Crop an image using coordinate

60 次查看（过去 30 天）
hu
hu 2014-10-2
评论： Image Analyst 2022-5-16
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 个评论
显示 1更早的评论
nisa sarfraz
nisa sarfraz 2020-1-14
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
Image Analyst
Image Analyst 2020-1-14
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);

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Image Analyst
Image Analyst 2014-10-2
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 个评论
Aparna Sivakumar
Aparna Sivakumar 2022-5-16
Hi how could you get the follwing values>? imcrop(grayImage,[115, 89, 161-115,142-89]);
Image Analyst
Image Analyst 2022-5-16
I don't know. It looks like the original poster changed/edited his post. It no longer mentions 151 and he removed the image.

请先登录，再进行评论。

更多回答（3 个）

John BG
John BG 2017-3-4
编辑：John BG 2017-3-4
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
<mailto:jgb2012@sky.com jgb2012@sky.com>
Nalini Vishnoi
Nalini Vishnoi 2014-10-2
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 个评论
显示 1更早的评论
Geoff Hayes
Geoff Hayes 2014-10-2
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 2014-10-2
编辑：hu 2014-10-2
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

请先登录，再进行评论。

Amruta Talreja
Amruta Talreja 2017-3-4
I2 = imcrop(B1,[12659 11677 11281 12661]); imshow(I2); I wrote this code and i am getting a blank screen. plz help.
  4 个评论
显示 2更早的评论
Warid Islam
Warid Islam 2019-7-12
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?
Image Analyst
Image Analyst 2019-7-13
If you use imcrop() without any arguments, you'll be in interactive drawing mode.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Image Processing and Computer Vision 的更多信息

标签

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by