Can we do polyfit on matrix?
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Hi,
I have two matrix, A and B, each of 1000 rows and 100 columns. I need to do 100 polyfit on the columns. I can loop through the columns. But I am just wondering if there is any simple way to plug in A, B without the loop and return the result in an other matrix C.
Thanks,
Jennifer
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Respuestas (3)
Jan
el 27 de Oct. de 2015
Editada: Jan
el 27 de Oct. de 2015
As far as I understand all columns are processed by polyfit independently. So you can at least omit the expensive checking of the inputs:
A = rand(1000, 100);
B = rand(1000, 100);
n = 3;
V = ones(1000, n + 1);
for k = 1:100
x = A(:, k);
y = B(:, k);
% Vandermonde matrix:
V(:, n+1) = 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(:)));
...
end
I fyou need further outputs of polyfit and e.g. a normalization of the input values, explain this explicitly here. Posting your existing code is always a good idea to reduce the need to guess, what you exactly need.
3 comentarios
Sarah
el 5 de Dic. de 2018
what if these two matrices were not of the same size? ohw would the solution change then?
Jos (10584)
el 27 de Oct. de 2015
Editada: Jos (10584)
el 27 de Oct. de 2015
A loop is the most obvious choice. You can hide the loop using arrayfun
FitFH = @(k) polyfit(X(:,k), Y(:,k), 1)
P = arrayfun(FitFH, 1:size(X,2), 'un',0)
P{X} will hold the fit for the X-th columns.
4 comentarios
Jos (10584)
el 28 de Oct. de 2015
help arrayfun
will give you the answer. The output is non-uniform.
Jos (10584)
el 28 de Oct. de 2015
And by the way, you can write your own (anonymous) polyfit function that skips the input checks as Jan suggested, but this might be over your head right now.
Namrata Badiger
el 28 de Mayo de 2020
A = rand(1000, 100);
B = rand(1000, 100);
n = 3;
V = ones(1000, n + 1);
for k = 1:100
x = A(:, k);
y = B(:, k);
% Vandermonde matrix:
V(:, n+1) = 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(:)));
...
endB = rand(1000, 100);n = 3;V = ones(1000, n + 1);for k = 1:100 x = A(:, k); y = B(:, k); % Vandermonde matrix: V(:, n+1) = 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(:))); ... end
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