lsqcurvefit help - Field assignment to a non-structure array object

조회 수: 4 (최근 30일)
Mar
Mar 2019년 1월 14일
댓글: Torsten 2019년 1월 16일
Hi,
I am getting the error "Field assignment to a non-structure array object" in the line I call the lsqcurvefit function and I don't understand why.
I try to fit 2 different models. When I fit model_1 everything is fine. When I fit model_2 I got the error.
a, b - both vectors of size [1,41].
I have the following code:
opts = optimset('Display' ,'off');
for i=1:size(Image,1)
x(i,:) = lsqcurvefit (@model_1, init, a(1:end-1), Image(i, 1:end-1), zeros(size(init)), init*10, opts)
end
for i=1:size(Image,1)
x(i,:) = lsqcurvefit (@model_2, init, a(1:end-1), b(1:end-1), Image(i, 1:end-1), zeros(size(init)), init*10, opts)
end
Why do I get the error when I have one more entry in model_2? i.e. the vector b?
function y = model_2(p, a, b)
A = p(1);
B = p(2);
C = p(3);
D = p(4);
E = p(5);
y = A*exp(-a*C-b/D)+(1-A)*exp(-a*B).*(E*exp(-b/100)+(1-E)*exp(-b/40));
end
function y = model_1(p,a)
A = p(1);
B = p(2);
C = p(3);
y = A*exp(-B*a) + (1-A)*exp(-C*a);
end

답변 (1개)

Walter Roberson
Walter Roberson 2019년 1월 14일
편집: Walter Roberson 2019년 1월 15일
lsqcurvefit must have parameter order
  1. objective. @model_1 or @model_2
  2. x0. init in both cases
  3. xdata. a(1:end-1) in both cases
  4. ydata. Image(1:end-1) in the first case and b(1:end-1) in the second case
  5. lb. zeroes in the first case and Image(1:end-1) in the second case
  6. ub. init*10 in the first case and zeroes in the second case
  7. options. opt (a struct) in the first case and zeroes in the second case
  8. no documented parameter . absent in the first case and opt (a struct) in the second case.
Internally the code attempts to add additional fields to the struct expected in the 7th position and fails when the parameter is numeric zeroes .
  댓글 수: 16
Mar
Mar 2019년 1월 16일
I evaluated it and It does return sensful values.
Torsten
Torsten 2019년 1월 16일
Then you should do as "lsqcurvefit" suggests: Set a larger value for the maximum number of iterations:
options.MaxIterations and/or options.MaxFunctionEvaluations

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