# finding mean for values

7 views (last 30 days)
FIR on 28 Nov 2012
Commented: Image Analyst on 7 Feb 2020
I have values named in variabla Block in the my values are as
val(:,:,1)
8x8 matrix values
val(:,:,2)
;
;
;
val(:,:,3)
8x8 matrix values
for these all i have to find mean,for val(:,:,1) to vali(:,:,3)
Jan on 29 Nov 2012
Please, FIR, explain the input data in valid Matlab syntax. I cannot guess what "val(:,:,1) 8x8 matrix values" and the three semicolons exactly mean. We try to help you, but it would be much easier if you care about clear questions also.

Jan on 28 Nov 2012
Edited: Jan on 28 Nov 2012
help mean
doc mean
It can be solved in one single line. Either a reshape is helful, or remember that the mean value of a set of variables is the mean value of the mean values of subsets, when they all have the same size.
Image Analyst on 28 Nov 2012
He talks about "each block" so I'm guessing that he's processing a much larger array in "chunks" of 8 by 8. So he processes one block, then "jumps" over to the adjacent block and processes that, and so on for each block. Thus I thought blockproc() would be perfect for that. But he doesn't seem to know how to extract a plane from the N dimensional array, like
plane1 = val(:, :, 1);
plane2 = val(:, :, 2);
and so on.

Image Analyst on 28 Nov 2012
Use blockproc(). Here's three grayscale examples that you can adapt to color very easily.
% 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);
% 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);
Image Analyst on 7 Feb 2020
Take each output from processing each color channel individually and concatenate them together:
blockyImageSDR = blockproc(rgbImage(:, 1), blockSize, StDevFilterFunction); % Process red channel.
blockyImageSDG = blockproc(rgbImage(:, 2), blockSize, StDevFilterFunction); % Process green channel.
blockyImageSDB = blockproc(rgbImage(:, 3), blockSize, StDevFilterFunction); % Process blue channel.
rgbSD = cat(3, blockyImageSDR, blockyImageSDG, blockyImageSDB);

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