Why do number of features are different for same person's retinal image

I am doing a project "Personal Identification on the basis of Retina". I have made feature matrix that comprise of distance and angles between a candidate a feature point and its 4 nearest neighbors. In my data set I have two images per person. When I calculate my feature matrix then for both the images of same person the number of extracted features is different. Will this difference effect the results in matching step between test data and training data?? Or there is some way through which I can correct this? Below are two images of same person.

Answer 1

[Image Analyst]

It depends on what you call a feature. For a simple example, let's say that your features were the optic disc diameter, and the number of vessel bifurcations. So that's just two features and it will be two features for all possible images, even though the values of those features may be different.

Or lets say the features were the optic disc diameter, the number of vessel bifurcations, and the (x,y) locations of vessel bifurcations. If you consider the list of locations as a single feature, since they can be stored in an N by 2 array, then you just have 3 features. However if you consider the list to be N features, then you have N+2 features total and that will vary from image to image because the number of bifurcations, N, will vary from image to image.

[Sidra Aleem]

@Image Analyst my feature points are Bifurcation points, end points and crossing number. And in feature vector I am storing the angle and distance between a feature point and its 4 nearest neighbours

[Image Analyst]

OK. Is that what a published paper did?

http://www.visionbib.com/bibliography/contentsmedical.html#Medical%20Applications,%20CAT,%20MRI,%20Ultrasound,%20Heart%20Models,%20Brain%20Models

[Sidra Aleem]

Yes i am doing all this on the basis of published papers. But that paper is not in this link. Also i contacted the doctor who provided me with data set. She said one image is of roght eye and other is of left eye.

[Image Analyst]

Again, the number of features could be different or the same depending on how you're defining features. And if they're different, it's not surprising since the eyes are different physical eyes. Heck, even if it were the same eye it's possible the features could be different from one snapshot to the next. So I'm not sure anything here really needs "fixing."

[Sidra Aleem]

@ Image Analyst I am calculating feature points through crossing number technique. In which a 3 x 3 window is formed. After that difference between adjacent pixels pairs that are surrounding the central pixel of window is taken. Depending on the final value ( 3 for bifurcation, 4 for crossing number and 1 for end point) pixel is differentiated in to a specific feature out of the three. Below is the link for this technique.

https://www.mathworks.com/matlabcentral/answers/331737-how-to-implement-crossing-number-technique-for-detection-of-retinal-bifurcation-point

Below are some of the published papers that I followed.

http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=6965041 http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=6473297 http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=7853731

[Image Analyst]

You can get crossings with bwmorph(binaryImage, 'branchpoints');

[Sidra Aleem]

@Image Analyst I have Implemented CN technique on my own. Because bwmorph(binaryImage, 'branchpoints'); gives 1 to only crossing number points. Whereas my feature points are "end points, bifurcation and crossing")

I am also able to get end points with bwmorph(binaryImage, 'endpoints');

However I need bifurcation points as well. Below is the example of bifurcation point.

[1  0  1
0   1 0
0  1  0] 
it should become 

[0 0 0 
 0 1 0
 0  0 0]
is there anyway to get this by bwmorph as well?