finding interior region

2 vues (au cours des 30 derniers jours)
Mohammad Golam Kibria
Mohammad Golam Kibria le 21 Juin 2011
Hi I have the following code and have some output as follows:
I =
0 1 1 1 1 1 0
1 0 0 0 0 0 1
1 0 1 1 1 0 1
1 0 1 1 1 0 1
1 0 0 0 0 0 1
0 1 1 1 1 1 0
K>> boundary = bwboundaries(I,8,'noholes'); K>> boundary
boundary =
[19x2 double]
[ 7x2 double]
here is two boundary, I need to know how to determine which boundary is interior, Here [ 7x2 double] is in interior region of [19x2 double]. How to determine that easily?
Can anybody help?
Thanks

Réponse acceptée

Andrei Bobrov
Andrei Bobrov le 21 Juin 2011
>> [B,L] = bwboundaries(I,8,'noholes')
B =
[19x2 double]
[ 7x2 double]
L =
0 1 1 1 1 1 0
1 0 0 0 0 0 1
1 0 2 2 2 0 1
1 0 2 2 2 0 1
1 0 0 0 0 0 1
0 1 1 1 1 1 0
>>
B{1} -> label 1 (L == 1)
B{2} -> label 2 (L == 2)
EDIT 2
>> [B,L,N,A] = bwboundaries(I,8,'noholes')
B =
[19x2 double]
[ 7x2 double]
L =
0 1 1 1 1 1 0
1 0 0 0 0 0 1
1 0 2 2 2 0 1
1 0 2 2 2 0 1
1 0 0 0 0 0 1
0 1 1 1 1 1 0
N =
2
A =
(2,1) 1
>> full(A)
ans =
0 0
1 0
Having label 2 (number row A = 2) inside label 1(column number 1of matrix A)
  3 commentaires
Andrei Bobrov
Andrei Bobrov le 21 Juin 2011
Dear Mohammad! Please reading help function 'bwboundaries'.
Mohammad Golam Kibria
Mohammad Golam Kibria le 22 Juin 2011
Thanks, Now It is clear

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Plus de réponses (1)

Sean de Wolski
Sean de Wolski le 21 Juin 2011
You could also do an
Ibwnoholes = imfill(I,'holes');
which will fill the holes and then that boundary won't show up.

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