# Summing anonymous functions in a for loop: adding penalty terms for optimization

조회 수: 5(최근 30일)
Evan Hemingway 2022년 6월 28일
댓글: Evan Hemingway 2022년 6월 28일
I have the following piece of code for passing into lsqnonlin:
baseFun = @(p) ((x-p(1))*cos(p(3))+(y-p(2))*sin(p(3))).^2/a^2+...
((x-p(1))*sin(p(3))+(y-p(2))*cos(p(3))).^2/b^2-1;
totalFun = baseFun;
% Penalty stiffness
lam = 1e5;
j = 1;
for k = whichVis
totalFun = @(p) totalFun(p) + lam*(x(j) - p(1) - xeye(1,k)*cos(p(3)) + xeye(2,k)*sin(p(3)))^2+lam*(y(j)-p(2)-xeye(1,k)*sin(p(3)) -xeye(2,k)*cos(p(3)))^2;
j = j + 1;
end
I am trying to rigidly fit an ellipse of known size in the plane to a set of markers. I know how the markers should relate to the local x-y coordinates of the ellipse (x aligned along semi-major, y aligned along semi-minor). Sometimes, these markers are invisible. Hence, I only add penalty terms for markers indexed in whichVis.
p is a 3X1 array of ellipse center x, ellipse center y, and alpha, the inclination of the ellipse. These are the parameters to be optimized.
baseFun is an anonymous function, the equation of an ellipse in the plane.
totalFun is the entire anonymous function to be passed into lsqnonlin, with penalty terms added.
lam is a penalty stiffness.
xeye is a 2X8 array of known marker positions in local x-y coordinates of the ellipse.
x and y are marker positions of the markers which are visible.
The indices in x and y are ordered appropriately to correspond to the indicies in whichVis.
Am I constructing my objective function properly? When I print its value the j and k indices are unevaluated. I expect p to be the only thing to remain symbolic. I would have expected MATLAB to find the values in the arrays xeye, x, and y and fill them in as numbers.
Sample arrays to run code:
xeye = [31.6016 -15.3975 24.3983 42.1441 -35.7844 0.0596 -27.5122 5.8206
20.2474 23.6097 -22.9670 -3.5785 17.3939 -26.7238 -20.1340 26.0083]
whichVis = [1 2 3 4 5 6 7 8]
x = [-237.4388
-283.9550
-219.5598
-218.5919
-291.4497
-238.4623
-265.7021
-263.3386]
y = [-141.6660
-160.0548
-181.3641
-158.6064
-178.2309
-196.2995
-200.7612
-147.0627]
a = 44.5499
b = 26.5382
The code executes but I am skeptical that it's set up properly.
Thank you.

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### 채택된 답변

Torsten 2022년 6월 28일
편집: Torsten 2022년 6월 28일
basefun returns a vector of size (8x1) . Now to each component of this vector you add the sum of the following scalar values
lam*(x(1) - p(1) - xeye(1,1)*cos(p(3)) + xeye(2,1)*sin(p(3)))^2+lam*(y(1)-p(2)-xeye(1,1)*sin(p(3)) -xeye(2,1)*cos(p(3)))^2;
lam*(x(2) - p(1) - xeye(1,2)*cos(p(3)) + xeye(2,2)*sin(p(3)))^2+lam*(y(2)-p(2)-xeye(1,2)*sin(p(3)) -xeye(2,2)*cos(p(3)))^2;
...
lam*(x(8) - p(1) - xeye(1,8)*cos(p(3)) + xeye(2,8)*sin(p(3)))^2+lam*(y(8)-p(2)-xeye(1,8)*sin(p(3)) -xeye(2,8)*cos(p(3)))^2;
Of course, x, y and xeye have to be defined before setting up the function handle.
If this is what you want, your code is correct.
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Evan Hemingway 2022년 6월 28일
Ironically, the results were better with my original code. I think what may have been happening before was that maybe my objective function included only one spring associated with the final k and j value. Now it has a tough time converging properly. Perhaps I should lower my penalty stiffness. Anyway, thanks for your help.

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