matrix dimension error for interpolation in loop

Question

Melissa 2012-12-14

0
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https://ww2.mathworks.cn/matlabcentral/answers/56630-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

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Melissa 2012-12-14

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 2012-12-14

编辑：Matt J 2012-12-14

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.

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