Matrix operations without loop.

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Junaid
Junaid 2011-11-27
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.
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Walter Roberson
Walter Roberson 2011-11-27
Or is it that K and L will always be the same and the answer would go in R(K,M) ?
Junaid
Junaid 2011-11-27
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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采纳的回答

Walter Roberson
Walter Roberson 2011-11-27
I think this should work:
R = A(sub2ind(size(A), repmat(Q(:),1,size(B,2), B))
  1 个评论
Junaid
Junaid 2011-11-28
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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更多回答（1 个）

N Madani SYED
N Madani SYED 2011-11-27
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
  1 个评论
Junaid
Junaid 2011-11-27
Thanks Syed. I prefer to have solution without loops as I having very big dataset.

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