How would one write their own polyfit function using only mldivide and for loops?
I have a basic idea:
function [A,e] = MyPolyRegressor(x, y, n)
c=ones(n,1);
for i=1:n;
c(:,i)=x.^(i-1);
end
A=c\y
e=c*A-y
But it doesnt quite work.

3 Kommentare

Jan
Jan am 26 Okt. 2012
Bearbeitet: Jan am 26 Okt. 2012
Please explain what "doesn't quite work" mean explicitly. Do you get an error message or do the results differ from your expectations? This forum is fine for solving problem, but bad for guessing them.
Notes: The POWER operation is very expensive. Better multiply x cummulatively.
SB
SB am 26 Okt. 2012
Well, there's a dimension mismatch in line 4. Even when I switch c to c=ones(size(X)) to fix that issue, there are too many coefficients, none of which are correct.
Jan
Jan am 26 Okt. 2012
Bearbeitet: Jan am 26 Okt. 2012
Because you didn't format your code properly (please learn how to do this...), it is not possible to find out, which one is the "line 4".
But with some guessing: "ones(n,1)" and even "ones(size(x))" create vectors, while the required Vandermonde-matrix needs the dimensions [length(x), n+1].

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 Akzeptierte Antwort

Jan
Jan am 26 Okt. 2012
Bearbeitet: Jan am 26 Okt. 2012

4 Stimmen

function p = fPolyFit(x, y, n)
% Construct Vandermonde matrix:
x = x(:);
V = ones(length(x), n + 1);
for j = n:-1:1
V(:, j) = V(:, j + 1) .* x;
end
% Solve least squares problem:
[Q, R] = qr(V, 0);
p = transpose(R \ (transpose(Q) * y(:)));
% Equivalent: (V \ y)'

1 Kommentar

SB
SB am 26 Okt. 2012
Bearbeitet: SB am 26 Okt. 2012
For a weighted Least Squares problem, would you do function [A, e] = WeightedLeastSquares(X, Y, w,n)
X=diag(w)*X
Y=diag(w)*Y
X = X(:);
V = ones(length(X), n + 1);
for j = n:-1:1
V(:, j) = V(:, j + 1) .* X;
end
[Q, R] = qr(V, 0);
A= (R \ (transpose(Q) * Y(:)));
e= V*A-Y;
?

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Weitere Antworten (1)

Vrushabh Bhangod
Vrushabh Bhangod am 19 Mai 2018
Bearbeitet: Walter Roberson am 10 Jun. 2018

1 Stimme

function [p,mu,f] = plofit(x,y,n)
% x = input samples
% y = output function,n = order
m = length(x); %number of rows in the Vandermonde Matrix
V = zeros(m,n);
a = n;
for i = 1:m
v = zeros(1,n);
for j = a:-1:1
v(n+1-j) = realpow(x(i),j);
end
V(i,:) = v;
end
V(:,n+1)=ones(m,1);% adding 1 column to ones to the vandermonde matrix
%%QR method to compute the least squares solution for the coefficients,'p'
[Q,R] = qr(V,0);
p = transpose(R \ (transpose(Q) * y'));
f = polyval(p,x);
%%to find mean
mean = sum(x)/length(x);
sq = 0;
for i =1:length(x)
sq = sq + (x(i)-mean)^2;
end
sd = (sq/length(x))^0.5;
mu = [mean;sd];
end

Gefragt:

SB
am 26 Okt. 2012

Kommentiert:

Max
am 20 Aug. 2022

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