Stenosis Detection of Arteries
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I have data set of CTA of heart in 2D images. I segmented the coronary arteries by FCM clustering algorithm. For detecting stenos, i implement center line algorithm and get the skeleton of arteries.Now i want to get the width of arteries. Can anyone help me in this regard that how i find width or discontinuity of arteries in MATLAB.The IMAGE OF SEGMENTED ATERIES WITH CENTRELINE IS ATTACHED HEREBY

2 comentarios
Fareena
el 27 de Mzo. de 2017
can you please share your code for fcm clustering.
Asadullah Mumtaz
el 3 de Mayo de 2017
can you please share the code this output image
Respuestas (2)
Jeff E
el 30 de Abr. de 2015
1 voto
Not sure what you mean by discontinuity, but on way to obtain an average width would be the sum of the artery mask divided by the total length of the midline skeleton.
5 comentarios
Image Analyst
el 30 de Abr. de 2015
I second that. That should probably be good enough, particularly since the result might be a fair amount different had you just taken the radiograph from a viewpoint rotated a few degrees or at a different time in the heartbeat cycle.
beenish Aziz
el 30 de Abr. de 2015
Editada: beenish Aziz
el 1 de Mayo de 2015
Image Analyst
el 1 de Mayo de 2015
Brings back memories. Many years ago I got my doctorate in tracking vessels in angiograms and finding stenoses. Of course it's not easy and not just some simple chunk of a hundred lines of code that I can give you. People (like you) are still working on the problem today so that tells you it's not easy. We can't really do complicated algorithm development here because you can find such image processing algorithms here: VisionBib. Go there and see how other people have done it. But if you have smaller, more specific MATLAB questions, we can try to answer those.
Since this project will require that you get good at programming, please see this link: http://www.mathworks.com/matlabcentral/answers/8026-best-way-s-to-master-matlab
beenish Aziz
el 1 de Mayo de 2015
beenish Aziz
el 8 de Mayo de 2015
Harsha Jawale
el 21 de Dic. de 2015
0 votos
please share your code
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