MATLAB: Indexing Error – can’t get for loop to function because indices are incorrect

for loopfuctionsindexingleast squares regressionMATLAB

function basis = regbas(CorrectionType)
basisN0 = @(x)[ 1 ];
basisN1 = @(x)[ 1, x ];
basisN2 = @(x)[ 1, x, x^2 ];
basisN3 = @(x)[ 1, x, x^2, x^3 ];
if strcmp(CorrectionType,'Constant')
    basis = basisN0
elseif strcmp(CorrectionType,'Linear')
    basis = basisN1
elseif strcmp(CorrectionType,'Quadratic')
    basis = basisN2
elseif strcmp(CorrectionType,'Cubic')
    basis = basisN3
else
    disp("Error - CorrectionType should be either 'Constant', 'Linear', 'Quadratic' or 'Cubic'.")
end
end
This function defines "basis"
function coef = alsqr(arr,basis)
x = arr(:,1); % time is column 1
y = arr(:,2); % change second value for column based on which acceleration data you're using
if length(x) ~= length(y); disp('Error - The dimensions of the x and y value lists in the data are not the same')
end
order = length(basis(1));
n = length(x);
A = ones(n,order);
 for ORD = 1:order
        vec = basis(ORD);
        for dat = 1:n
            A(dat, ORD) = vec(x(dat)); % ERROR OCCURS HERE
        end
 end
AT = A';
AA = AT*A;
Ay = AT*y;
w = AA\Ay;% same as inv(AA)*Ay;
coef = round(w,9);
end
I am looking to create a matrix "A" that has length(x) rows and the same number of columns as basis. When I run this function (alsqr), I get an error inside the second for loop telling me "Array indices must be positive integers or logical values." I have tried many different formats and can't get it to run correctly. These functions are to solve for a least squares regression.

Best Answer

regbas returns a function handle. Presumably it is that function handle that is being passed into alsqr 
x = arr(:,1); % time is column 1
So x is expected to be a vector of times.
order = length(basis(1));
Invoking the function handle on scalar input 1 returns a vector, and you are calculating the length of the vector returned. Reasonable code.
 for ORD = 1:order
        vec = basis(ORD);
You are passing the current order, such as 1, 2, 3, to basis(), which is accepting those as x values and returning a vector of values. ORD = 1 might return [1, 1, 1^2], ORD = 2 might return [1, 2, 2^2] and so on. So vec will be a numeric vector of length order 
        for dat = 1:n
            A(dat, ORD) = vec(x(dat)); % ERROR OCCURS HERE
x is a vector of times. x(dat) is a scalar time, such as 0.013 . You then use that scalar time, which will not generally be a positive integer, to try to index vec, which is numeric vector of integers like [1, 3, 9] . But those times are not valid subscripts.
Reminder that your basis is not a vector of function handles to be indexed and individually applied to data: your basis is a function handle to something that returns a vector. As in
    A = cell2mat(arrayfun(basis, x(:), 'uniform', 0));
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