finding interior or exterior points

8 次查看（过去 30 天）
Mohammad Golam Kibria
Hi, I have a matrix as follows: I =
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 1 1 1 0 0 1
0 0 0 0 1 0 1
0 0 0 1 0 1 1
0 0 0 0 0 0 0
L=bwlabel(I)
L =
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 1 1 1 0 0 1
0 0 0 0 1 0 1
0 0 0 2 0 1 1
0 0 0 0 0 0 0
Here label 1 makes a continuous curve line. I need to know where point (5,4) belongs to, upper portion or lower portion. Thanks

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采纳的回答

Andrei Bobrov
Andrei Bobrov 2011-6-28
Have matrix
I =
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 1 1 0
0 0 1 0 0 1
0 0 1 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
1 0 0 0 1 0
0 1 0 0 1 0
0 0 1 0 1 0
0 0 1 0 1 0
0 0 1 0 1 0
0 0 0 1 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
We need to know where point (5,4) belongs to, upper portion or lower portion.
Variant of solution.
idx = {5,4};
BW2 = bwmorph(I,'diag');
L = bwlabel(~BW2);
Id1 = bwdist(I,'chessboard') == 1;
for i1 = 1:max(L(:))
L(bwdist(L == i1,'chessboard')==1 & Id1) = i1;
end
if L(idx{:}) == L(1,1), disp('pixel in upper portions' );
elseif L(idx{:}) == L(end,1), disp('pixel in lower portions' );
else disp('pixel on boundaries of portions' );
end
_*best way, as in http://www.mathworks.com/matlabcentral/answers/10501-finding-region*_
L = bwlabel(~BW2,4);
if L(idx{:}) == max(L(1,:)), disp('pixel in upper portions' );
elseif L(idx{:}) == max(L(end,:)), disp('pixel in lower portions' );
else disp('pixel on boundaries of portions' );
end

更多回答（1 个）

Walter Roberson
Walter Roberson 2011-6-28
The upper and lower portions are the same, as your curve of label 1 does not partition the space: you can go around by way of (3,1) to pass between "upper" and "lower" without encountering any "1".

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