finding neighbor of a position

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Hi, I have a matrix.
I =
1 0 0
2 5 0
0 0 3
0 0 0
I know the position of 5 in I is 6 linear index.
is there any easy function to have the 8 other neighbors of 5. Thanks

Accepted Answer

Oleg Komarov
Oleg Komarov on 26 Jun 2011
EDITED: should be fine now
I =[ 1 0 0
2 5 0
0 0 3
0 0 0];
l = 8;
sz = size(I);
% row, col subs of center
[r,c] = ind2sub(sz,l); % c = ceil(l/4); r = mod(l,4)+ c*sz(1);
% Calculate 8 neighbors
neigh(1:8,1:2) = [r+[-1;0;1;-1;1;-1;0;1] c+[-1;-1;-1;0;0;1;1;1] ];
% Only those in the range
neigh = neigh(all(neigh,2) & neigh(:,1) <= sz(1) & neigh(:,2) <= sz(2),:);
% Convert to position
idx = (neigh(:,2)-1)*sz(1) + neigh(:,1);
  3 Comments

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More Answers (3)

Sean de Wolski
Sean de Wolski on 27 Jun 2011
idx = find(conv2(double(I==5),ones(3),'same'))
%This includes the 6, but that could easily be taken care of with setdiff.

Wolfgang Schwanghart
Wolfgang Schwanghart on 26 Jun 2011
  3 Comments
Mohammad Golam Kibria
Mohammad Golam Kibria on 28 Jun 2011
thanks,this also works fine for me

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Andrei Bobrov
Andrei Bobrov on 26 Jun 2011
idl = 6;
idxs = ...
nonzeros(bsxfun(@plus,idl - [1 0 -1]',size(I,1)*[-1 0 1]).*[1 1 1;1 0 1;1 1 1])
CORRECTED 06/27/2011 10:05 MSK
idl = 6;
s = size(I);
I0 = zeros(s+2);
I0(2:end-1,2:end-1) = reshape(1:numel(I),s);
idxs = nonzeros(I0(bsxfun(@plus,find(I0==idl) - [1 0 -1]',(s(1)+2)*[-1 0 1])).*[1 1 1;1 0 1;1 1 1])
MORE variant (06/27/2011 11:12 MSK)
s = size(I);
[ii jj] = ind2sub(s,idl);
v = [-1 -1 -1;0 0 0;1 1 1];
R=ii+v;
C=jj+v';
loc = (R<=s(1) & R>=1&C<=s(2) & C>=1&[1 1 1;1 0 1;1 1 1])>0;
idxl = sub2ind(s,R(loc),C(loc));
MORE variant 2 (06/27/2011 11:35 MSK) with idea of Oleg
s = size(I);
[ii jj] = ind2sub(s,idl);
R = ii + [-1 0 1 -1 1 -1 0 1];
C = jj + [-1 -1 -1 0 0 1 1 1];
loc = (R<=s(1) & R>=1&C<=s(2) & C >= 1 )>0;
idxl = sub2ind(size(I),R(loc),C(loc));
LAST variant (06/27/2011 13:43 MSK)
I1 = zeros(size(I));
I1(idl)=1;
idx = find(bwdist(I1,'chessboard')==1)
or
idx = find(bwdist(I==5,'chessboard')==1)
  3 Comments
Mohammad Golam Kibria
Mohammad Golam Kibria on 28 Jun 2011
Thanks this also works fine for me

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