How to reshape matrix by column number?

Hi, I have a 163 x 13 matrix of numerical values that I'd like to turn into a longer matrix. I'd like to keep the first two columns the same, but output the column number in a new column (starting with column 3 assigned the key 1), and then the value for that column in the next column. Then I'd like to remove rows with zero in the fourth column.

For example, if "demand" is my matrix, I want to produce "demandlong"

demand = [1	78	0	0	0	0	0	0	0	0	8	45	0
         1	79	0	0	0	0	31	8	0	0	0	0	0
         1	80	456	0	4	0	39	0	0	0	16	0	0]

demandlong = 
[1 78 10 45
 1 79 5  31
 1 80 1  456
 1 80 3  4
 1 80 5  39
 1 80 9  16]

Answer 1

[Stephen23]

D = [1,78,0,0,0,0,0,0,0,0,8,45,0;1,79,0,0,0,0,31,8,0,0,0,0,0;1,80,456,0,4,0,39,0,0,0,16,0,0]

D = 3×13

1 78 0 0 0 0 0 0 0 0 8 45 0 1 79 0 0 0 0 31 8 0 0 0 0 0 1 80 456 0 4 0 39 0 0 0 16 0 0

M = D(:,3:end).';
[C,R] = find(M);
Z = [D(R,1:2),C,nonzeros(M)]

Z = 8×4

1 78 9 8 1 78 10 45 1 79 5 31 1 79 6 8 1 80 1 456 1 80 3 4 1 80 5 39 1 80 9 16

[Emma Kuttler]

Thanks! How would you modify the code if you wanted to also return the rows with zero in them?

[Stephen23]

"How would you modify the code if you wanted to also return the rows with zero in them?"

Do you mean something like this?:

D = [1,78,0,0,0,0,0,0,0,0,8,45,0;1,79,0,0,0,0,31,8,0,0,0,0,0;1,80,456,0,4,0,39,0,0,0,16,0,0]

D = 3×13

1 78 0 0 0 0 0 0 0 0 8 45 0 1 79 0 0 0 0 31 8 0 0 0 0 0 1 80 456 0 4 0 39 0 0 0 16 0 0

M = D(:,3:end).';
S = size(M);
[C,R] = ndgrid(1:S(1),1:S(2));
Z = [D(R,1:2),C(:),M(:)]

Z = 33×4

1 78 1 0 1 78 2 0 1 78 3 0 1 78 4 0 1 78 5 0 1 78 6 0 1 78 7 0 1 78 8 0 1 78 9 8 1 78 10 45

Answer 2

[KSSV]

demand = [1	78	0	0	0	0	0	0	0	0	8	45	0
    1	79	0	0	0	0	31	8	0	0	0	0	0
    1	80	456	0	4	0	39	0	0	0	16	0	0] ;

n = size(demand,1) ; 
iwant = zeros(n,4) ;
iwant(:,1:2) = demand(:,1:2) ;
for i = 1:n
    id = find(demand(i,:)) ;
    iwant(i,3) = id(3)-2 ; 
    iwant(i,4) = demand(i,id(3)) ;
end
iwant

iwant = 3×4

1 78 9 8 1 79 5 31 1 80 1 456