How to find neighbor matrix?

2 vues (au cours des 30 derniers jours)
Tan Wen Kun
Tan Wen Kun le 3 Déc 2015
Commenté : Walter Roberson le 7 Déc 2015
This is some code i write:
for i=1:h %h=height of the image
for j=1:w %w=width of image
if (i,j)==1 %1 is the border
if (i,j-1) ~=1 && if(i,j+1) ~=1
table(j-1,j+1) =1
table(j+1,j-1) =1
for j=1:w %w=width of image
for i=1:h %h=height of the image
if (j,1)==1 %1 is the border
if (j,i-1) ~=1 && if(j,i+1) ~=1
table(i-1,i+1) =1
table(i+1,i-1) =1
I hope loop horizontal and vertical to find the neighbor matrix dont care of diagonal situation.
As long as it read x then read 1 then read y then (x,y) and(y,x)=1
How to solve this?
A=
1 1 1 1 1 1 1 1 1 1
1 2 2 1 3 3 1 4 4 1
1 2 2 1 3 1 5 5 5 1
1 1 1 1 1 1 1 1 1 1
result=
2 3 4 5
2 1 1 0 0
3 1 1 1 1
4 0 1 1 1
5 0 1 1 1
  1 commentaire
Walter Roberson
Walter Roberson le 3 Déc 2015
table((i-1),(j+1)) attempts to index table(0,2) because i starts at 1.

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Walter Roberson
Walter Roberson le 3 Déc 2015
table = (A == 1);
s = max(size(table,1),size(table,2));
if s ~= size(table,1) || s ~= size(table,2)
table(s,s) = false; %make it square
end
table = table | table.'; %make it symmetric
  2 commentaires
Tan Wen Kun
Tan Wen Kun le 7 Déc 2015
Modifié(e) : Tan Wen Kun le 7 Déc 2015
Your solution just extracted 1 from the labelimg, that not I want.
What I want is when is label= 1 2 1 3 1
I made table(2,3)=1 and table(3,2)=1
Attempted to access labelimg(214,0); index must be a positive integer or logical.
for i=1:max(labelimg(:))
for j=1:max(labelimg(:))
if labelimg(i,j)==1 %1 is the border
if labelimg(i,j-1) ~=1
if labelimg(i,j+1) ~=1
table(j-1,j+1) =1 ;
table(j+1,j-1) =1 ;
how to solve this when I using j-1 when loop first element got problem and also j+1 when loop last element?How to solve the index out of bound?
Walter Roberson
Walter Roberson le 7 Déc 2015
maxlab = max(labelimg(:));
for i = 1 : maxlab
for j = 2 : maxlab - 1
Now you can test labelimg(i,j-1) and labelimg(i,j+1)

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