How can I change the intensities of a certain pixel in an image along with 3*3 surrounding neighbor pixels?

1 回表示 (過去 30 日間)
Lets say I have a image(matrix) as following:
a =
1 1 1 1 1
1 1 1 1 1
1 2 2 2 1
1 2 0 2 1
1 2 2 2 1
1 1 1 1 1
1 1 1 1 1
Finding that 0 in the middle and changing it is easy. I can do this :
a(a==0)=4;
and the result will be this:
a =
1 1 1 1 1
1 1 1 1 1
1 2 2 2 1
1 2 4 2 1
1 2 2 2 1
1 1 1 1 1
1 1 1 1 1
Now I want a code that can find the zero, convert that zero to 4, and also convert the 3*3 neighbors of zero to 4 as well. So after the code that you're suggesting, my output should look something like this:
b =
1 1 1 1 1
1 1 1 1 1
1 4 4 4 1
1 4 4 4 1
1 4 4 4 1
1 1 1 1 1
1 1 1 1 1
thanks in advance, Sina

採用された回答

Matt J
Matt J 2016 年 9 月 21 日
a( imdilate(a==0, true(3)) ) = 4;
  3 件のコメント
Image Analyst
Image Analyst 2016 年 9 月 21 日
Use the strel() function to generate different shaped kernels.
Changoleon
Changoleon 2016 年 9 月 21 日
That is very true and quick response, thank you

サインインしてコメントする。

その他の回答 (1 件)

Adam
Adam 2016 年 9 月 21 日
[x, y] = find( a == 0 );
a( x-1:x+1, y-1:y+1 ) = 4;
Obviously you would need to deal with when you are on the border and there is no x-1 or y+1 or whatever, but that is easy enough for you to adapt.
  1 件のコメント
Matt J
Matt J 2016 年 9 月 21 日
Changoleon commented:
Adam,
Fast and accurate answer. Thanks you. I just realized that I need to change my values circularly which is impossible to make a perfect circle inside a matrix ( since you can not cover half of an element). How to change it so it look like a hexagon or octagon?

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeConvert Image Type についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by