how to find mean for the values in an image

Sharen H (view profile)

on 20 Jul 2013

i have to divide my image into n number of small 2X2 matrix and find the mean value and replace that 2x2 matrix with the mean value . example

x=

`   [2 3  4 5`
`   2 3  4 5`
`   1 2  3 4`
`   1 2  3 4]`

i have to find the mean of

`                           [2 3`
```                            2 3]
similarly for others and replace that concerned matrix with that mean value
the output should get is```
```[2 2 4 4
```
` 2 2 4 4`
` 1 1 3 3`
` 1 1 3 3]`

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Wayne King (view profile)

on 20 Jul 2013
Edited by Wayne King

Wayne King (view profile)

on 20 Jul 2013

Do you have the Image Processing Toolbox? You can use blockproc(), but I'm not sure how you are getting your expected output above. You say you need the mean of 2x2 chunks submatrices like [2 3; 2 3]. Do you mean that you want the column means of those, which are 2 and 3, or you want the mean of all 4 elements which is 2.5? How are you taking the mean and ending up with a matrix the same size as the original?

I'll show you how to use blockproc() to return the column means for each 2x2 submatrix in x. That means for each 2x2 submatrix, you'll get back a 1x2 with the two column means. Accordingly, a 4x4 matrix reduces to a 2x4.

```fun = @(block_struct) mean(block_struct.data);
Y = blockproc(x,[2 2],fun);
```

If you want the overall mean of each 2x2 submatrix, giving you a 2x2 matrix at the end:

```fun = @(block_struct) mean(block_struct.data(:));
Y = blockproc(x,[2 2],fun);
```

Sharen H

Sharen H (view profile)

on 20 Jul 2013

i am using matlab 7.6 .i have image processing tool box the command here is blkproc()

Wayne King

Wayne King (view profile)

on 20 Jul 2013

I think you do not have the blockproc() function from the Image Processing Toolbox, blockproc() was only added to the Image Processing Toolbox in R2009b, which is 7.9

Sharen H

on 20 Jul 2013

okok thank you

Andrei Bobrov (view profile)

on 20 Jul 2013
Edited by Andrei Bobrov

Andrei Bobrov (view profile)

on 23 Jul 2013

for youe case without blockproc (without Image Processing Toolbox):

```kron(cellfun(@mean2,mat2cell(x,2*[1 1],2*[1 1])),[1 1; 1 1])
```

or use Peter J. Acklam' idea. Thanks to the cyclist for this link.

```m = [2 2]; % size your small block
s = size(x);
n = s./m;
```
```a = permute(reshape(x,m(1),n(1),n(2),[]),[1 3 2 4]);
out = kron(reshape(mean(reshape(a,[],prod(m))),n(1),[]),ones(n))
```

use function blkproc

```sizeblock = [2 2];
Y = blkproc(x,sizeblock,@mean2);
out = kron(Y,ones(size(x)./sizeblock));
```

Sharen H

on 23 Jul 2013

Sharen H

Sharen H (view profile)

on 23 Jul 2013

when i try for a [3,3] small block i am not able to process...

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