finding interior region

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Mohammad Golam Kibria
Mohammad Golam Kibria el 21 de Jun. de 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

Respuesta aceptada

Andrei Bobrov
Andrei Bobrov el 21 de Jun. de 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 comentarios
Andrei Bobrov
Andrei Bobrov el 21 de Jun. de 2011
Dear Mohammad! Please reading help function 'bwboundaries'.
Mohammad Golam Kibria
Mohammad Golam Kibria el 22 de Jun. de 2011
Thanks, Now It is clear

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Más respuestas (1)

Sean de Wolski
Sean de Wolski el 21 de Jun. de 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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