1);[A(2:end,1),reshape(b0,[],m-1)']];

or with bsxfun

[m,n] = size(A);
n1 = n - 1;
A0 = A(2:end,2:end)';
b0 = reshape( bsxfun(@eq,1:n1,reshape(A0,n1,[],m-1)),[],m-1)';
B = [nan,repmat(A(1,2:end),1,n-1);[A(2:end,1),b0]];

[Moe]

Thanks Andrei,

The output of your code is:

B = [NaN  240  245  250  240  245  250  240  245  250
25  1  0  0  0  1  1  0  0  1
63  1  0  0  0  1  0  0  0  0];

While it's different with the output that I wanted in main question. Can you edit your code?

[Andrei Bobrov]

corrected