Crop an image using coordinate

Hi,
How can I crop an image by having the top left coordinate (X1,Y1) and bottom right (X2,Y2).
I2 = imcrop(im,[x1 y1 x2 y2])
does not seem to work?
Thanks

3 Comments

Hi Hu
this is John BG ( <mailto:jgb2012@sky.com jgb2012@sky.com> ) please check my answer, image cropping without imcrop.
If you find it useful would you please mark my answer as accepted answer?
thanks in advance
John BG
I=imread(images(1).name); %crop one to get rect
this line of code give error of index matrix dimension exeeds. kindly suggest some solution
Nisa, start a new question, and give the code that you used to load up your images structure (was it the dir function?). Your images structure is evidently empty, meaning it found not images.
filePattern = fullfile(pwd, '*.png');
images = dir(filePattern);
if isempty(images)
message = sprintf('No PNG images found in folder %s\n', pwd)
uiwait(errordlg(message))
return;
end
thisImage=imread(images(1).name);
imshow(thisImage)
% Crop one to get rect
croppedImage = imcrop(thisImage);

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 Accepted Answer

You used 151 instead of 115 in your width. Here, try this:
grayImage = imread('circuit.tif');
subplot(1,2,1);
imshow(grayImage);
h = impixelinfo();
axis on;
I2 = imcrop(grayImage,[115, 89, 161-115,142-89]);
subplot(1,2,2);
imshow(I2)
axis on;

2 Comments

Hi how could you get the follwing values>? imcrop(grayImage,[115, 89, 161-115,142-89]);
I don't know. It looks like the original poster changed/edited his post. It no longer mentions 151 and he removed the image.

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More Answers (3)

John BG
John BG on 4 Mar 2017
Edited: John BG on 4 Mar 2017
Hi Hu
A=imread('im1.jpg');imshow(A)
p=ginput(2)
p1max=max(p(:,1));p2max=max(p(:,2));p1min=min(p(:,1));p2min=min(p(:,2));
A(:,[1:p1min],:)=[];
A([1:p2min],:,:)=[];
A(:,[uint64(p1max-p1min+1):end],:)=[];
A([uint64(p2max-p2min+1):end],:,:)=[];
imshow(A)
EXPLANATION
1.
as example let's start with this image
A=imread('im1.jpg');imshow(A)
2. Selecting 2 diagonal points that define the borders you want to remove. I use command ginput but you already have the values
p=ginput(2)
3. Calculating border values from ginput points
p1max=max(p(:,1));p2max=max(p(:,2));p1min=min(p(:,1));p2min=min(p(:,2));
4. Removing on borders, one side of the frame at a time
A(:,[1:p1min],:)=[];
A([1:p2min],:,:)=[];
A(:,[uint64(p1max-p1min+1):end],:)=[];
A([uint64(p2max-p2min+1):end],:,:)=[];
5.
Check
imshow(A)
Hu
if you find these lines useful would you please mark my answer as Accepted Answer?
To any other reader, if you find this answer of any help please click on the thumbs-up vote link,
thanks in advance for time and attention
John BG
You can try using the following command
I2 = imcrop(im,[x1 y1 x2-x1 y2-y1]);
imcrop uses the following syntax:
I2 = imcrop(I,RECT);
where RECT is a 4-element vector with the form [XMIN YMIN WIDTH HEIGHT];

3 Comments

The I2 = imcrop(im,[x1 y1 x2-x1 y2-y1]) does not work.
How can I "convert" my coordinates to be represented in the RECT?
hu - please describe what you mean by ...does not work. Perhaps provide an example pair of coordinates with what you expect to be the desired outcome.
hu
hu on 2 Oct 2014
Edited: hu on 2 Oct 2014
Hi,
If I wish to crop the image (get the square in the middle) and based on the code sugested, It does not work:
I=imread('circuit.tif');
I2 = imcrop(I,[115, 89, 161-151,142-89]);
imshow(I2)
The coordiantes are (115,89,161,142) (Xtop left,Ytop left, Xbottom right, Ybottom right).
Thanks

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Amruta Talreja
Amruta Talreja on 4 Mar 2017
I2 = imcrop(B1,[12659 11677 11281 12661]); imshow(I2); I wrote this code and i am getting a blank screen. plz help.

4 Comments

What is size(B1)?
Try this:
I2 = imcrop(B1,[12659 11677 11281 12661]);
whos I2;
whos B1;
fprintf('Max=%f, min=%f\n', max(I2(:)), min(I2(:)));
imshow(I2, []);
Hi,
Is it possible to crop the image manually without using a rectangle? For example, can I create a boundary manually with a mouse on the Matlab image and display the cropped image separately?
If you use imcrop() without any arguments, you'll be in interactive drawing mode.

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Asked:

hu
on 2 Oct 2014

Commented:

on 16 May 2022

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