Filter löschen
Filter löschen

Matrix operations without loop.

3 Ansichten (letzte 30 Tage)
Junaid
Junaid am 27 Nov. 2011
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.
  3 Kommentare
Walter Roberson
Walter Roberson am 27 Nov. 2011
Or is it that K and L will always be the same and the answer would go in R(K,M) ?
Junaid
Junaid am 27 Nov. 2011
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*

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 27 Nov. 2011
I think this should work:
R = A(sub2ind(size(A), repmat(Q(:),1,size(B,2), B))
  1 Kommentar
Junaid
Junaid am 28 Nov. 2011
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.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

N Madani SYED
N Madani SYED am 27 Nov. 2011
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 Kommentar
Junaid
Junaid am 27 Nov. 2011
Thanks Syed. I prefer to have solution without loops as I having very big dataset.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Creating and Concatenating Matrices finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by