how to segment (divide) an image into 4 equal halves?
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Am working on medical imaging. i have to segment an image into 4 equal parts like 4 quadrants and each segmented image should get displayed separately. can anyone help me with the code?
9 个评论
Sivakumaran Chandrasekaran
2012-8-25
编辑：Walter Roberson
2013-5-1
if size(I)=256 * 256, follow this..
I1=I(1:127,1:127);figure,imshow(I1);
I2=I(1:127,128:256);figure,imshow(I2);
I3=I(127:256,1:127);figure,imshow(I3);
I4=I(127:256,127:256);figure,imshow(I4);
Image Analyst
2012-8-25
That is not the definition of image segmentation (it's cropping), so I removed that tag. Image segmentation is when you try to identify individual regions, usually of some irregular shape that represents some object (such as a person or a care). For example, find the lungs in a CT image.
Nandan M S
2016-4-22
编辑：Nandan M S
2016-4-22
Hey, try this ...just copy, paste & run..
<<
/matlabcentral/answers/uploaded_files/50473/d.jpg>>
clc;
clear all;
close all;
I=imread('d.jpg');
subplot 334
imshow(I);
[r c p]= size(I);%r-rows,c-columns,p-planes
A=I(1:r/2,1:c/2,:);
B=I(1:r/2,c/2+1:c,:);
C=I(r/2+1:r,1:c/2,:);
D=I(r/2+1:r,c/2+1:c,:);
subplot 332
imshow(A);
title('Image part 1');
subplot 333
imshow(B);
title('Image part 2');
subplot 335
imshow(C);
title('Image part 3');
subplot 336
imshow(D);
title('Image part 4');
Asma Askri
2016-8-21
thanks a lot :) :) , the code is so useful
kautsar rusydi
2017-1-14
hey Nanda, may i ask about your code? what meaning of 'p' variable, tell me more about p/plane please? Nandan's code
Image Analyst
2017-1-14
kautsar, p is the number of planes in the 3D array. For example in a color image it's the number of color channels, like p=3 for an RGB image. Or for a volumetric image, like for CT or MRI, p is the number of slices through the body or subject. It's always good to give at least 3 outputs of size because sometimes people might think an image is gray scale (because it looks like it) but they saved it to a BMP or JPG format file and it got converted into color. If you only give two numbers, like
[rows, columns] = size(imageArray);
then columns will be the number of color channels times the actual number of columns, which would be 3 times the number of columns if imageArray is an RGB color image. See Steve's blog for more information.
SATISH KUMAR
2017-3-18
Hi, Nandan, your code is working so nicely, i need one small help can you please merge those divided images into original size. thanks in advance
Image Analyst
2017-3-18
SATISH, to reassemble, do this:
fullImage = [A, C; B, D];
mehrdad bahadori
2023-2-15
you can use the function that I have posted on matlab exchange. It divides the image into MxN equal sized images, then you can display them in your desired way.
采纳的回答
Dishant Arora
2012-8-25
I1=I(1:size(I,1)/2,1:size(I,2)/2,:);
I2=I(size(I,1)/2+1:size(I,1),1:size(I,2)/2,:);
I3=I(1:size(I,1)/2,size(I,2)/2+1:size(I,2),:);
I4=I(size(I,1)/2+1:size(I,1),size(I,2)/2+1:size(I,2),:);
try this out, it will do
2 个评论
Dishant Arora
2012-8-25
I am assuming the image u want to segment is color image(3-D)
Sivakumaran Chandrasekaran
2012-8-25
Hi Dishant, can u explain the line...The input image is divided into many sub images . all are 2D images.Does the line indicates the same which i typed above
更多回答（5 个）
Image Analyst
2021-3-7
编辑：Image Analyst
2021-3-7
Since everyone seems to want a different number, I've created this general purpose demo where you can specify how many strips vertically and horizontally you want the image to be divided into.
clc; % Clear the command window.
close all; % Close all figures (except those of imtool.)
clear; % Erase all existing variables. Or clearvars if you want.
workspace; % Make sure the workspace panel is showing.
