arrayfun() mistakes a vector optimization parameter for a scalar optimization parameter

Hey guys, I have a problem with the function arrayfun not recognizing the dimensionality of the variable I'm trying to extract from it. The code is here:

options = optimset('MaxFunEvals',10,'MaxIter',10);
*c_hat = [c_hat; arrayfun(@(idx) fminsearch(@(c) 1/2*sumsqr(SST(idx,1)-Phi(1,idx)*(c.*rC(idx,1))) + lambda*sum(c.*rC(idx,1)),1,options),1:size(SST,2))];* %broke up the line into two
%size(SST) = 413x1
%size(Phi) = 413x413
%size(rC) = 413x1
%size(c_hat) = 413x1

Background: This is a minimization problem, I'm using multivariable nonlinear regression to find the value of 413 coefficients, to be contained in the vector c. However, the way MATLAB interprets it, c is a scalar. This isn't surprising because looking closely at the code, fact that c is a scalar is just as plausible as the fact that c is vector of size 413 (in both cases, the minimization function line

c_hat = [c_hat; arrayfun(@(idx) 
fminsearch(@(c) 1/2*sumsqr(SST(idx,1)
-Phi(1,idx)*(c.*rC(idx,1))) 
+ lambda*sum(c.*rC(idx,1)),1,options)
,1:size(SST,2))];

is valid and will run). That being said, how do I let MATLAB know that c is not a scalar, but a vector?

Thanks in advance!

[Stephen23]

Duplicate:

https://www.mathworks.com/matlabcentral/answers/351753-having-trouble-with-array-dimensions-in-an-arrayfun-call-with-fminsearch-nested-in-it

Answer 1

[Jiro Doke]

Isnt't it because you are passing in "1" (scalar) as the initial guess to fminsearch? Perhaps you can pass in a 1x413 vector of ones:

c_hat = [c_hat; arrayfun(@(idx) fminsearch(@(c) 1/2*sumsqr(SST(idx,1)- ...
    Phi(1,idx)*(c.*rC(idx,1))) + lambda*sum(c.*rC(idx,1)),ones(1,413),options), ...
    1:size(SST,2))];

[Walter Roberson]

Right. As I posted in your other Question,

"If you want c to be a vector of length 100, then you need to pass a vector of length 100 in the position where you have passed 100. The first parameter to fminsearch is a function handle; the second parameter is the starting point vector, not the length of the starting point vector."

[Jiro Doke]

Ah, thanks Walter. Didn't see his other question (and your answer).