What does this error mean? (lsqcurvefit)
2 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
I have no idea what line291 is saying here. Seems pretty much advanced stuff. It came up when I was trying running a bunch of lsqcurve fit using Trust region algorithm (due to lb and ub, otherwise would have preferred Levenberg algorithm). I am iterating my input parameters myself too via for loop, as it seemed that iniital parameters were affecting final outcome too much just to see where it goes. and give best solution. But after 5 mins this came up.
Error using trdog>quad1d (line 291)
Square root error in trdog/quad1d.
Error in trdog (line 180)
[nss,tau] = quad1d(nss,ssssave,delta);
Error in snls (line 316)
[sx,snod,qp,posdef,pcgit,Z] = trdog(x,g,A,D,delta,dv,...
Error in lsqncommon (line 150)
[xC,FVAL,LAMBDA,JACOB,EXITFLAG,OUTPUT,msgData]=...
Error in lsqcurvefit (line 253)
[xCurrent,Resnorm,FVAL,EXITFLAG,OUTPUT,LAMBDA,JACOB] = ...
Error in lambda7 (line 211)
[phiLSQ,resnorm,fval,output]= lsqcurvefit( @(phi,x) lambdafn7(phi,x),initial_param,x,y,lb,ub,options);
1 commentaire
Star Strider
le 15 Jan 2017
‘What does this error mean?’
We have no idea. We don’t have your code to refer to.
Réponse acceptée
dpb
le 15 Jan 2017
I'd venture it got into a negative parameter region and hence got an imaginary value from a square root that was trapped by the internal routine quad1d. Well, let's just look and see...
Indeed,
function[nx,tau] = quad1d(x,ss,delta)
%QUAD1D 1D quadratic zero finder for trust region step
% [nx,tau] = quad1d(x,ss,delta) tau is min(1,step-to-zero)
% of a 1-D quadratic ay^2 + b*y + c.
...
% Algorithm:
% numer = -(b + sign(b)*sqrt(b^2-4*a*c));
% root1 = numer/(2*a);
% root2 = c/(a*root1); % because root2*root1 = (c/a);
...
numer = -(b + sign(b)*sqrt(b^2-4*a*c));
r1 = numer/(2*a);
r2 = c/(a*r1);
tau = max(r1,r2);
tau = min(1,tau);
if tau <= 0,
error(message('optimlib:trdog:SqrRt'));
end
nx = tau*x;
So indeed that's the cause. As Star says, not much we can say other than that w/o actual code and data to go with it but looks like your modifications didn't help in this case, anyway. What if you remove those?
5 commentaires
dpb
le 15 Jan 2017
"Btw which part did you want me to remove?"
The part that is related to ..." I am iterating my input parameters myself too". Wondered if that were buggering up things.
Plus de réponses (1)
satya kothapalli
le 18 Jan 2018
Modifié(e) : Walter Roberson
le 18 Jan 2018
HI Vipul,
I also facing the same problem with my model code, as you have described in the first comment. It is interesting that you have sorted out multiple loops and identified the unexpected initial parameters in the end. Can you describe exactly what changes you have made to run smoothly. It would help me a lot.
Thank you, Regards, Satya.
0 commentaires
Voir également
Catégories
En savoir plus sur Downloads dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!