compare matrices with different dimension
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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
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채택된 답변
Stephen23
2015년 1월 10일
편집: Stephen23
2015년 1월 11일
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.
추가 답변 (1개)
Image Analyst
2015년 1월 10일
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.
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