compare matrices with different dimension

Hi,

A = [100123 1 1 50; 
     100123 1 2 53; 
     100123 1 3 55; 
     100456 2 1 78; 
     100456 2 2 75;
     100789 1 1 80]

B = [100123 1 56; 
     100456 2 76]

Comparing the 1st and 2nd columns of A and B, I've to compute the difference between each value in the 4th comlumn of A and the values in the 3th column of B.

In the case to compute (A - B), the expected result is a Matrix as

C = [100123 1 1 -6; 
     100123 1 2 -3; 
     100123 1 3 -1; 
     100456 2 1  2; 
     100456 2 2 -1]

Tnx for any suggestion

Answer 1

[Stephen23]

You could try this:

>> X = bsxfun(@eq,A(:,1).',B(:,1));
>> Y = bsxfun(@minus,A(:,4).',B(:,3));
>> Z = [A(any(X,1),1:3),Y(X)]

Z =

        100123           1           1          -6
        100123           1           2          -3
        100123           1           3          -1
        100456           2           1           2
        100456           2           2          -1

It returns only the rows of A for which the first element matches any first row element in B, and in the last column gives the difference of corresponding end elements of A and B.

[gianluca]

Thanks Stephen,

your code worked well. As my original data are a little different than the example I've posted, I've added to your code a further logical operation to select the data that have the same key (1st col) and the same number (2nd col). Finally the following code works well for my purpose:

X1 = bsxfun(@eq,A(:,1).',B(:,1));
X2 = bsxfun(@eq,A(:,2).',B(:,2));
X = X1.*X2;
X = logical(X);
Y = bsxfun(@minus,A(:,4).',B(:,3));
Z = [A(any(X,1),1:3),Y(X)];

Your advices have been very useful!

Answer 2

[Image Analyst]

Try this:

A = [100123 1 1 50; 
     100123 1 2 53; 
     100123 1 3 55; 
     100456 2 1 78; 
     100456 2 2 75;
     100789 1 1 80]

B = [100123 1 56; 
     100456 2 76]

[rowsA, colA] = size(A);
[rowsB, colB] = size(B);
B_rep = [repmat(B(1,:), [rowsA/rowsB, 1]); repmat(B(2,:), [rowsA/rowsB, 1])]
C = A; % Initialize
C(:, 4) = A(:, 4) - B_rep(:, 3)

However I get:

C =
    100123           1           1          -6
    100123           1           2          -3
    100123           1           3          -1
    100456           2           1           2
    100456           2           2          -1
    100789           1           1           4

Since you, for some reason , want the last row cropped off of C, you can then do this:

C = C(1:end-1,:)

to get exactly what you listed.