How to crop image with nonlinear cropping shape?
3 ビュー (過去 30 日間)
古いコメントを表示
Dear all,
I have following image:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/147156/image.jpeg)
And would like to crop it as follows:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/147157/image.jpeg)
Here is the code I'm using right now is:
clc;close all;clear all;
corn=imread('After_phi_o-phi-f.jpg');
mm_corn=imerode(corn,strel('disk',1));
bw_corn=im2bw(mm_corn, graythresh(mm_corn));
cc_corn = bwconncomp(bw_corn);
aba_corn = [cellfun(@numel,cc_corn.PixelIdxList)];
[mv_corn,ind] = sort(aba_corn,'descend');
L_corn=labelmatrix(cc_corn);
ki_corn = find(aba_corn >= mv_corn(2));
mbi_corn = ismember(L_corn, ki_corn);
bw_corn(~mbi_corn) = 0;
Ibw = imfill(bw_corn,'holes');
Ilabel = bwlabel(Ibw);
stat = regionprops(Ilabel,'centroid');
imshow(bwconvhull(im2bw(corn, graythresh(corn)))); hold on;
plot([stat(1).Centroid(1),stat(2).Centroid(1)], [stat(1).Centroid(2),stat(2).Centroid(2)], 'r');1)
And result is:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/147158/image.jpeg)
So I want cut information from the left until the red line, how I can do it?
Thanks for any help.
4 件のコメント
採用された回答
Image Analyst
2015 年 2 月 16 日
Get the horizontal and vertical profiles using sum(), and get the top and bottom row and left and right column using find()
horizontalProfile = sum(binaryImage, 1);
verticalProfile = sum(binaryImage, 2);
topRow = find(verticalProfile, 1, 'first');
bottomRow = find(verticalProfile, 1, 'last');
leftColumn= find(horizontalProfile, 1, 'first');
rightColumn = find(horizontalProfile, 1, 'last');
croppedImage = binaryImage(topRow:bottomRow, leftColumn:rightColumn);
2 件のコメント
Image Analyst
2015 年 2 月 16 日
Is the line always red? And does it never extend to the boundaries of the blob? If so I'd suggest you extract the red line, then find the endpoints of it with bwmorph(), then draw a black line in the image using imline(), demo attached. Then extract the biggest blob, demo attached.
その他の回答 (0 件)
参考
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!