Matrix operations without loop.

Hi everyone,
Let say A is the matrix of size 100 x 100 , A is kind of lookup table.
and There is an other matrix B that has the size let say 10 x 100. Where each col is histogram. Now for Query col vector Q size of 10 x 1. I want to get resultant matrix R of size(B), where each cell value in R, is obtained by looking the values of Q and B in A. let say:
R(1,1) = A( Q(1), B(1,1));
R(5,5) = A(Q(5)), B(5,5));
For sure, the values in B and Q are in range of A indexes. I hope you understand the scenario. Thanks a lot in advance.

댓글 수: 3

Walter Roberson
Walter Roberson 2011년 11월 27일
Either I do not understand or else there is a problem with your specification.
Suppose you are indexing A(Q(K), B(L,M)) then apparently the answer is to get stored in R(K, something), but I cannot tell whether it is supposed to go in R(K,L) or in R(K,M), or whether L and M will always be the same making the two possible destinations equal ?
Walter Roberson
Walter Roberson 2011년 11월 27일
Or is it that K and L will always be the same and the answer would go in R(K,M) ?
Junaid
Junaid 2011년 11월 27일
Dear Walter, Thank for your kind consideration. I think my specification couldn't deliver full meaning.
Matrix *A* is weight Matrix. and B is the Dataset of 100 Histograms with each 10 bins. Q is the Query. For each value of bin there is already computed weight in Matrix A. So rather then Computing that again I want to retrieve that value from A.
So R(i, j) = A ( Q(i), B(i,j) )
So the resultant *R* is the matrix of Query *Q*, each col *i* of *R* is the weigth of *Q* with colum in *B*

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 채택된 답변

Walter Roberson
Walter Roberson 2011년 11월 27일

1 개 추천

I think this should work:
R = A(sub2ind(size(A), repmat(Q(:),1,size(B,2), B))

댓글 수: 1

Junaid
Junaid 2011년 11월 28일
Dear Walter, Could you please tell me what if Q is an other matrix and now I want to get R_new such that each cell is the sum(minimum(R)).
Case is simple. For Q( where each col is histogram) I want to get first nearest neighbor from B.

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추가 답변 (1개)

N Madani SYED
N Madani SYED 2011년 11월 27일

0 개 추천

I think the following should work
for i = 1:10
for j = 1:100
m = Q(i);
n = B(i,j);
R(i,j)= A(m,n);
end
end

댓글 수: 1

Junaid
Junaid 2011년 11월 27일
Thanks Syed. I prefer to have solution without loops as I having very big dataset.

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