Jonas的尼斯解决办法。 由于我看到这番话,我也想到我的解决办法。 希望这比你原来的法典更加相似。

你所犯的一个错误是假定这一联系,只能用一个通行证。 一般来说,由于对一些“snake分子”的探测取决于你如何通过矩阵进行检索。 除此之外,我外加一.。

 bw=[  1     1     0     2     2     2     0     3;
        1     1     0     2     0     2     0     3;
        1     1     1     1     0     0     0     3;
        0     0     0     0     0     0     0     3;
        4     4     4     4     0     5     0     3;
        0     0     0     4     0     5     0     3;
        6     6     0     4     0     0     0     3;
        6     6     0     4     0     7     7     7 ];


 %Set up matrix   
 [r,c] = size(bw);   

 %Zero pad border
 bwZ = [zeros(1,c+2);[zeros(r,1) bw zeros(r,1)];zeros(1,c+2)];

 %Iterate over all elements within zero padded border
 done=0;%Done flag 
 cc=1;  %Infinite loop protection
 while not(done) && cc<r*c
     done=1;cc=cc+1;
     for i=2:r+1
         for j=2:c+1
            %Point should be evaluated
            p = bwZ(i,j);
            if p >= 1
                %Pick out elements around active elements
                if bwZ(i-1,j)==0;ue=inf;else ue = bwZ(i-1,j); end;%Up element
                if bwZ(i+1,j)==0;de=inf;else de = bwZ(i+1,j); end;%Down element
                if bwZ(i,j-1)==0;le=inf;else le = bwZ(i,j-1); end;%Left element
                if bwZ(i,j+1)==0;re=inf;else re = bwZ(i,j+1); end;%Right element

                bwZ(i,j) = min([ue de le re]);
                %Set flag, if something has changed 
               if bwZ(i,j) ~= p
                    done = 0;
               end
            end
         end
     end
 end

 %Remove zero padding
 bw= bwZ(2:end-1,2:end-1)

Output:

bw=

 1     1     0     1     1     1     0     3
 1     1     0     1     0     1     0     3
 1     1     1     1     0     0     0     3
 0     0     0     0     0     0     0     3
 4     4     4     4     0     5     0     3
 0     0     0     4     0     5     0     3
 6     6     0     4     0     0     0     3
 6     6     0     4     0     3     3     3

如果数字不需要保留,你可以简单地把<条码>bwlabel<>/条码>放在你原来的形象上:

newImage = bwlabel(originalImage>0);

http://www.ohchr.org。

Here s another version. It checks each connected component to see whether there s another connected component touching it. If yes, that connected component is relabelled.

%# nCC: number of connected components
nCC = max(originalImage(:));

for cc = 1:nCC
%# check whether the component exists
myCC = originalImage==cc;
if any(any(myCC))

   %# create a mask to check for neighbors
   %# by creating a border of 1 pixel
   %# around the original label
   msk = imdilate(myCC,true(3)) & ~myCC;

   %# read all the pixel values under the mask
   neighbours = originalImage(msk);

   %# we re not interested in zeros, remove them
   neighbours = neighbours(neighbours > 0);

   if ~isempty(neighbours)
   %# set the label of all neighbours to cc
   originalImage( ismember(originalImage,neighbours) ) = cc;
   end
end
end