Change the order of matrix

I want to change the order of B in accordance with Index.

Index means that 'row 2' in B should change to 'row 5' in New_B.

I wrote this code, but I don't know how to add remain elements in B to New_B.

B = [1 2; 1 4; 1 6; 1 8; 2 3; 2 7; 3 4; 3 6; 3 8; 4 6; 4 7; 6 7; 6 8; 7 8];
Index = [2,5; 4,1; 6,6; 8,2; 9,7; 10,3; 11,9; 12,4; 14,8]; 

New_B = zeros(size(B));
for i=1:length(Index)
    temp = Index(i,:);
    
        if temp(1)~=temp(2)
          New_B(temp(2),:)  = B(temp(1),:);
        end
     
end

result should be

New_B = [1,8; 3,6; 4,6; 6,7; 1,4; 1,2; 3,8; 7,8; 4,7; 1,6; 2,3; 2,7; 3,4; 3,4; 6,8]

[Stephen23]

"result should be..."

Given that the second column of Index only has indices going up to 9, how does your example output New_B contain 15 rows?

[Guillaume]

Your index matrix tells us what happens to some of the rows of B. What about the others? What should be done about them?

Also, what if the instructions in Index conflict? e.g. Index sends 2 rows to the same destination?

[NA]

''Given that the second column of Index only has indices going up to 9, how does your example output New_B contain 15 rows?''

B and New_B should be the same size.

''What about the others? What should be done about them?''

Orders of others are not impotant and coming after the arranged index.

''what if the instructions in Index conflict? e.g. Index sends 2 rows to the same destination?''

I have not encountered with mentioned case yet.

Answer 1

[Ameer Hamza]

Try this:

B = [1 2; 1 4; 1 6; 1 8; 2 3; 2 7; 3 4; 3 6; 3 8; 4 6; 4 7; 6 7; 6 8; 7 8];
Index = [2,5; 4,1; 6,6; 8,2; 9,7; 10,3; 11,9; 12,4; 14,8];

B_temp = B;
B_temp(Index(:,1),:) = [];
B_new(Index(:,2),:) = B(Index(:,1),:);
B_new = [B_new; B_temp];

[Ameer Hamza]

Can you describe the rule here? I think it does not match the rule in your question.

[NA]

B = [1 2; 2 3; 3 4; 4 5; 4 6; 6 8; 7 8];
Index = [3 2; 4 5; 5 7];
B_temp = B;
B_temp(Index(:,1),:) = [];
B_new(Index(:,2),:) = B(Index(:,1),:);
B_new =

     0     0  --->  B_temp(1,:)       1   2
     3     4                          3   4
     0     0  --->  B_temp(2,:)       2   3 
     0     0  --->  B_temp(3,:)       6   8
     4     5                          4   5
     0     0  --->  B_temp(4,:)       7   8
     4     6                          4   6

[Ameer Hamza]

Now the rule is a bit clear. Is this the required result?

B = [1 2; 2 3; 3 4; 4 5; 4 6; 6 8; 7 8];
Index = [3 2; 4 5; 5 7];

idx1 = Index(:,1);
idx2 = Index(:,2);
idx3 = setdiff(1:size(B,1), idx2)';
idx4 = setdiff(1:size(B,1), idx1)';
B_new(idx2, :) =  B(idx1,:);
B_new(idx3, :) = B(idx4,:);

[NA]

Thanks.