Re-index a vector based on the indices of another vector

Hi all,

I have a vector A:

A = [1	1	1	1
1	1	1	1
1	1	1	1
1	1	2	2
1	1	2	2
1	1	3	3
1	1	3	3
1	1	3	3
1	1	3	3
1	2	1	4];

where the final column corresponds to the index of each row.

I have another vector B:

B =[1	2	1
    1	2	1
    1	1	3
    1	1	3
    1	1	1
    1	1	1
    1	1	3
    1	1	2
    1	1	2
    1	1	2];

I want to add an index column (a fourth column) to vector B where it's decided based on the value of each element in the row and then comparing it with vector A. For example, if all elements equal to 1 in vector B, then the index is 1 as shwon in A. As a result, I should get the following:

B_new = =[1	2	1	4
    1	2	1	4
    1	1	3	3
    1	1	3	3
    1	1	1	1
    1	1	1	1
    1	1	3	3
    1	1	2	2
    1	1	2	2
    1	1	2	2
    ];

My attempt to do this seems complicated and it doesn't work and I'm sure there is a simpler one, but here it is anyway:

B_new = [];
for i = 1:size(A,1)
    if (B(i,1)==A(i,1) && ...
            B(i,2)==A(i,2) && ...
                B(i,3)==A(i,3))
        Bnew = [B(i,:) A(i,4)];
        B_new = [B_new ; Bnew];
    end
end

Any help would be appreicted.

Thanks

Answer 1

[Voss]

[ism,idx] = ismember(B,A(:,[1 2 3]),'rows');
assert(all(ism))
B_new = [B A(idx,4)];

[LH]

Thanks.

[Voss]

You're welcome!

Answer 2

[Anjaneyulu Bairi]

Hi,

To create another column in 'B' based on matching of its row values with vector 'A', refer the below code:

B =[1	2	1
    1	2	1
    1	1	3
    1	1	3
    1	1	1
    1	1	1
    1	1	3
    1	1	2
    1	1	2
    1	1	2];

A = [1	1	1	1
1	1	1	1
1	1	1	1
1	1	2	2
1	1	2	2
1	1	3	3
1	1	3	3
1	1	3	3
1	1	3	3
1	2	1	4];

% Initialize the new B with an additional column for the index
B_new = [B zeros(size(B, 1), 1)];

% Find the index for each row in B by comparing with A
for i = 1:size(B, 1)
    for j = 1:size(A, 1)
        if isequal(B(i, :), A(j, 1:3))
            B_new(i, 4) = A(j, 4);
            break; 
        end
    end
end

% print the vector
disp(B_new);

     1     2     1     4
     1     2     1     4
     1     1     3     3
     1     1     3     3
     1     1     1     1
     1     1     1     1
     1     1     3     3
     1     1     2     2
     1     1     2     2
     1     1     2     2

I hope it helps to resolve your query.