Least Squares Regression, Help me find the error in my code!

1 Ansicht (letzte 30 Tage)
SB
SB am 26 Okt. 2012
Suppose the curve you're fitting can follow the form
y = an*x^n+ an-1*x^(n-1)+...+ a1x + a0
x and y are two column vectors of the same size and n is the degree of the polynomial. A is an array of the coefficients, in the order from an to a0.
Solve for A, and the e, the array of the residuals.
Here's what I have: function [A,e] = MyPolyRegressor(x, y, n)
for i= 1:n
x(i)=x.^(n-1);
end;
A=x/y;
e=A*x-y;
Whenever I run it, it says, "In an assignment A(I) = B, the number of elements in B and I must be the same." Please help!

Antworten (2)

Matt J
Matt J am 26 Okt. 2012
Bearbeitet: Matt J am 26 Okt. 2012
In
x(i)=x.^(n-1);
the left hand side is a scalar location and the right hand side is not a scalar. So there's no room for it. I don't really understand why you're trying to overwrite x with powers of itself like x.^(n-1).
You should consider the VANDER command. I'm assuming you're not allowed to use POLYFIT.
  3 Kommentare
SB
SB am 26 Okt. 2012
I'd try that, but I'm trying to avoid using any commands beyond basic array ones. I was thinking something like
for i = 1:n
b(i) = x(i)^(n-i) ;
end
A=b\y';
e=A*b-y;
I can't quite get the answer though.
Matt J
Matt J am 26 Okt. 2012
Bearbeitet: Matt J am 26 Okt. 2012
b=ones(length(x),n+1);
for i = 0:n-1
b(:,i+1) = x(:).^(n-i) ;
end

Melden Sie sich an, um zu kommentieren.


Azzi Abdelmalek
Azzi Abdelmalek am 26 Okt. 2012
There is a problem in your x array size, it should be a matrix (nxm) where m is the length of y and m is the lenngth of unknown variables
  1 Kommentar
SB
SB am 26 Okt. 2012
I fixed that in my current code (shouldve been b*A rather than A*b), but I'm still getting the wrong answer ( the correct answer and my new code is posted above).

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Performance and Memory finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by