format long g;
format compact;
fprintf('Beginning to run %s.m ...\n', mfilename);
% Read in image
grayImage = imread('pout.tif');
[rows, columns, numColorChannels] = size(grayImage)
imshow(grayImage);
axis on;
impixelinfo
numBandsVertically = 4;
numBandsHorizontally = 3;
topRows = round(linspace(1, rows+1, numBandsVertically + 1))
leftColumns = round(linspace(1, columns+1, numBandsHorizontally + 1))
% Draw lines over image
for k = 1 : length(topRows)
yline(topRows(k), 'Color', 'y', 'LineWidth', 2);
end
for k = 1 : length(leftColumns)
xline(leftColumns(k), 'Color', 'y', 'LineWidth', 2);
end
% Extract into subimages and display on a new figure.
hFig2 = figure();
plotCounter = 1;
for row = 1 : length(topRows) - 1
row1 = topRows(row);
row2 = topRows(row + 1) - 1;
for col = 1 : length(leftColumns) - 1
col1 = leftColumns(col);
col2 = leftColumns(col + 1) - 1;
subplot(numBandsVertically, numBandsHorizontally, plotCounter);
subImage = grayImage(row1 : row2, col1 : col2, :);
imshow(subImage);
caption = sprintf('Rows %d-%d, Columns %d-%d', row1, row2, col1, col2);
title(caption);
drawnow;
plotCounter = plotCounter + 1;
end
end
hFig2.WindowState = 'Maximized';
fprintf('Done running %s.m.\n', mfilename);
3 个评论
Quoc-Bao Ta
2021-5-16
编辑：Quoc-Bao Ta
2021-5-16
Hi Image Analyst,
After implementing the algorithm for some purposes, then i want to know how to recombine all those to a original image of actual size.
Please help me with code.
Thank you so much!
Walter Roberson
2021-5-16
This demonstration does not store the subImage(), so they cannot be recombined later.
Walter Roberson
2021-5-16
Consider using mat2cell() to split the array apart, and using cell2mat() to put it back together.
For the case where the image is to be divided into a fixed number of blocks as evenly as practical, but the blocks not all being exactly the same size (e.g, you cannot divide 512 pixels into 3 equal partitions):
Nxblk = 3; Nyblk = 5;
yblksizes = diff(round(linspace(1, size(TheImage,1)+1, Nyblk+1)));
xblksizes = diff(round(linspace(1, size(TheImage,2)+1, Nxblk+1)));
ImageCell = mat2cell(TheImage, yblksizes, xblksizes, size(TheImage,3));
ImageReconstructed = cell2mat(ImageCell);
Image Analyst
2012-8-25
You can simply call imcrop 4 times to get the 4 images. Here, try this full demo. Just copy, paste, and run:
clc; % Clear the command window.
workspace; % Make sure the workspace panel is showing.
format longg;
format compact;
fontSize = 20;
% Read in a standard MATLAB color demo image.
folder = fullfile(matlabroot, '\toolbox\images\imdemos');
baseFileName = 'peppers.png';
% Get the full filename, with path prepended.
fullFileName = fullfile(folder, baseFileName);
if ~exist(fullFileName, 'file')
% Didn't find it there. Check the search path for it.
fullFileName = baseFileName; % No path this time.
if ~exist(fullFileName, 'file')
% Still didn't find it. Alert user.
errorMessage = sprintf('Error: %s does not exist.', fullFileName);
uiwait(warndlg(errorMessage));
return;
end
end
rgbImage = imread(fullFileName);
% Get the dimensions of the image. numberOfColorBands should be = 3.
[rows columns numberOfColorBands] = size(rgbImage);
% Display the original color image.
subplot(2, 3, 1);
imshow(rgbImage, []);
title('Original Color Image', 'FontSize', fontSize);
% Enlarge figure to full screen.
set(gcf, 'units','normalized','outerposition',[0 0 1 1]);
% Get the rows and columns to split at,
% Taking care to handle odd-size dimensions:
col1 = 1;
col2 = floor(columns/2);
col3 = col2 + 1;
row1 = 1;
row2 = floor(rows/2);
row3 = row2 + 1;
% Now crop
upperLeft = imcrop(rgbImage, [col1 row1 col2 row2]);
upperRight = imcrop(rgbImage, [col3 row1 columns - col2 row2]);
lowerLeft = imcrop(rgbImage, [col1 row3 col2 row2]);
lowerRight = imcrop(rgbImage, [col3 row3 columns - col2 rows - row2]);
% Display the images.
subplot(2, 3, 2);
imshow(upperLeft);
subplot(2, 3, 3);
imshow(upperRight);
subplot(2, 3, 5);
imshow(lowerLeft);
subplot(2, 3, 6);
imshow(lowerRight);
5 个评论
safa
2013-5-1
how to divide an image into 9 (3X3) equal halves too?((2X2) or (3X3)or (5X3))
Walter Roberson
2013-5-1
Is the size of the image promised to be an exact multiple of 3 in both x and y directions?
