finding neighbor of a position

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

Answer 1

[Oleg Komarov]

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);

[Mohammad Golam Kibria]

Im not sure,If l is the center position then in case l=1 it is not giving proper output.

[Oleg Komarov]

Hopefuly now is ok. Tested initial and final position.

[Mohammad Golam Kibria]

Thanks

Answer 2

[Sean de Wolski]

 idx = find(conv2(double(I==5),ones(3),'same'))

%This includes the 6, but that could easily be taken care of with setdiff.

[Mohammad Golam Kibria]

Thanks

Answer 3

[Wolfgang Schwanghart]

Hi,

you might find this function useful.

http://www.mathworks.com/matlabcentral/fileexchange/16991-neighbor-indexing

Cheers, W.

[Mohammad Golam Kibria]

I am very weak in matlab.would you please write how to call the function ixneighbors for index 8

[Wolfgang Schwanghart]

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]

thanks,this also works fine for me

Answer 4

[Andrei Bobrov]

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)

[Oleg Komarov]

Boundary conditions not satisfied

[Andrei Bobrov]

Thanks Oleg! Corrected...

[Mohammad Golam Kibria]

Thanks this also works fine for me