fitting an ellipse to my binary iris image ?

1 回表示 (過去 30 日間)
nedaa
nedaa 2012 年 11 月 12 日
コメント済み: Image Analyst 2016 年 10 月 31 日
I want to fit an ellipse to the boundary of the iris in my edge map eye image in order to segment the iris from the background.i really appretiate your help. the image is :
thanks inadvance.
na

採用された回答

Matt J
Matt J 2012 年 11 月 12 日
function report=ellipsefit(XY)
%ELLIPSEFIT - form 2D ellipse fit to given x,y data
%
% report=ellipsefit(XY)
%
%in:
%
% XY: Input matrix of 2D coordinates to be fit. Each column XY(:,i) is [xi;yi]
%
%out: Finds the ellipse fitting the input data parametrized both as
% A*x^2+B*x*y C*y^2+D*x+E*y=1 and [x-x0,y-y0]*Q*[x-x0;y-y0]=1
%
% report: a structure output with the following fields
%
% report.Q: the matrix Q
% report.d: the vector [x0,y0]
% report.ABCDE: the vector [A,B,C,D,E]
% report.AxesDiams: The minor and major ellipse diameters
% report.theta: The counter-clockwise rotation of the ellipse.
%
%NOTE: The code will give errors if the data fit traces out a non-elliptic or
% degenerate conic section.
%
%See also ellipsoidfit
X=XY(1,:).';
Y=XY(2,:).';
M= [X.^2, X.*Y, Y.^2, X, Y, -ones(size(X,1),1)];
[U,S,V]=svd(M,0);
ABCDEF=V(:,end);
if size(ABCDEF,2)>1
error 'Data cannot be fit with unique ellipse'
else
ABCDEF=num2cell(ABCDEF);
end
[A,B,C,D,E,F]=deal(ABCDEF{:});
Q=[A, B/2;B/2 C];
x0=-Q\[D;E]/2;
dd=F+x0'*Q*x0;
Q=Q/dd;
[R,eigen]=eig(Q);
eigen=eigen([1,4]);
if ~all(eigen>=0), error 'Fit produced a non-elliptic conic section'; end
idx=eigen>0;
eigen(idx)=1./eigen(idx);
AxesDiams = 2*sqrt(eigen);
theta=atand(tand(-atan2(R(1),R(2))*180/pi));
report.Q=Q;
report.d=x0(:).';
report.ABCDE=[A, B, C, D, E]/F;
report.AxesDiams=sort(AxesDiams(:)).';
report.theta=theta;
  13 件のコメント
Matt J
Matt J 2013 年 3 月 28 日
Matt J, how to read the image i want using your code
You were probably really talking to ImageAnalyst.
Image Analyst
Image Analyst 2016 年 10 月 31 日
Except that I didn't post any code.

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeSpline Postprocessing についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by