Is there a faster way to run my code?

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Gn Gnk
Gn Gnk el 23 de Sept. de 2020
Comentada: Gn Gnk el 29 de Sept. de 2020
Hello ,
i wrote the code bellow . Is there any faster and more efficient way to run this code (not using for-loop for example or something like that):
for count1=1:length(r)
for count2=1:length(C)
distance(count2,:)=abs(r(count1,:)-C(count2,:));
dist(count2)=sum(distance(count2,:),2);
end
[dist_hard index_hard(count1)]=min(dist);
end
The problem here is that when r or C contain many elements the code is slow and i its more than obvious that i dont want that .
Any help would be valuable .

Respuesta aceptada

Walter Roberson
Walter Roberson el 23 de Sept. de 2020
[dist_hard, index_hard] = min( pdist2(r, distance, 'cityblock'), [], 2);
Note: the distance measure you are using is the L1-norm, also known as "city block".
  11 comentarios
Walter Roberson
Walter Roberson el 27 de Sept. de 2020
Does C consist of only values 0 and 1? If so then the distance from r(count1,:) to C(count2,:) is
nnz(r(count1,:) ~= C(count2,:))
and you could vectorize over all C entries as
sum(r(count1,:) ~= C,2)
providing you are using R2016b or later.
If C does consist entirely of 0 and 1, then you can do your entire calculation as
[dist_hard, index_hard] = max(r*C.',[],2);
Note that in the case of ties in the distance, this code will pick the first of them.
Gn Gnk
Gn Gnk el 29 de Sept. de 2020
Thank you so much for your effort !

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Más respuestas (2)

Bruno Luong
Bruno Luong el 23 de Sept. de 2020
Editada: Bruno Luong el 23 de Sept. de 2020
Use knnsearch if you have the right toolbox (I don't so the code is untested)
[index_hard, dist_hard] = knnsearch(C,r,'K',1,'Distance','cityblock')

Bruno Luong
Bruno Luong el 27 de Sept. de 2020
Editada: Bruno Luong el 27 de Sept. de 2020
For binary arrays
r=rand(50,8)>0.5;
C=rand(60,8)>0.5;
[dmin, index_hard] = min(sum(xor(r,permute(C,[3 2 1])),2),[],3);
index_hard'

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