Connecting points on opposite end of mask

3 次查看（过去 30 天）
kunal
kunal 2025-6-19
评论： Image Analyst 2025-6-24
Given an mask image i have certain points as seen in the image, I want to connect the point to the point at the opposite side of it given the list of points/array of point.Well the output would look something like.
  9 个评论
显示 7更早的评论
kunal
kunal 2025-6-23
The red points are the corner points where there is a drastic change in the angles and i am later using using connected point to divide this mask area in different 4 point segments that will be similar to rectangle.
Image Analyst
Image Analyst 2025-6-24
I don't think this addressed any of the numbered questions. I don't see how that polygon would be divided into 4 rectangles. Anyway, if you go to the corners, the tips of the thick lines will be pointy arrowheads, not rectangles. But let's say you want to split that thick white line/polygon into 4 or more smaller rectangles. Why? And then what are you going to do with those rectangular masks?
I'm just trying to help but want to make sure you don't go on a wild goose chase for some solution you think will work but might not work.
https://en.wikipedia.org/wiki/XY_problem

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Matt J
Matt J 2025-6-20
编辑：Matt J 2025-6-23
Ran in:
load('coping_data (1).mat')
img=logical(img);
n=height(points);
f=@(z,n) circshift(z,-n);
mod1=@(z) mod(z-1,n) + 1;
x=points(:,1)'; dx=diff([x,x(1)]);
y=points(:,2)'; dy=diff([y,y(1)]);
i=find(dy<-5 & f(dx,1)<-10 & f(dy,2)>+5 & abs(f(dx,1))<50 ,1);
I=mod1( i+2:i+n/2+1 );
J=flip( mod1( I(end)+(1:n/2) ) );
X=[x(I);x(J)]; Y=[y(I);y(J)];
imshow(img,[])
line(X,Y,'Color','r','LineWidth',3)

更多回答（1 个）

Image Analyst
Image Analyst 2025-6-20
If you can get the inner boundary and outer boundary as separate arrays, then it would be easy to find the boundary point closest to the red points by using pdist2. If you can't figure out pdist2 then I can do it if you give me the inside boundary and outside boundary.

类别

在 Help Center 和 File Exchange 中查找有关 Region and Image Properties 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