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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

Risposta accettata

Oleg Komarov
Oleg Komarov il 26 Giu 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 Commenti
Oleg Komarov
Oleg Komarov il 27 Giu 2011
Hopefuly now is ok. Tested initial and final position.

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Più risposte (3)

Sean de Wolski
Sean de Wolski il 27 Giu 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 il 26 Giu 2011
  3 Commenti
Wolfgang Schwanghart
Wolfgang Schwanghart il 27 Giu 2011
I = [ 1 0 0;
2 5 0;
0 0 3;
0 0 0];
% find the neighbors of the elements where I = 5
I5 = I==5;
[ix,ixn] = ixneighbors(I,I5)
ix =
6
6
6
6
6
6
6
6
ixn =
10
7
2
5
9
1
11
3
% thus, ixn are the linear indices of the neighbors of the indices ix.
% You'll find the values associated with the neighbors by
I(ixn)
ans =
0
0
2
0
0
1
3
0
Mohammad Golam Kibria
Mohammad Golam Kibria il 28 Giu 2011
thanks,this also works fine for me

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Andrei Bobrov
Andrei Bobrov il 26 Giu 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 Commenti
Andrei Bobrov
Andrei Bobrov il 27 Giu 2011
Thanks Oleg! Corrected...
Mohammad Golam Kibria
Mohammad Golam Kibria il 28 Giu 2011
Thanks this also works fine for me

Accedi per commentare.

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