matrix dimension error for interpolation in loop

Good Afternoon All,

I seem to be having a bit of trouble with matrix dimensions agreeing (I know rookie mistake). I am using radial basis functions and have created a model which I want to evaluate. I seem to be having the error here:

for j = 1:nSites
   r(j) = norm(x - rbfmodel.X(j,:)');
end

Error in evalRBF (line 21)
   r(j) = norm(x - rbfmodel.X(j,:)');

where Nsites is the number of sites in the original data points (rbfmodel.X) and x is the vector of points to be evaluated.

I think that I need some sort of indexing for x, maybe something along the lines of:

r = zeros(nSites,nSites);
for i = 1:nSites
   for j = 1:i-1
      r(i,j) = norm(x(i,:) - rbfmodel.X(j,:));
   end
   r(1:i-1,i) = r(i,1:i-1);
end

Anyone have suggestions?

Here is the code for evalRBF

function fx = evalRBF(x,rbfmodel)

% EVALRBF(X,RBFMODEL)Evaluates a radial basis function surrogate at a given point

% Input:
%   x           = the point to be evaluated
%   rbfmodel    = structure of all parameters that define the RBF surrogate
%     .X        =   matrix of data sites
%     .kernel   =   string indicating the choice of kernel function
%     .coeff    =   coefficients that define the RBF
% Output:
%   fx          = RBF value at x

% nSites is the number of data sites provided in Original Model
[nSites] = size(rbfmodel.X,1);

% r is the vector of distances between data sites
r = zeros(nSites,1);
for j = 1:nSites
   r(j) = norm(x - rbfmodel.X(j,:)');
end

% Adjust x as necessary
if strcmp(rbfmodel.poly,'regpoly0')
   x = [];
else
   x = feval(typePoly,x');
   x = x';
   x(1,:) = [];
end   

% Evaluate RBF at x
y  = [kernelRBF(rbfmodel.kernel,r,rbfmodel.c); 1; x(:)];
fx = rbfmodel.coeff'*y;
if ~isfinite(fx)
   fx = 1/eps;
end
return

[Matt J]

How is it that you don't know the dimensions of x? Is it someone else's code?

When you use the debugger to stop the program at the line where the error occurs, what do size(x) and size(rbfmodel.X) return?

[Melissa]

it is someone elses, yes. I size(x) depends on what data I wanted evaluated and rbfmodel.X will return original size matrix (ie for my exaple is 2x79). is there a way to get it for work say I want to evaluate x and its only at one point [50,50].

[Matt J]

Well first of all, it should be clear to you now where your bug is coming from. If x is always a 1x2 row vector (as with [50,50]) you will clearly not be able to subtract it from a 1x79 row vector (as rbfmodel.X(j,:) will be).

If you really want x to be a 2x1 column vector and you want it to be subtracted from every column of rbfmodel.X, change the relevant line as in my Answer below.

Answer 1

[Matt J]

A first guess

r(j) = norm(x(:) - rbfmodel.X(:,j));