Multiple use of polyfit - could I get it faster?
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I have a more general problem in the sense that I am supposed to do polyfit more than 10000 times on a given problem. Is there any way I can improve the speed of this instead of looping more than 10000 times through all of this? I do assume that the polynoms are all of the same degree.
Basically, I want to do matrixP = polyfit(matrixX, matrixY, n)
1 件のコメント
Bruno Luong
2011 年 2 月 23 日
There might be some way if the numbers of data remain unchanged. Can we assume that?
回答 (3 件)
Jan
2011 年 2 月 23 日
If you are sure, that your X-values are properly scaled (perferred: mean(x)==0, std(x)==1) and do not need the 2nd and 3rd output of POLYFIT, you can omit the nice and secure error checks of POLYFIT:
P = polyfitM(matrixX, matrixY, n)
[sx1, sx2] = size(x);
P = zeros(sx1, n+1);
for Index = 1:sx1
x = matrixX(Index, :);
y = matrixY(Index, :);
x = x(:);
y = y(:);
q = ones(sx1, 1);
V(:, n+1) = q;
for j = n:-1:1
q = q .* x;
v(:, j) = q;
end
[Q, R] = qr(V, 0);
P(Index, :) = transpose(R \ (transpose(Q) * y));
end
If your X-values are equidistant, you could save 99.9% processing time by calculating the Vandermonde matrix V and the QR decomposition once only. But even the reduced error checks in the posted method can run a factor of 2 to 3 faster compared to calling POLYFIT in a loop - please check this on your computer.
0 件のコメント
Matt Tearle
2011 年 2 月 23 日
In the manner of Jan's answer, just stripping down to a brute-force least-squares fit can give a lot of speed-up. I'm assuming that you have a matrix of x and a matrix of y, with each column representing a different data set. And also, therefore, assuming that each data set has a different set of x values. Here's a function to compare them:
function [c1,c2] = multpolyfit(x,y,n)
m = size(x,2);
c1 = zeros(n+1,m);
tic
for k=1:m
c1(:,k) = polyfit(x(:,k),y(:,k),n)';
end
toc
c2 = zeros(n+1,m);
tic
for k = 1:m
M = repmat(x(:,k),1,n+1);
M = bsxfun(@power,M,0:n);
c2(:,k) = M\y(:,k);
end
toc
My results:
>> x = rand(20,10000);
>> y = pi + x - 0.5*x.^2 + 0.1*rand(size(x));
>> [c1,c2] = multpolyfit(x,y,2);
Elapsed time is 9.531862 seconds.
Elapsed time is 0.610350 seconds.
3 件のコメント
Matt Tearle
2011 年 3 月 22 日
@Ian: something like this?
function [c,R2] = multpolyfit(x,y,n)
m = size(x,2);
c = zeros(n+1,m);
r = zeros(size(y));
for k = 1:m
M = repmat(x(:,k),1,n+1);
M = bsxfun(@power,M,0:n);
c(:,k) = M\y(:,k);
r(:,k) = M*c(:,k)-y(:,k);
end
sserr = sum(r.^2);
sstot = sum(bsxfun(@minus,y,mean(y)).^2);
R2 = 1 - sserr./sstot;
It seems to be about 25% faster than doing the individual calculation every time through the loop. Caveat: I *think* the formula/implementation is correct... but I wouldn't bet my house on it.
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