how to find the location of element in matrix?

조회 수: 2 (최근 30일)
sheno39
sheno39 2013년 10월 18일
댓글: Azzi Abdelmalek 2013년 10월 28일
my code:
I=[2 3 10 4 6; 1 4 7 5 3; 5 2 8 4 3;8 2 1 7 3;1 9 8 3 4;];
D = padarray(I,[1 1],0,'both');
[x y]=size(D);
m=1;
n=1;
for i=2:x-1
for j=2:y-1
A=[D(i-1,j-1:j+1),D(i,j-1),D(i,j+1),D(i+1,j-1:j+1)];
I1(m,n)=max(A(:));
%[maxval(m,n) index(m,n)]=max(A(:));
n=n+1;
end
m=m+1;
n=1;
end
[tc locatn]=ismember(I1,I);
s=[5,5];
[R C]=ind2sub(s,locatn');
Required output
locatn=[7 11 12 11 17; 3 11 11 11 21; 4 4 12 13 19; 10 10 10 13 19; 10 4 10 15 19]
i got the output as,
locatn=[7 11 12 11 3; 3 11 11 11 21; 4 4 12 4 12; 10 10 10 4 12; 10 4 10 4 12]
can anyone help me to get the required output? how to correct this code?

채택된 답변

Andrei Bobrov
Andrei Bobrov 2013년 10월 18일
편집: Andrei Bobrov 2013년 10월 18일
I=[2 3 10 4 6; 1 4 7 5 3; 5 2 8 4 3;8 2 1 7 3;1 9 8 3 4;];
D = padarray(I,[1 1],0,'both');
s = size(I);
N = padarray(reshape(1:numel(I),s),[1 1],0,'both');
pt = [1 1 1;1 0 1;1 1 1] > 0;
s1 = s+2;
out = zeros(s);
for ii=2:s1(1)-1
for jj=2:s1(2)-1
D1 = D(ii-1:ii+1,jj-1:jj+1);
K = N(ii-1:ii+1,jj-1:jj+1);
K = K(pt);
[~,i0] = max(D1(pt));
out(ii-1,jj-1) = K(i0);
end
end
or
I=[2 3 10 4 6; 1 4 7 5 3; 5 2 8 4 3;8 2 1 7 3;1 9 8 3 4;];
D = padarray(I,[1 1],0,'both');
s = size(I);
pt = [1 1 1;1 0 1;1 1 1] > 0;
a = -1:1;
j1 = [1;1;1]*a;
i1 = j1';
i1 = i1(pt);
j1 = j1(pt);
s1 = s+2;
R = zeros(s);
C = zeros(s);
for ii=2:s1(1)-1
for jj=2:s1(2)-1
D1 = D(ii-1:ii+1,jj-1:jj+1);
[~,i0] = max(D1(pt));
R(ii-1,jj-1) = ii + i1(i0) - 1;
C(ii-1,jj-1) = jj + j1(i0) - 1;
end
end
or
I=[2 3 10 4 6; 1 4 7 5 3; 5 2 8 4 3;8 2 1 7 3;1 9 8 3 4;];
Ip = padarray(I,[1 1],0,'both');
s = size(Ip);
s0 = s - 2;
idxi = padarray(reshape(1:numel(I),s0),[1 1],0,'both');
ptr = reshape(bsxfun(@plus,(0:2)',(0:2)*s(1)),1,[]);
p2 = bsxfun(@plus,(1:s0(1))',(0:s0(2)-1)*s(1));
idx = bsxfun(@plus,p2(:),ptr([1:4,6:end]));
[~,ii] = max(Ip(idx),[],2);
s1 = size(idx);
out = reshape(idxi(idx(sub2ind(s1,(1:s1(1))',ii))),s0);
[R,C] = ind2sub(size(out),out);

추가 답변 (1개)

Azzi Abdelmalek
Azzi Abdelmalek 2013년 10월 18일
편집: Azzi Abdelmalek 2013년 10월 18일
I think there is an error in your expected locatn, try this
I2=unique(I1(:));
ii=histc(I1(:),I2);
idx1=zeros(size(I));
for k=1:numel(I2)
a=I2(k)
jdx=ismember(I1,a);
j1=find(jdx,ii(k))';
idx=ismember(I,a);
i1=find(idx,ii(k));
i1(end:numel(j1))=i1(end);
idx1(j1)=i1;
end
locatn=idx1
  댓글 수: 2
sheno39
sheno39 2013년 10월 28일
Sir 'I1' represents what?
Azzi Abdelmalek
Azzi Abdelmalek 2013년 10월 28일
If you look for the location of 5 from I1 in the matrix I, there are two 5 in I1, we should find the locations of the two first 5 in I, that's what i1 represent.

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Colormaps에 대해 자세히 알아보기

태그

Community Treasure Hunt

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

Start Hunting!

Translated by