Selecting data within a particular shape from a matrix.
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Hi,
I want to select data around point [169,547] in the attached v variable. Attached is the figure of the shape in which I want the coordinate values, their distance and angle with respect to the center point [169,547]. The case represents a dynamic crack which moves forward in the next frame and thus the center position will change in the next frame and the whole process will be repeated again. Any guidance regarding the implementation would be great.
Cheers
Waqas![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/236503/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/236503/image.png)
4 个评论
Image Analyst
2019-8-31
His coordinate is probably (x, y) not (row, column), so the matrix should exist at row 169, column 547 in a matrix of 304 rows and 640 columns.
Matt J
2020-7-22
waqas commented:
Hi,
I started a new question which is on the same lines as this question. It would be interesting to hear from you guys about that! I tried to figure it out by my own but am unable to get exactly what I want to.
采纳的回答
Matt J
2019-8-31
编辑:Matt J
2019-8-31
As an example,
xc=547; yc=169; %center coordinates
rint=10; rext=50; %internal and external radii
[X,Y]=meshgrid((1:size(v,2))-xc, (1:size(v,1))-yc );
crack= Y<=rint & Y>=-rint & X<=sqrt(rint^2-Y.^2);
region= X.^2+Y.^2<=rext^2 & ~crack;
selected_data = v(region);
imshow(region);
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/236530/image.png)
10 个评论
Matt J
2019-10-7
编辑:Matt J
2019-10-7
If you know the inclineAngle, it should be a 2-line modification,
[X,Y]=meshgrid((1:size(v,2))-xc, (1:size(v,1))-yc );
[Theta,Rho]=cart2pol(X,Y);
[X,Y]=pol2cart(Theta+inclineAngle, Rho);
crack= Y<=rint & Y>=-rint & X<=sqrt(rint^2-Y.^2);
region= X.^2+Y.^2<=rext^2 & ~crack;
selected_data = v(region);
imshow(region);
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