Is there a faster way to run my code?

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Gn Gnk
Gn Gnk am 23 Sep. 2020
Kommentiert: Gn Gnk am 29 Sep. 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 .

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 23 Sep. 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 Kommentare
Walter Roberson
Walter Roberson am 27 Sep. 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 am 29 Sep. 2020
Thank you so much for your effort !

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Weitere Antworten (2)

Bruno Luong
Bruno Luong am 23 Sep. 2020
Bearbeitet: Bruno Luong am 23 Sep. 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 am 27 Sep. 2020
Bearbeitet: Bruno Luong am 27 Sep. 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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