How do I write a function for fmincon using a for loop?

6 visualizaciones (últimos 30 días)
Warren Boschen
Warren Boschen el 11 de Oct. de 2022
Editada: Matt J el 21 de Oct. de 2022
Hello,
I am looking to minimize two parameters, and , with the following minimization:
I was recommended to do use the function fmincon to do this, but I'm having issues figuring out how to write the function "fun" as written on the documentation. Here is the function I have written:
function S0 = signal_min(atti, bvals, lambdas)
S0 = 0;
for b = 1:b_num_unique % For every unique b...
% This loop determines the indices of a particular b-value.
b_set = [];
for i = 1:length(bvals)
if bvals(i, :) == b_unique(b)
b_set = [b_set; i];
end
end
% This set of loops finds the spherical mean of the signal for a particular
% b-value by first creating an array of the average values at each
% point (x,y,z) and then finding the mean of that array.
bset_avg = zeros(X, Y, Z, length(b_set));
for i = 1:length(b_set)
bset_avg = bset_avg + atti(:, :, slices(z), b_set(i));
end
S_avg = bset_avg./length(b_set);
S0 = S0 + (S_avg(x, y, slices(z)) - sm_signal(bvals(b), lambdas(0), lambdas(1)))^2;
end
function [signal] = sm_signal(bvalue, lambda_parallel, lambda_perpendicular)
signal = exp(cast(-bvalue*lambda_perpendicular, 'double'))...
*sqrt(pi/(cast(4*bvalue*(lambda_parallel - lambda_perpendicular), 'double')))...
*erf(sqrt(cast(bvalue*(lambda_parallel - lambda_perpendicular), 'double')));
end
end
And here is the error produced:
Not enough input arguments.
Error in smt_fmincon>signal_min (line 122)
for b = 1:bvals % For every b-value...
Error in fmincon (line 552)
initVals.f = feval(funfcn{3},X,varargin{:});
Error in smt_fmincon (line 89)
[lambda, sigval] = fmincon(@signal_min, lambda0, C, d, Ceq, deq, lb, ub);
Caused by:
Failure in initial objective function evaluation. FMINCON cannot continue.
From how I have seen other functions for fmincon written, they seem like they're just supposed to be one line. What I can't figure is how to include the b-dependence in just one singular expression, as the purpose of the minimization is to minimize the squared residuals across all b's. That is to say, I don't want to minimize and then take the sum which is what I think I would be doing if I were to remove the for loop from inside signal_min, along the lines of below:
for b = 1:bvals
fmincon(@signal_min, x0, C, d, Ceq, deq, lb, ub);
end
What do you suggest I do to work around this issue?
Many thanks,
Warren

Respuestas (1)

Matt J
Matt J el 11 de Oct. de 2022
Editada: Matt J el 21 de Oct. de 2022
Forget the loop. Also, use lsqcurvefit instead:
params0=[lambda0(1), lambda0(1)-lambda0(2) ]; %re-parametrize
[params, sigval] = lsqcurvefit(@sm_signal, params0, double(bvalues),Savg(:),[0;0]);
lambda_parallel=params(1); %undo re-parametrization
lambda_perpendicular=params(1)-params(2);
function [signal] = sm_signal( params,bvalues)
bvalues=bvalues(:);
[lambda_parallel, delta_lambda]=deal(params(1),params(2));
signal = exp(-bvalues*lambda_perpendicular)...
.*sqrt(pi./(4*bvalues*delta_lambda))...
.*erf(sqrt(bvalues*delta_lambda));
end
  7 comentarios
Torsten
Torsten el 21 de Oct. de 2022
You set this condition in C and d in your call to "fmincon" ?
Matt J
Matt J el 21 de Oct. de 2022
Editada: Matt J el 21 de Oct. de 2022
How does the loop that you wrote account for the parameter ?
Loop? There are no loops in the code I posted for you.
However, the constraint that you mention is enforced by imposing (using the 5th input argument to lsqcurvefit) a lower bound of 0 on both unknown variables. One of these variables is deltaLambda = and so deltaLambda>=0 is equivalent to your original constraint.

Iniciar sesión para comentar.

Categorías

Más información sobre Programming en Help Center y File Exchange.

Productos


Versión

R2019b

Community Treasure Hunt

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

Start Hunting!

Translated by