How I can repeatedly divide a binary image ?

2 visualizaciones (últimos 30 días)
Zara Khan
Zara Khan el 13 de Abr. de 2019
Comentada: Zara Khan el 23 de Abr. de 2019
I want to divide a binary image into 4 equal parts first. Then each parts again to 4 parts and so on. I want to repeat this work upto certain no of times . How to do that?

Respuestas (3)

darova
darova el 13 de Abr. de 2019
Use mat2cell()
clc,clear
A = imread('fig3.png');
I = im2bw(A);
k = 128; % divide matrix into 128 parts
[m,n] = size(I);
rows_rest = mod(m,k);
cols_rest = mod(n,k);
nrows = (m-rows_rest)/k *ones(1,k);
ncols = (n-cols_rest)/k *ones(1,k);
nrows(1:rows_rest) = nrows(1:rows_rest) + 1;
ncols(1:cols_rest) = ncols(1:cols_rest) + 1;
C = mat2cell(I,nrows,ncols);
  7 comentarios
Zara Khan
Zara Khan el 14 de Abr. de 2019
Image Analyst: what will be the easiest way ??
darova
darova el 14 de Abr. de 2019
The best way is to pay someone

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Image Analyst
Image Analyst el 14 de Abr. de 2019
Your image is not a power of 2.
Try this code:
clc; % Clear the command window.
clear all;
close all; % Close all figures (except those of imtool.)
workspace; % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 16;
%===============================================================================
% Get the name of the image the user wants to use.
baseFileName = 'fig3.png';
folder = pwd;
fullFileName = fullfile(folder, baseFileName);
% Check if file exists.
if ~exist(fullFileName, 'file')
% The file doesn't exist -- didn't find it there in that folder.
% Check the entire search path (other folders) for the file by stripping off the folder.
fullFileNameOnSearchPath = baseFileName; % No path this time.
if ~exist(fullFileNameOnSearchPath, 'file')
% Still didn't find it. Alert user.
errorMessage = sprintf('Error: %s does not exist in the search path folders.', fullFileName);
uiwait(warndlg(errorMessage));
return;
end
end
%=======================================================================================
% Read in demo image.
grayImage = imread(fullFileName);
% Get the dimensions of the image.
[rows, columns, numberOfColorChannels] = size(grayImage)
if numberOfColorChannels > 1
grayImage = grayImage(:, :, 3);
end
% Make sure it's square
if rows ~= columns
grayImage = imresize(grayImage, [rows, rows]);
% Get the dimensions of the image.
[rows, columns, numberOfColorChannels] = size(grayImage)
end
% Display image.
subplot(2, 2, 1);
imshow(grayImage, []);
impixelinfo;
axis('on', 'image');
caption = sprintf('Original image: %d rows by %d columns', rows, columns);
title(caption, 'FontSize', fontSize, 'Interpreter', 'None');
hp = impixelinfo(); % Set up status line to see values when you mouse over the image.
% Set up figure properties:
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0.05 1 0.95]);
% Get rid of tool bar and pulldown menus that are along top of figure.
% set(gcf, 'Toolbar', 'none', 'Menu', 'none');
% Give a name to the title bar.
set(gcf, 'Name', 'Demo by ImageAnalyst', 'NumberTitle', 'Off')
drawnow;
% 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.
nnzFunction = @(theBlockStructure) nnz(theBlockStructure.data(:));
% Now do a loop to divide by half each time and count the number of non-zeros.
counter = 1;
newRows = rows;
newCols = columns;
while newRows >= 1
% Round sizes.
newRows = floor(newRows);
newCols = floor(newCols);
% Resize the image.
grayImage = imresize(grayImage, [newRows, newCols], 'nearest');
% Get the dimensions of the image.
[rows, columns, numberOfColorChannels] = size(grayImage)
% Show the image
subplot(2, 2, 2);
imshow(grayImage);
axis('on', 'image');
caption = sprintf('%d rows by %d columns', newRows, newCols);
title(caption, 'FontSize', fontSize);
% Count the number of non-zero pixels.
whitePixelCount(counter) = nnz(grayImage);
% Display counts so far.
subplot(2, 2, 3:4);
plot(whitePixelCount, 'b.-', 'MarkerSize', 30);
grid on;
title('White Pixel Count', 'FontSize', fontSize);
xlabel('Image number', 'FontSize', fontSize);
ylabel('Count (Number of White Pixels)', 'FontSize', fontSize);
drawnow;
% Compute half the number of rows and columns.
% May be a fractional number if the rows and columns are not a power of 2.
newRows = (rows / 2);
newCols = (columns / 2);
% Show user the results.
promptMessage = sprintf('%d white pixels in this image.\nDo you want to Continue processing,\nor Quit processing?', whitePixelCount(counter));
titleBarCaption = 'Continue?';
buttonText = questdlg(promptMessage, titleBarCaption, 'Continue', 'Quit', 'Continue');
if contains(buttonText, 'Quit')
return;
end
% Increment index for whitePixelCount array.
counter = counter + 1;
end
% Show in command window.
whitePixelCount
0001 Screenshot.png
  4 comentarios
Image Analyst
Image Analyst el 18 de Abr. de 2019
Start with a small image of 16 by 16 pixels and explain step by step what you want as the output at each step.
Zara Khan
Zara Khan el 19 de Abr. de 2019
I want to set a label like :
Label 1: 16X16 will be divided into 4 equal quadrants ,now will store number of white pixels from 4 quadrants
Label 2: Each 8X8 will be divided into 4 equal quadrants again , now will store number of white pixels from 4(previous) + 32(now)
This will continue as many labels we want.
note: I am taking the complement of the 'fig3.png'

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Walter Roberson
Walter Roberson el 19 de Abr. de 2019
  7 comentarios
Zara Khan
Zara Khan el 20 de Abr. de 2019
can I store white pixels counts from each blocks gradully in an array format?
Like said
step 1: 16X16 will be divided into 4 equal quadrants ,now will store number of white pixels from 4 quadrants
step 2: Each 8X8 will be divided into 4 equal quadrants again , now will store number of white pixels from 4(previous) + 32(now)
For one image it will be 1X32
for 10 image ---10X32 in this format.
white pixels count from 32 blocks .
Zara Khan
Zara Khan el 23 de Abr. de 2019
Image Analyst : No

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