Is there a faster way to run my code?

조회 수: 1 (최근 30일)
Gn Gnk
Gn Gnk 2020년 9월 23일
댓글: Gn Gnk 2020년 9월 29일
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 .

채택된 답변

Walter Roberson
Walter Roberson 2020년 9월 23일
[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".
  댓글 수: 11
Walter Roberson
Walter Roberson 2020년 9월 27일
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 2020년 9월 29일
Thank you so much for your effort !

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추가 답변 (2개)

Bruno Luong
Bruno Luong 2020년 9월 23일
편집: Bruno Luong 2020년 9월 23일
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 2020년 9월 27일
편집: Bruno Luong 2020년 9월 27일
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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