Image Analyst
2013-5-1
You wouldn't call them "halves", which by definition is 50% of the pixels. But anyway, to divide them into other rectangles is the same concept. Did you see how I figured out the starting and ending row and starting and ending column and then used imcrop()? Do you think you could do that? That you could figure out the rows and columns by yourself if you wanted to divide the image by 3 or 5? Just make sure you use floor(), ceil(), round(), or int32() to get rid of any fractional part.
NAGARAJA JUJARE
2019-3-29
how do i save the individual croped image separately?
Image Analyst
2019-3-29
imshow(upperLeft, 'upperLeft.png');
imshow(upperRight, 'upperRight.png');
imshow(lowerLeft, 'lowerLeft.png');
imshow(lowerRight, 'lowerRight.png');
Biza Ferreira
2013-5-1
编辑：Walter Roberson
2013-5-1
I=imread('images/fig1.tif');
[r c]= size(I);
A=I(1:r/2,1:c/2);
B=I(1:r/2,c/2+1:c);
C=I(r/2+1:r,1:c/2);
D=I(r/2+1:r,c/2+1:c);
L=([B C;D A]);
figure, imshow(L), title('Image changed');
7 个评论
Ahmet Ipek
2016-8-8
Can we again combine these pieces?
Walter Roberson
2016-8-25
The line
L=([B C;D A]);
is recombining the pieces, in a different order.
shafaa abdullah
2017-2-7
编辑：shafaa abdullah
2017-2-7
thank you but why the orginal image converted to grayscale i want C piece remain colored piece please help me i have an exam
Walter Roberson
2017-2-7
I = imread('images/fig1.tif');
[r c p]= size(I);
A = I(1:r/2, 1:c/2, :);
B = I(1:r/2, c/2+1:c, :);
C = I(r/2+1:r, 1:c/2, :);
D = I(r/2+1:r, c/2+1:c, :);
L = [B C;D A];
figure
imshow(L)
title('Image changed');
Mustafa Ahmed
2021-3-6
i need to divide the image into 8 sections
Image Analyst
2021-3-6
Just make the obvious modifications. It's not that hard is it, even for a beginner?
Mustafa Ahmed
2021-3-7
tried , only shows the four parts
clc;
clearvars;
close all;
a=imread('peppers.png');
[x, y, z]=size(a);
x1=x/2;
y1=y/2;
A=a(1:x1,y1:end,:);
subplot(321)
imshow(a);
title('Original Image')
subplot(322)
imshow(A);
title('Cropped Img of 1st quad')
B=a(1:x1,1:y1,:);
subplot(323)
imshow(B);
title('Cropped Img of 2nd quad')
C=a(x1:end,1:y1,:);
subplot(324)
imshow(C);
title('Cropped Img of 3rd quad')
D=a(x1:end,y1:end,:);
subplot(325)
imshow(D);
title('Cropped Img of 4th quad')
L=([B A;C D]);
subplot(326)
imshow(L)
title('Re-Construct Orig Image')
DGM
2022-12-17
If you have MIMT, this becomes incredibly simple. MIMT imdetile() requires no special consideration of geometry divisibility, number of channels, etc.
Note that in this example [384 512] is not integer-divisible by [5 6]. How that's resolved is a matter of the selected options. Here, I'm just using the defaults. The ordering of the tiles (the direction) can be specified. In this case, I'm using row-wise detiling, since that's all that montage() supports.
% read an image
inpict = imread('peppers.png'); % 384x512x3
% split the image into 30 tiles in a 5x6 pattern
tiling = [5 6]; % [y x]
subimages = imdetile(inpict,tiling,'direction','row'); % 77x85x3x30
% show the result, adding padding for display clarity
montage(subimages,'size',tiling,'bordersize',[5 5],'backgroundcolor','w')
The subimages all have the same geometry and are returned as a 4D image instead of an unmanageable pile of named variables. If it's preferred to handle the output as a cell array, that can easily be done by using num2cell() on the output of imdetile().
subimages = imdetile(inpict,tiling,'direction','col'); % detile columnwise for simplification
subimages = squeeze(num2cell(subimages,[1 2 3])); % convert to 30x1 cell
subimages = reshape(subimages,tiling); % reshape into 5x6 cell for easy indexing
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