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
4 个评论
Matt J
2019-8-31
The attached v variable is a matrix of size 304x640, so the coordinate [547,169] does not exist.
waqas
2019-8-31
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
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);
10 个评论
waqas
2019-8-31
It seems to work perfectly in this configuration. I am trying to flip the crack but am getting following shape with the changes in the code.
Moreover, eventually I want to extract the coordinates in the white region and convert them to polar coordinates with respect to initial point. Should I add xc and yc into the X and Y we are getting from the above code?
Matt J
2019-8-31
To flip the crack:
crack= Y<=rint & Y>=-rint & X>=-sqrt(rint^2-Y.^2);
To get polar coordinates:
[rho,theta]=cart2pol(X(region), Y(region));
waqas
2019-8-31
waqas
2019-8-31
Matt J
2019-9-2
You would need to apply a coordinate rotation to the X and Y before generating the mask. This can be easily done by converting X,Y to polar coordinates. Or, you could use my AxelRot utility,
Image Analyst
2019-9-2
If the slotted shape is from an actual image that may vary from one snapshot to the next, rather than some theoretical computer graphics thing, then you'll have to use image analysis to find that slotted shape.
waqas
2019-9-3
The data is infact coming from image analysis using digital image correlation but since the crack is not visible and the data is quite noisy so I am trying to validate the displacements using asymptotic series (william's solution). End goal is to get the crack tip accuratly. Attached is an image to give a better picture.
waqas
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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