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From: Ayan on 3 Jun 2010 02:20
Walter Roberson <roberson(a)hushmail.com> wrote in message <AOtNn.21341$7d5.5805(a)newsfe17.iad>...
> Ayan wrote:
>
> > I am looking for some help in dividing an image to 8x8 block and access
> > each block separately. My image I is of size 640x480. I need to divide
> > the image into subblocks of size 8x8 and have to find histogram of each
> > block separately. blkproc is not helping as I cannot access each block
> > separately.
>
> You posted the same question before, but then and now I do not know what
> you mean by "access each block separately" ? blkproc() would pass the
> provided function the complete matrix that is the current block. Doing a
> histogram of a single block would seem to only require the values in the
> block. Is there something you have not mentioned that would require the
> block coordinates?
Hi Walter,

I dont thing my problem is very much complex. The only thing is I am very much new to Matlab for which I am facing the problem. I tried to break the main image into subblocks of 8x8 as I need to find the local max of the histogram of each block. But as Jos have mentioned, my image size is not divisible absolutely by 8 (245x133 is my image size) and hence I need to do something to pass the arguments into mat2cell. I am trying to handle this issue.

Regards,
Ayan
From: Laxminarayana on 2 Jul 2010 04:00
"Ayan " <a.chaki(a)lycos.com> wrote in message <hu7hj5$31d$1(a)fred.mathworks.com>...
> Walter Roberson <roberson(a)hushmail.com> wrote in message <AOtNn.21341$7d5.5805(a)newsfe17.iad>...
> > Ayan wrote:
> >
> > > I am looking for some help in dividing an image to 8x8 block and access
> > > each block separately. My image I is of size 640x480. I need to divide
> > > the image into subblocks of size 8x8 and have to find histogram of each
> > > block separately. blkproc is not helping as I cannot access each block
> > > separately.
> >
> > You posted the same question before, but then and now I do not know what
> > you mean by "access each block separately" ? blkproc() would pass the
> > provided function the complete matrix that is the current block. Doing a
> > histogram of a single block would seem to only require the values in the
> > block. Is there something you have not mentioned that would require the
> > block coordinates?
> Hi Walter,
>
> I dont thing my problem is very much complex. The only thing is I am very much new to Matlab for which I am facing the problem. I tried to break the main image into subblocks of 8x8 as I need to find the local max of the histogram of each block. But as Jos have mentioned, my image size is not divisible absolutely by 8 (245x133 is my image size) and hence I need to do something to pass the arguments into mat2cell. I am trying to handle this issue.
>
> Regards,
> Ayan

Hi Ayan,

Though my process look lengthy, but it may be helpful to you.
As u said that ur image size is 640x480 and u want to divide it into 8x8 blocks (i'm assuming NON OVERLAPPING) try the below code.

%////////////////////////////////////////////////////////////////////////

I=imread('imagefile.extension');
[r c]=size(I);
bs=8; % Block Size (8x8)

nob=(r/bs)*(c/bs); % Total number of 8x8 Blocks

% Dividing the image into 8x8 Blocks
kk=0;
for i=1:(r/bs)
for j=1:(c/bs)
Block(:,:,kk+j)=I((bs*(i-1)+1:bs*(i-1)+bs),(bs*(j-1)+1:bs*(j-1)+bs));
end
kk=kk+(r/bs);
end

% Accessing individual Blocks

figure;imshow(Block(:,:,1)) % This shows u the fist 8x8 Block in a figure window
figure;imshow(Block(:,:,2)) % This shows u the second 8x8 Block (i.e as per my %coding rows from 1:8 and col from 9:16) in a figure window and so on.....

% Looking at Histograms of Individual Blocks

figure;imhist(Block(:,:,1)) % Displays histogram window of first Block
figure;imhist(Block(:,:,2)) % Displays histogram window of second Block andso on.....

% \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
Hope this will be helpful to You.....

ALL THE BEST...
From: Jan Simon on 2 Jul 2010 04:19
Dear Laxminarayana,

> I=imread('imagefile.extension');
> [r c]=size(I);
> bs=8; % Block Size (8x8)
>
> nob=(r/bs)*(c/bs); % Total number of 8x8 Blocks

Then do not forget to pre-allocate:
Block = zeros(r/bs, c/bs, nob);

> % Dividing the image into 8x8 Blocks
> kk=0;
> for i=1:(r/bs)
> for j=1:(c/bs)
> Block(:,:,kk+j)=I((bs*(i-1)+1:bs*(i-1)+bs),(bs*(j-1)+1:bs*(j-1)+bs));
> end
> kk=kk+(r/bs);
> end

This is equivalent to:
Im = imread('imagefile.extension');
[r, c]=size(I);
Im = reshape(Im, r/8, 8, c/8, 8);
Im = permute(Im, [1, 3, 2, 4]);
Im = reshape(Im, 8, 8, []);
You can access also matrix sizes, which are not multiple of 8:
r8 = 8 * floor(r / 8);
c8 = 8 * floor(c / 8);
Im = reshape(Im(1:r8, 1:c8), r8/8, 8, c8/8, 8);
Now Im(:, :, n) is the n.th block.

Good luck, Jan
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