[help] Error using ==> fmincon Too many input arguments

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Kang Wang
Kang Wang 2011-3-23
回答： Ruslan Dautkhanov 2015-3-16
When I run a simulation of local constant cross validation for multivariables, I encountered the problem of too many input arguments.
function cv_m = lccvm(z,xc,xd,b,n) sum1 = 0 for i = 1:n dxc = (xc-xc(i,1))/(z(1)*n^(-1/5)) kc=exp(-0.5*dxc.^2); % continous kernel l=(xd==xd(i,1))+z(2)*(xd~=xd(i,1)); % discrete kernel k=kc.*l; % mixed kernel k(i,0)=0; % leave-one-out gx1=sum(b.*k)/sum(k); sum1=sum1+(b(i,1)-gx1)^2 end cv_m = 1/n*sum1;
and the bounds of z are z(1) from 0 to 20 z(2) from 0 to 1 there is no other constraint.
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Kang Wang
Kang Wang 2011-3-24
here is the fmincon file:
clear all;
close all;
n = 25;
loop = 100;
alpha = 1;
beta = 0.1;
gamma = 0.01;
for looop = 1:loop
xd = unidrnd(2,n,1)-1;
xc = randn(n,1);
u = randn(n,1);
y = alpha + beta*xd + gamma*xc + u;
z0 = [0.5;0.5];
lb = [0;0]
ub = [20;1]
[z(looop:1), cv(looop,1)]=fmincon(@lccvm,z0,lb,ub,@cons);
end
Walter Roberson
Walter Roberson 2011-3-24
Refer to my Answer: it is exactly what is going on in your situation.

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采纳的回答

Walter Roberson
Walter Roberson 2011-3-24
The objective function cannot accept that many variables directly. Please see the documentation of how to pass extra parameters.
  1 个评论
Kang Wang
Kang Wang 2011-3-24
Thanks for your help. The problem is not solved yet. I tried to minimize the objective function by choosing z=[z1;z2]. When I tried to choose a scaler z = z to minimize the function, it worked by running [z_star(looop,1), lccv(looop,1)]=fminbnd(@(z) lccvm(z,xd,xc,y,n),0.1,2). Then I went back the original problem, I wrote code like the following:
(1)The cross validation file:
clear all;
close all;
n = 25;
loop = 100;
alpha = 1;
beta = 0.1;
gamma = 0.01;
for looop = 1:loop
xd = unidrnd(2,n,1)-1;
xc = randn(n,1);
u = randn(n,1);
y = alpha + beta*xd + gamma*xc + u;
z0 = [0.5;0.5];
lb = [0;0];
ub = [20;1];
lccv = @(z)lccvm(xc,xd,z,y,n); %Anonymous Function?
[z_star(looop,1), lccv(looop,1)]=fmincon(lccv,z0,[],[],[],[],lb,ub);
end
(2) the function file
function cv_m = lccvm(c,d,e,b,N)
sum1 = 0;
for i = 1:N
dxc = (c-c(i,1))/(e(1)*N^(-1/5));
kc=exp(-0.5*dxc.^2); % continous kernel
l=(d==d(i,1))+e(2)*(d~=d(i,1)); % discrete kernel
k=kc.*l; % mixed kernel
k(i,1)=0; % leave-one-out
gx1=sum(b.*k)/sum(k);
sum1=sum1+(b(i,1)-gx1)^2;
end
cv_m = 1/N*sum1;
I hope that you can show me the details about my problem. Sincerely.

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Ruslan Dautkhanov
Ruslan Dautkhanov 2015-3-16

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