How can I create a matrix based on a double vector lookup?

I have three vectors of same length:

a=[5;10;10;15;15;15;20;20;20]

b=[3;3;5;3;5;10;3;5;10]

c=[0.2;0.3;-0.1;0.4;0.5;0.6;-0.2;-0.3;0.7]

The entries in a, b and c are related to each other (e.g. first row a=5, b=3 --> c=0.2).

I would like to translate this into a matrix with unique(a) being the row lookup value, unique(b) the column lookup value and "c" the values resulting from the row and column lookup. Making "a" and "b" unique is not a problem but putting this into a matrix with cross-values of "c" seems to be difficult. For all combinations where there is no match, the value should be zero (=0). Final matrix should look like:

[0,   3,    5,   10; ...
 5,   0.2,  0,   0;...
 10,  0.3, -0.1, 0;...
 15,  0.4,  0.5, 0.6;...
 20, -0.2, -0.3, 0.7]

Note: The matrix doen't need to have the some row and column dimension. First row and column are the unqiue lookup values from vector "a" and "b"; vector "c" values are in the matrix (2:end,2:end).

[Jan]

I do not understand the relation between the inputs and outputs exactly.

[Pascal]

Think of "a" and "b" as an integer that defines a specific combination and "c" is the cumulated return of this combination. So maybe it helps to visualize if you put the three vectors into a matrix like this:

x=[a,b,c]

x =

    5.0000    3.0000    0.2000
   10.0000    3.0000    0.3000
   10.0000    5.0000   -0.1000
   15.0000    3.0000    0.4000
   15.0000    5.0000    0.5000
   15.0000   10.0000    0.6000
   20.0000    3.0000   -0.2000
   20.0000    5.0000   -0.3000
   20.0000   10.0000    0.7000

Meaning that the combination 5~3 leads to a total return of 20% (row 1) the combination 10~3 results in a total return of 30% and so on...

Now, I aim to make "a" and "b" unique and creating a new matrix looking up the total returns in "c" corresponding the values in "a" and "b".

If you check the result matrix:

result =

         0    3.0000    5.0000   10.0000
    5.0000    0.2000         0         0
   10.0000    0.3000   -0.1000         0
   15.0000    0.4000    0.5000    0.6000
   20.0000   -0.2000   -0.3000    0.7000

You will find combination 5~3 in result(2,2), combination 10~3 in result(3,2) and so on.

Answer 1

[Cedric]

You probably want something like:

 [un_a,~,ic_a] = unique(a) ;
 [un_b,~,ic_b] = unique(b) ;
 result = zeros(length(un_a), length(un_b)) ;
 result(sub2ind(size(result), ic_a, ic_b)) = c ;

and if you want to add unique values as row/col 1, you can do it as

 result = [0, un_b.'; un_a, result] ;

There are plenty of variants of this solution; what you probably missed when you tried to develop a solution, is that UNIQUE outputs three vectors, the last two being indices that you can use for lookup operations.