comparing 3x3 block with center pixel
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i have a matrix as shown below

i wanted to take 3x3 pixel and take 3 pixels at a time along with the center pixel,
and compare the selected 3 pixel value with the center pixel,
and if 2 or morepixel has value greater than the center pixel and i wanted to assign 1 to it else 0
So in this case i will get 0-1-1-0 and then convert the binary 0110 to its corresponding decimal value = 6
Then the next 3x3 pixel

4 commentaires
Matt J
le 20 Nov 2019
Do you take the 3x3 blocks in non-overlapping tiles or is it a sliding window?
Elysi Cochin
le 20 Nov 2019
KALYAN ACHARJYA
le 20 Nov 2019
You did not answered the @Matt J question.
Also is the start pixel at 70 or 200?
What about boundary pixels?
Elysi Cochin
le 20 Nov 2019
Réponse acceptée
Plus de réponses (1)
KALYAN ACHARJYA
le 20 Nov 2019
Modifié(e) : KALYAN ACHARJYA
le 20 Nov 2019
Self Declaration: Odd Way but simpler
for i=2:Row-1
for j=2:Col-1
if (mat(i,j)<mat(i-1,j-1)+mat(i,j)<mat(i,j-1)+mat(i,j)<mat(i+1,j-1))=>2
data(1)=1;
else
data(1)=0;
end
if (mat(i,j)<mat(i-1,j-1)+mat(i,j)<mat(i-1,j)+mat(i,j)<mat(i-1,j+1))=>2
data(2)=1;
else
data(2)=0;
end
if (mat(i,j)<mat(i+1,j-1)+mat(i,j)<mat(i_1,j)+mat(i,j)<mat(i+1,j+1))=>2
data(3)=1;
else
data(3)=0;
end
if (mat(i,j)<mat(i-1,j+1)+mat(i,j)<mat(i,j+1)+mat(i,j)<mat(i+1,j+1))=>2
data(4)=1;
else
data(4)=0;
end
mat(i,j)=bin2dec(logical(data));
end
end
7 commentaires
Elysi Cochin
le 20 Nov 2019
Modifié(e) : Elysi Cochin
le 21 Nov 2019
KALYAN ACHARJYA
le 20 Nov 2019
mat = [70 20 100 24 89; 230 200 220 78 55; 210 80 10 57 34; ...
110 94 67 88 28; 32 36 89 78 58; 59 73 54 68 48];
[Row, Col] = size(mat);
for i=2:Row-1
for j=2:Col-1
if (mat(i,j)<mat(i-1,j-1)+mat(i,j)<mat(i,j-1)+mat(i,j)<mat(i+1,j-1))>=2
data(1)=1;
else
data(1)=0;
end
if (mat(i,j)<mat(i-1,j-1)+mat(i,j)<mat(i-1,j)+mat(i,j)<mat(i-1,j+1))>=2
data(2)=1;
else
data(2)=0;
end
if (mat(i,j)<mat(i+1,j-1)+mat(i,j)<mat(i-1,j)+mat(i,j)<mat(i+1,j+1))>=2
data(3)=1;
else
data(3)=0;
end
if (mat(i,j)<mat(i-1,j+1)+mat(i,j)<mat(i,j+1)+mat(i,j)<mat(i+1,j+1))>=2
data(4)=1;
else
data(4)=0;
end
mat(i,j)=bi2de(logical(data))
end
end
KALYAN ACHARJYA
le 20 Nov 2019
Modifié(e) : KALYAN ACHARJYA
le 20 Nov 2019
use Logical, no need to combine separately
Jan
le 20 Nov 2019
Easier:
for i=2:Row-1
data = zeros(1, 4);
for j=2:Col-1
if (mat(i,j)<mat(i-1,j-1)+mat(i,j)<mat(i,j-1)+mat(i,j)<mat(i+1,j-1))>=2
data(1) = 8;
end
if (mat(i,j)<mat(i-1,j-1)+mat(i,j)<mat(i-1,j)+mat(i,j)<mat(i-1,j+1))>=2
data(2) = 4;
end
if (mat(i,j)<mat(i+1,j-1)+mat(i,j)<mat(i-1,j)+mat(i,j)<mat(i+1,j+1))>=2
data(3) = 2;
end
if (mat(i,j)<mat(i-1,j+1)+mat(i,j)<mat(i,j+1)+mat(i,j)<mat(i+1,j+1))>=2
data(4) = 1;
end
mat(i,j) = sum(data);
end
end
I think Jan's modification would really need to be as follows,
out=zeros(size(mat));
data = zeros(1, 4);
for i=2:Row-1
for j=2:Col-1
if (mat(i,j)<mat(i-1,j-1)+mat(i,j)<mat(i,j-1)+mat(i,j)<mat(i+1,j-1))>=2
data(1) = 8;
end
if (mat(i,j)<mat(i-1,j-1)+mat(i,j)<mat(i-1,j)+mat(i,j)<mat(i-1,j+1))>=2
data(2) = 4;
end
if (mat(i,j)<mat(i+1,j-1)+mat(i,j)<mat(i-1,j)+mat(i,j)<mat(i+1,j+1))>=2
data(3) = 2;
end
if (mat(i,j)<mat(i-1,j+1)+mat(i,j)<mat(i,j+1)+mat(i,j)<mat(i+1,j+1))>=2
data(4) = 1;
end
out(i,j) = sum(data); data(:)=0;
end
end
Jan
le 21 Nov 2019
@Matt J: You are right, modifying mat inside the loop, which uses mat as input also is not wanted. If you declare data before the loops and reset the values to 0 by data(:)=0 or if it is recreated by zeros(1,4) does not really matter. For large arrays one of the methods might be faster than the other.
Matt J
le 21 Nov 2019
@Jan: It does matter however, that you were recreating data outside the j-loop. It needs to be reset inside both loops after each (i,j) is processed.
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