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Asked by troy on 18 May 2013

Assume i have 8x8 blocks of image. Here my code:

[L L]=size(img); L2=C/8; imgblk=cell(L2,L2); for i=1:L2 for j=1:L2 imgblk{i,j}=img( (i-1)*4+1:i*4 , (j-1)*4+1:j*4); end end

Now, how to apply coding below into imgblk{i,j} since imgblk{i,j} is in 'cell'??

A (i,j) = [Ʃ(fi,j - fi,j+1 ) - Ʃ(fi,j-1 - fi,j )] and B (i,j) = [Ʃ(fi,j - fi,j+1 ) - Ʃ(fi,j+1 - fi,j+2 )]

Also, how to do in horizontal first, then vertical? Thanks for those who help me =)

Answer by Image Analyst on 19 May 2013

Edited by Image Analyst on 19 May 2013

I don't understand this code at all. Some questions

- What is C?
- What is L2 for?
- What is imgblk intended to be used for? Why is it necessary? Why are you messing around with cells anyway?
- Do you know that sum(I2(i,j) - I2(i,j+1)) will clip to zero if I2 is uint8? Cast them to double before subtracting to avoid that.
- What are A and B supposed to represent?

Show 2 older comments

troy on 20 May 2013

Its ok. I know everyone have their works. I have tried blockproc before and could't understand it. But I will try to understand blockproc if ypu give me blokproc demo. I hope you will read the paper someday. I have tried to do this almost 1 year but get like nothing.lol

1 more thing. Why are you said that I can't use the imgblk?

Thanks.

Image Analyst on 20 May 2013

I didn't say you can't use it. I said you don't use it.

Here's the first blockproc demo:

% Demo code to divide the image up into 16 pixel by 16 pixel blocks % and replace each pixel in the block by the median, mean, or standard % deviation of all the gray levels of the pixels in the block. % clc; clearvars; close all; workspace; fontSize = 16;

% Read in a standard MATLAB gray scale demo image. folder = fullfile(matlabroot, '\toolbox\images\imdemos'); if ~exist(folder, 'dir') % If that folder does not exist, don't use a folder % and hope it can find the image on the search path. folder = []; end baseFileName = 'cameraman.tif'; fullFileName = fullfile(folder, baseFileName); grayImage = imread(fullFileName); % Get the dimensions of the image. numberOfColorBands should be = 1. [rows, columns, numberOfColorBands] = size(grayImage) % Display the original gray scale image. subplot(2, 2, 1); imshow(grayImage, []); title('Original Grayscale Image', 'FontSize', fontSize);

% Enlarge figure to full screen. set(gcf, 'Position', get(0,'Screensize')); set(gcf,'name','Image Analysis Demo','numbertitle','off')

% Define the function that we will apply to each block. % First in this demo we will take the median gray value in the block % and create an equal size block where all pixels have the median value. % Image will be the same size since we are using ones() and so for each block % there will be a block of 8 by 8 output pixels. medianFilterFunction = @(theBlockStructure) median(theBlockStructure.data(:)) * ones(size(theBlockStructure.data), class(theBlockStructure.data));

% Block process the image to replace every pixel in the % 8 pixel by 8 pixel block by the median of the pixels in the block. blockSize = [8 8]; % Quirk: must cast grayImage to single or double for it to work with median(). % blockyImage8 = blockproc(grayImage, blockSize, medianFilterFunction); % Doesn't work. blockyImage8 = blockproc(single(grayImage), blockSize, medianFilterFunction); % Works. [rows, columns] = size(blockyImage8);

% Display the block median image. subplot(2, 2, 2); imshow(blockyImage8, []); caption = sprintf('Block Median Image\n32 blocks. Input block size = 8, output block size = 8\n%d rows by %d columns', rows, columns); title(caption, 'FontSize', fontSize);

% Block process the image to replace every pixel in the % 4 pixel by 4 pixel block by the mean of the pixels in the block. % The image is 256 pixels across which will give 256/4 = 64 blocks. % Note that the size of the output block (2 by 2) does not need to be the size of the input block! % Image will be the 128 x 128 since we are using ones(2, 2) and so for each of the 64 blocks across % there will be a block of 2 by 2 output pixels, giving an output size of 64*2 = 128. % We will still have 64 blocks across but each block will only be 2 output pixels across, % even though we moved in steps of 4 pixels across the input image. meanFilterFunction = @(theBlockStructure) mean2(theBlockStructure.data(:)) * ones(2,2, class(theBlockStructure.data)); blockSize = [4 4]; blockyImage64 = blockproc(grayImage, blockSize, meanFilterFunction); [rows, columns] = size(blockyImage64);

% Display the block mean image. subplot(2, 2, 3); imshow(blockyImage64, []); caption = sprintf('Block Mean Image\n64 blocks. Input block size = 4, output block size = 2\n%d rows by %d columns', rows, columns); title(caption, 'FontSize', fontSize);

% Block process the image to replace every pixel in the % 8 pixel by 8 pixel block by the standard deviation % of the pixels in the block. % Image will be smaller since we are not using ones() and so for each block % there will be just one output pixel, not a block of 8 by 8 output pixels. blockSize = [8 8]; StDevFilterFunction = @(theBlockStructure) std(double(theBlockStructure.data(:))); blockyImageSD = blockproc(grayImage, blockSize, StDevFilterFunction); [rows, columns] = size(blockyImageSD);

