Remove inner part of the Image
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Hi All
I have multiple images and I want to remove the inner part the image but don't know how I can remove this.by using mask or any other approach ? Can anybody help me with this
these are some image . and I want a unique approach that work for all the images,some of the images are attached
1st one in input and second one is like output
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Réponses (3)
Oindri
le 4 Juil 2018
Hi, You may try, first detecting the edge of the inner circle because its abrupt change in intensity, which will be easily detectable. Once you have defined that boundary, replace all internal pixels with black pixels. Given the varied nature of boundary, apply automatic detection of boundary.
KSSV
le 4 Juil 2018
Modifié(e) : KSSV
le 4 Juil 2018
files = dir('*JPG') ;
nx = 512 ; ny = 512 ;
N = length(files) ;
I = cell(N,1) ;
for i = 1:N
I{i} = imread(files(i).name);
I{i} = imresize(I{i},[nx ny]) ;
end
C = round([nx ny]/2) ; % center of the image
R = 60. ; % radius of circle
th = linspace(0,2*pi,500) ;
xc = C(1)+R*cos(th) ;
yc = C(1)+R*sin(th) ;
%
I2 = cell(N,1) ;
x = 1:ny ; y = 1:nx ;
[X,Y] = meshgrid(x,y) ;
idx = inpolygon(X(:),Y(:),xc,yc) ;
for i = 1:N
T = I{i}(:,:,1) ;
T(idx) = 10 ;
I2{i} = repmat(T,1,1,3) ;
imshow(I2{i})
end
3 commentaires
Image Analyst
le 4 Juil 2018
The attached algorithm seem to work with both your images. The function findcircles() didn't seem to work, so I thresholded followed by repeated erosions until the background and inner zone separated into two regions, then took the inner region and expanded it back out and used it as a mask to erase the inner zone of the gray scale image. You can adjust stepSize to erase a little closer to the inner boundary if you want.
Results for the first image:
And for the other image:
6 commentaires
Image Analyst
le 6 Juil 2018
You can use imfreehand() to lasoo the occasional photo that it didn't work for. See my attached demo. Otherwise it will take you quite a bit of work to make it more automatic. It will be tricky and could take a bit of time to develop the algorithm. You could consider hiring someone if you can't do it yourself. Or else contact the Mathworks and they'll do it for you.
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