findding neares label

2 vues (au cours des 30 derniers jours)
Mohammad Golam Kibria
Mohammad Golam Kibria le 20 Juin 2011
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

Réponse acceptée

Andrei Bobrov
Andrei Bobrov le 20 Juin 2011
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

Plus de réponses (1)

Walter Roberson
Walter Roberson le 20 Juin 2011
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.
  1 commentaire
Mohammad Golam Kibria
Mohammad Golam Kibria le 20 Juin 2011
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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