% Display the block standard deviation filtered image. subplot(2, 2, 4); imshow(blockyImageSD, []); title('Standard Deviation Filtered Image', 'FontSize', fontSize); caption = sprintf('Block Standard Deviation Filtered Image\n32 blocks. Input block size = 8, output block size = 1\n%d rows by %d columns', rows, columns); title(caption, 'FontSize', fontSize);

Here's the second blockproc demo:

% Demo of blockproc in two different ways. % One uses an anonymous function to return a block of pixels % the same size as the sliding window block. % The other uses a custom written function to return a % single value for each sliding window position. function blockproc_demo() try clc; % Clear the command window. close all; % Close all figures (except those of imtool.) workspace; % Make sure the workspace panel is showing. fontSize = 20;

% Change the current folder to the folder of this m-file. if(~isdeployed) cd(fileparts(which(mfilename))); end

% Read in standard MATLAB demo image. grayImage = imread('cameraman.tif'); [rows, columns, numberOfColorChannels] = size(grayImage); % Display the original image. subplot(2, 2, 1); imshow(grayImage, []); caption = sprintf('Original Image\n%d by %d pixels', ... rows, columns); title(caption, 'FontSize', fontSize); % Enlarge figure to full screen. set(gcf, 'Position', get(0,'Screensize')); set(gcf, 'name','Image Analysis Demo', 'numbertitle','off')

% Block process the image. windowSize = 3; % Each 3x3 block will get replaced by one value. % Output image will be smaller by a factor of windowSize. myFilterHandle = @myFilter; blockyImage = blockproc(grayImage,[windowSize windowSize], myFilterHandle); [rowsP, columnsP, numberOfColorChannelsP] = size(blockyImage);

% Display the processed image. % It is smaller, but the display routine imshow() replicates % the image so that it looks bigger than it really is. subplot(2, 2, 2); imshow(blockyImage, []); caption = sprintf('Image Processed in %d by %d Blocks\n%d by %d pixels\nCustom Box Filter', ... windowSize, windowSize, rowsP, columnsP); title(caption, 'FontSize', fontSize);

% Now let's do it an alternate way where we use an anonymous function. % We'll take the standard deviation in the blocks. windowSize = 8; myFilterHandle2 = @(block_struct) ... std2(block_struct.data) * ones(size(block_struct.data)); blockyImageSD = blockproc(grayImage, [windowSize windowSize], myFilterHandle2); [rowsSD, columnsSD, numberOfColorChannelsSD] = size(blockyImageSD); subplot(2, 2, 4); imshow(blockyImageSD, []); caption = sprintf('Image Processed in %d by %d Blocks\n%d by %d pixels\nAnonymous Standard Deviation Filter', ... windowSize, windowSize, rowsSD, columnsSD); title(caption, 'FontSize', fontSize);

% Note: the image size of blockyImageSD is 256x256, NOT smaller. % That's because we're returning an 8x8 array instead of a scalar.

uiwait(msgbox('Done with demo')); catch ME errorMessage = sprintf('Error in blockproc_demo():\n\nError Message:\n%s', ME.message); uiwait(warndlg(errorMessage)); end return;

% Takes one 3x3 block of image data and multiplies it % element by element by the kernel and % returns a single value. function singleValue = myFilter(blockStruct) try % Assign default value. % Will be used near sides of image (due to boundary effects), % or in the case of errors, etc. singleValue = 0;

% Create a 2D filter. kernel = [0 0.2 0; 0.2 0.2 0.2; 0 0.2 0]; % kernel = ones(blockStruct.blockSize); % Box filter.

% Make sure filter size matches image block size. if any(blockStruct.blockSize ~= size(kernel)) % If any of the dimensions don't match. % You'll get here near the edges, % if the image is not a multiple of the block size. % warndlg('block size does not match kernel size'); return; end % Size matches if we get here, so we're okay.

% Extract our block out of the structure. array3x3 = blockStruct.data;

% Do the filtering. Multiply by kernel and sum. singleValue = sum(sum(double(array3x3) .* kernel));

catch ME % Some kind of problem... errorMessage = sprintf('Error in myFilter():\n\nError Message:\n%s', ME.message); % uiwait(warndlg(errorMessage)); fprintf(1, '%s\n', errorMessage); end return;

## 3 Comments

Direct link to this comment:http://mathworks.com/matlabcentral/answers/76272#comment_149678

What is "fi" and what is the sum over?

Direct link to this comment:http://mathworks.com/matlabcentral/answers/76272#comment_149696

fi,j is f subscript i,j . But that doesn't tell us what f is or what the sum is over.

Direct link to this comment:http://mathworks.com/matlabcentral/answers/76272#comment_149709

Sorry.Actually we denote the luminance value in the image f of size MxN or LxL, at the particular location (i,j) with f(i,j), i and j being the row and column index, respectively.

Calculation sum is for every pixel f(i,j) in the odd-numbered rows of the image two variables :A (i,j) and B (i,j)

In my case i denote f(i,j)=I2(i,j)

I have done the code.But get error.

and i got this error

>>Undefined function or method 'minus' for input arguments of type 'cell'.

Actually A (i,j) = [Ʃ(f(i,j) - f(i,j+1) ) - Ʃ(f(i,j-1) - f(i,j) )] and B (i,j) = [Ʃ(f(i,j) - f(i,j+1) ) - Ʃ(f(i,j+1) - f(i,j+2) )] .

where i€ (i, i+2, i+4) and i is odd-numbered row.