Can we do polyfit on matrix?

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JFz
JFz 2015 年 10 月 27 日
回答済み: Namrata Badiger 2020 年 5 月 28 日
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

回答 (3 件)

Jan
Jan 2015 年 10 月 27 日
編集済み: Jan 2015 年 10 月 27 日
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 件のコメント
donald adams
donald adams 2017 年 11 月 21 日
Great solution! Thanks
Sarah
Sarah 2018 年 12 月 5 日
what if these two matrices were not of the same size? ohw would the solution change then?

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Jos (10584)
Jos (10584) 2015 年 10 月 27 日
編集済み: Jos (10584) 2015 年 10 月 27 日
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 件のコメント
Jos (10584)
Jos (10584) 2015 年 10 月 28 日
help arrayfun
will give you the answer. The output is non-uniform.
Jos (10584)
Jos (10584) 2015 年 10 月 28 日
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.

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Namrata Badiger
Namrata Badiger 2020 年 5 月 28 日
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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