findding neares label

조회 수: 2 (최근 30일)
Mohammad Golam Kibria
Mohammad Golam Kibria 2011년 6월 20일
Hi I have the following 3 matrix. I =
0 0 1 1 0 0
0 1 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
I1 =
0 0 0 0 0 0
0 1 1 1 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
I2 =
0 0 0 0 0 0
0 0 0 0 0 0
1 0 0 0 0 1
0 1 0 0 1 0
0 0 1 1 0 0
0 0 0 0 0 0
I need to know the 1s in I1 is closer to the which matrices' 1s. Here If some one looks at the matrix then obviously 1s in I1 is closer to the 1s in I than in I2.
Is there any one to help?
Thanks

채택된 답변

Andrei Bobrov
Andrei Bobrov 2011년 6월 20일
so
D = bwdist(I)
P1 = sum(D(logical(I1)))
P2 = sum(D(logical(I2)))
if P1 < P2 , disp('I1 is closer to the 1s in I than in I2');
else disp('I2 is closer to the 1s in I than in I1'); end

추가 답변 (1개)

Walter Roberson
Walter Roberson 2011년 6월 20일
V = 1:size(I,1);
Ipos = V * I;
I1pos = V * I1;
I2pos = V * I2;
I1score = sum(abs(Ipos-I1pos));
I2score = sum(abs(Ipos-I2pos));
if I1score < I2score
%I1
elseif I1score > I2score
%I2
else
%equal
end
This is based on my arbitrary meaning of "closer", as you are very vague as to what "closer" means.
  댓글 수: 1
Mohammad Golam Kibria
Mohammad Golam Kibria 2011년 6월 20일
for I1 and I,pixel distance having 1 are 0,0,1,1,0,0 and for I1 and I2 are 0,2,root(3),root(3),2,0, so it may be consider that s in I1 is closer to the 1s in I than in I2.
perhaps now it is clear

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