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How do I create a structuring element of my own?

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oshawcole
oshawcole el 17 de Oct. de 2018
Editada: Matt J el 17 de Oct. de 2018
I want to create a structuring element with the center '0' (eg. [1 0 1])and I want to see the effect of erosion and dilation of it on a 3x3 binary matrix. How do I go about it? please explain. My code is giving me same answers for erosion and dilation.
aa=[1 1 1; 1 0 1; 1 1 1]
s=[1 1 1];
ero=imerode(aa,s)
dil=imdilate(aa,s)
pp=[1 0 1];
ero1=imerode(aa,pp)
dil1=imdilate(aa,pp)
  2 comentarios
Matt J
Matt J el 17 de Oct. de 2018
Editada: Matt J el 17 de Oct. de 2018
My code is giving me same answers for erosion and dilation.
As it should, for that choice of aa. What do you expect the result to be?
oshawcole
oshawcole el 17 de Oct. de 2018
Editada: oshawcole el 17 de Oct. de 2018
Mathematically, the result for aa eroded with s=[1 1 1] should be [0 1 0; 0 0 0; 0 1 0]. For s=[1 0 1]; I was curious as to what happens when I take the center of the structuring element to be 0; so I don't know the answer.

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Matt J
Matt J el 17 de Oct. de 2018
Elements outside the boundary of the matrix do not participate in the erosion/dilation. If you pad aa to a larger size, you will see different effects.
>> aaa=zeros(5); aaa(2:4,2:4)=aa
aaa =
0 0 0 0 0
0 1 1 1 0
0 1 0 1 0
0 1 1 1 0
0 0 0 0 0
>> ero=imerode(aaa,[1,1,1])
ero =
0 0 0 0 0
0 0 1 0 0
0 0 0 0 0
0 0 1 0 0
0 0 0 0 0
>> ero=imerode(aaa,[1,0,1])
ero =
0 0 0 0 0
1 0 1 0 1
1 0 1 0 1
1 0 1 0 1
0 0 0 0 0
>> ero=imdilate(aaa,[1,0,1])
ero =
0 0 0 0 0
1 1 1 1 1
1 0 1 0 1
1 1 1 1 1
0 0 0 0 0
  2 comentarios
oshawcole
oshawcole el 17 de Oct. de 2018
Awesome! Can you please explain why is this padding needed? Thank you so much. I really appreciate your help.
Matt J
Matt J el 17 de Oct. de 2018
Editada: Matt J el 17 de Oct. de 2018
It is needed so that the code can know that you are assuming zeros lie beyond the boundary of the original 3x3 matrix. Otherwise, it doesn't know what you are assuming.

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