I have many hundreds fundus images of the human eye.

Fundus image of human eye

The yellow arrow points to two spots which I I want to remove. The final result should be a perfect circle.

The removed area should be black.

I need this for image preprocessing for a machine learning project.

Any help would be great.


2 Answers 2


Find the largest morphological component:

m = SelectComponents[MorphologicalComponents[img, 0.1], "Area", -1];

enter image description here

Get the bounding circle (with radius multiplied by 0.98 to remove the blips)

circ = Flatten[{#[[1]], .98 #[[2]]} & /@ 
    ComponentMeasurements[m, {"Centroid", "BoundingDiskRadius"}][[All,2]], 1];
Show[img, Graphics[{Red, Thick, Circle[circ[[1]], circ[[2]]]}]]

enter image description here

mask = Show[Image[ConstantArray[0, Dimensions[ImageData[img]]]], 
  Graphics[{White, Disk[circ[[1]], circ[[2]]]}]]

enter image description here

newimg = ImageMultiply[img, mask]

enter image description here


I think it may be interesting to preprocess the image before using @Melago's approach.
This permits to get rid of the 0.98 coefficient.

One may say that the following preprocessing introduces a new "magic" coefficient (erosionRadius = 30), but this new coefficient is less critical than the 0.98.

img00 = Import["https://i.sstatic.net/kUebf.jpg"]  

enter image description here

img01 = MorphologicalBinarize[img00]  

enter image description here

erosionRadius = 30;
img02 = Erosion[img01, DiskMatrix[erosionRadius], Padding -> 0]  

enter image description here

m = Dilation[img02, DiskMatrix[erosionRadius]]    

enter image description here

Note that the top and the bottom are slightly flattened. This should not have effect on the extracted "BoundingDiskRadius"

Beyong this point, @Melago's approach is used :

circ = Flatten[{#[[1]], (* 0.98 : Removed *) #[[2]]} & /@ 
    ComponentMeasurements[m, {"Centroid", "BoundingDiskRadius"}][[All,
       2]], 1];
Show[img00, Graphics[{Red,Thick,Circle[circ[[1]], circ[[2]]]}]]  

enter image description here



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