I have a matrix y and a vector x, I need to find all the possible vectors resulted from the mapping of each value in x into each vector in y.

That is difficult to be understood; let's explain is with an example: Here is an example, I have the vector x = [0.7 0.7i; 0.7-0.7i]; The matrix y = [1 0; 2 0; 1 2]; the resulted matrix is supposed to be like this one Z = [0.7 0.7i 0; 0.7-0.7i 0; 0 0.7 0.7i; 0 0.7-0.7i; 0.7 0.7i 0.7 0.7i; 0.7 0.7i 0.7-0.7i; 0.7 - 0.7i 0.7-0.7i ; 0.7 - 0.7i 0.7 0.7i]; . That is equivalent into Z = [x_1 0; x_2 0; 0 x_1; 0 x_2; x_1 x_1; x_1 x_2; x_2 x_2; x_2 x_1];. That means it map each value in x into the row of Z according to the index value in y.

Here is my try code: clear all; clc;

y = [];
G = 2; 
v = 1 : G; 
for i = 1: G
    x=nchoosek(v,i);
    m = zeros(size(x,1),G-i);
y =[y ; x m];              % creat the matrix y 
end

x = [0.7   0.7i; 0.7-0.7i]; 

Z = []; s = zeros(G,1); 
for k=1:size(x,1)
for i=1:size(y,1) 
n=y(i,:);
n=n(n ~= 0);
s(n)=x(k); 
Z=[Z s];
s = zeros(G,1);
end
end

The problem in my code that matrix Z show the inverse, it means it takes the input x_1 from x and then map it into all possible values in y. For example the matrix Z starts with [x_1 0; 0 x_1; x_1 x_1 ….], however that should be the inverse, which means takes each values in x and map it as shown in the above example [x_1 0; x_2 0; x_3 0 …..]. The second issue, when y contains more than non-zeros values, my code cannot get all possible vectors, it can only get [x_1 x_1; x_2 x_2]; but I cannot map the other possibilities which are [x_1 x_2; x_2 x_1] and so on.

How can I solve that issue?

UPDATE

Here is the updated question with clear description. I have the vector x and matrix y, I need to fill the matrix z following the index taken from each row in matrix y. For example, if the first row in matrix y is [1 0] or [0 1]; then I will take all possible values from x and put it in z following the number taken from the row in y which is 1 in this case. Then, the same case for row 2 in matrix y which is [2 0] or [0 2]; it means that second column in z will be filled with all possible values in x. Then, the two columns in z can be filled which is equivalent into the case [1 2] in y, so it will take the first value from x and fill it with all other possible values from x, and so on. The rows in z should not be repeated.
The matrix Z is exactly as shown with below answer of AboAmmar below, but using the loop if with longer vector x and bigger matrix y will be little bit complicated.

CodePudding user response：

As you describe it, there are 4 distinct cases for each row of y and the corresponding output:

1. [0 1] or [1 0] => [x 0]
2. [0 2] or [2 0] => [0 x]
3. [1 2] => [x1 x1; x1 x2; x2 x2; x2 x1]
4. [2 1] => [x1 x1; x2 x1; x2 x2; x1 x2]

These don't seem to follow any obvious rule. So, the easiest (but not smartest) solution is to use if-else and select the suitable case from the above. We don't have all the information about the possible indices, or if rows like [1 1] and [2 2] might happen, so the following solution is by no means exhaustive; surprising errors might happen if other inputs are fed into y matrix.

y = [];
G = 2;
v = 1 : G;
for i = 1: G
    x = nchoosek(v,i);
    m = zeros(size(x,1),G-i);
    y = [y ; x m];              % creat the matrix y
end

Z = [];
x = [0.7   0.7i; 0.7-0.7i]
for i = 1:size(y,1)
    r = y(i,:);
    if ismember(r, [1 0; 0 1], 'rows')
        Z(end 1:end 2,:) = [x [0; 0]];
    elseif ismember(r, [2 0; 0 2], 'rows')
        Z(end 1:end 2,:) = [[0; 0] x];
    elseif ismember(r, [1 2], 'rows')
        Z(end 1:end 4,:) = [x(1) x(1); x(1) x(2); x(2) x(2); x(2) x(1)];
    elseif ismember(r, [2 1], 'rows')
        Z(end 1:end 4,:) = [x(1) x(1); x(2) x(1); x(2) x(2); x(1) x(2)];
    end
end

Z =
   0.7000   0.7000i   0.0000   0.0000i
   0.7000 - 0.7000i   0.0000   0.0000i
   0.0000   0.0000i   0.7000   0.7000i
   0.0000   0.0000i   0.7000 - 0.7000i
   0.7000   0.7000i   0.7000   0.7000i
   0.7000   0.7000i   0.7000 - 0.7000i
   0.7000 - 0.7000i   0.7000 - 0.7000i
   0.7000 - 0.7000i   0.7000   0.7000i

CodePudding user response：

Your code is valid if you have fix length in y, for example if each vector in y has one value and others are zeros, or two non-zeros values ...etc.

So you can do your code for each length separately and then build the matrix Z by combining all other matrices.