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# Replacing centre pixel value

Asked by FIR on 19 Dec 2012
`        I have a code ,of 7x7 natrix in that for each 3x3 i have found minimum value and if that minimum value is greater the centre pixel value ,the centre pixel value must be replaced by 5.have problem in replacing centre pixel value ,please help`
```for i=2:6
for j=2:6
K= A(i-1:i+1,j-1:j+1);
T=K(2,2);finding centre pixel value
```
```          B=min(K(:));
if B>T```
`              %%%please tell what must come here replacing centre pixel vaue by 5`
`              %%%%I DID   K(2,2)=5;`
```          else
%%%%%%%%%here also
%%I DID K=K;
end
end
end```
```BUT I GET 3x3 matrix only,how to get 7x7 matrix
```
`        please assist`

## 1 Comment

Image Analyst on 19 Dec 2012

You've been posting here long enough that you should already know how to format your questions. Please review the tutorial http://www.mathworks.com/matlabcentral/answers/13205-tutorial-how-to-format-your-question-with-markup. By the way, you can type Ctrl-A, Ctrl-I, Ctrl-C when your code is in MATLAB to fix indenting problems. Then come here and paste it in, highlight it, and click {}Code. Do not have any spaces before non-code text (like your first paragraph). It's not hard - give it a try.

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Answer by Image Analyst on 19 Dec 2012

The minimum value in a 7x7 window will never be greater than the center value. Please explain what you are thinking.

By the way, you can call imerode() to get the local min values in a 7x7 sliding window.

FIR on 19 Dec 2012

Just it is an example ,i want to try on my own ,the concept which i was given,the above explaination is just the concept,finallyy i need 7x7 but i get only 3x3

Image Analyst on 19 Dec 2012

But the concept is flawed! Anyway, you might play around and use imerode and subtract it from your original, or vice-versa. Then threshold and replace items meeting the threshold with 5. I think you already know how to do that without loops. Even if you did want to use loops, you'd have to have 4 loops, not 2. Two to scan rows and columns, and then at each pixel, another two loops (like you already have) to scan a window around that pixel. But not a very efficient way to do it. I suggest you review the code examples I gave you in your prior, related questions.

FIR on 20 Dec 2012

Can u please post an example here please,i could nor review the codes since there are many questions posted by me

Answer by Walter Roberson on 19 Dec 2012
```A(i,j) = 5;
```

Walter Roberson on 20 Dec 2012

In that case do not use K= A(i-3:i+3,j-3:j+3); and use the original instead, and use

```A(i,j) = 5;
```

like I originally indicated. The output will be A.

This is for replacing the center pixel like you asked. It probably isn't what you wanted, though, as replacing the center pixel as you proceed is going to affect the computations to the right and below. What you probably want is to output a new array that is like the original array but with the center pixels replaced. Such as

```newA = A;
```
```for i=2:6
for j=2:6
K= A(i-1:i+1,j-1:j+1);
T=K(2,2);finding centre pixel value
B=min(K(:));
if B>T
newA(i,j) = 5;
else
%I could not tell what you want done in this case
end
end
end
```
FIR on 20 Dec 2012

ok walter thanks

can u guide me in quanterion switching filter,please

Walter Roberson on 20 Dec 2012

I do not know anything about quaternion switching filters.