Indexing Error - can't get for loop to function because indices are incorrect

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.

Answer 1

[Walter Roberson]

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));
    

[Walter Roberson]

You would not even need the arrayfun if you were to vectorize your basis, such as

basisN3 = @(x)[ ones(size(x(:)), x(:), x(:).^2, x(:).^3 ];

[Caleb Nickens]

This worked perfectly! Thank you. I have not used arrayfun before, what does 'uniform' do?

Also, it seems that vectorizing the basis as such would be a bit less accurate. However, the arrayfun worked well.

[Walter Roberson]

Vectorizing the basis will not be less accurate.

https://www.mathworks.com/help/matlab/ref/arrayfun.html#d122e47424

uniform false: arrayfun returns the outputs of func in cell arrays. The outputs of func can have any sizes and different data types.

[Walter Roberson]

format long g
x = randn(50,1) * 1000;
basisN3 = @(x)[ 1, x, x^2, x^3 ];
basisN3v = @(x) [ones(size(x(:))), x(:), x(:).^2, x(:).^3];

A1 = cell2mat(arrayfun(basisN3, x, 'uniform', 0));
A2 = basisN3v(x);

size(A1)

ans = 1×2

50 4

size(A2)

ans = 1×2

50 4

max(abs(A1(:) - A2(:)))

ans = 
     0

isequal(A1, A2)

ans = logical
   1

Bit. For. Bit. Identical.

What led you to suspect that the vectorized version might be less accurate?

[Caleb Nickens]

I must have run it incorrectly. Following what you provided here, I see that it does work very well. Thank you again.