I have a matrix W of size ExF, and I have a matrix D of size NxFxNxE where D(n,:,n,:) = W'
Is there a fast way to replicate W to form D without having to loop through N
Thanks!

 Réponse acceptée

Mohammad Abouali
Mohammad Abouali le 28 Avr 2015
Modifié(e) : Mohammad Abouali le 28 Avr 2015

0 votes

w=[1 2; ...
3 4; ...
5 6];
N=5;
E=size(w,1);
F=size(w,2);
D=repmat(reshape(w',1,F,1,E),N,1,N,1);

Plus de réponses (2)

Jan
Jan le 28 Avr 2015

0 votes

You want to obtain W from D?
W = squeeze(D(k, :, k, :)).'

1 commentaire

Thang Bui
Thang Bui le 28 Avr 2015
Modifié(e) : Thang Bui le 28 Avr 2015
Nope, obtain D from W, the loop version looks roughly like this
D = zeros(N,F,N,E);
for n = 1:N
D(n,:,n,:) = W'; % perhaps need permute here
end

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Andrei Bobrov
Andrei Bobrov le 28 Avr 2015
Modifié(e) : Andrei Bobrov le 28 Avr 2015

0 votes

i1 = bsxfun(@plus,[0;N]+1,(0:N-1)*(2*N+1));
i2 = bsxfun(@plus,i1,reshape((0:F)*i1(end),1,1,[]));
D = zeros(N,F,N,E);
D(i2) = permute(repmat(W,[1,1,N]),[2,3,1]);

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