Faster method for polyfit along 3rd dimension of a large 3D matrix?

Question

Justin 2012-8-9

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评论： Jan 2022-2-11

I am currently working with large data cubes in the form of an MxNxP matrix. The P-dimension represents the signal at each (m,n) pixel. I must obtain a Z-order polynomial fit(where Z varies depending on the situation) for each signal in the data cube. Currently, I utilize a "for" loop to obtain the signal at each pixel, obtain the polynomial coefficients, then calculate the fitted curve. The code I use is fundamentally similar to the following:

dataCube = rand(1000,1000,300);
x = rand(300,1);
sizeCube = size(dataCube);
polyCube = zeros(sizeCube);
for ii = 1:sizeCube(1);
   for iii = 1:sizeCube(2);
        signal = squeeze(dataCube(ii,iii,:));
        a = polyfit(x,signal,z)
        y = polyval(a,x);
        polyCube(ii,iii,:) = y;
     end
  end

Because of the quantity of iterations in the for loop, this operation takes a considerable amount of time for each data cube. Is there a faster way to obtain the polynomial fitting, without having to resort to the iterative process I use here. Perhaps, something similar to the filter function where you can apply the filter to a specific dimension of a matrix, rather than extracting each signal?

filteredCube = filter(b,a,dataCube,[],3)

Thanks, Justin

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Walter Roberson 2012-8-9

Have you considered re-ordering your data, at least during the processing, so that the dimension you are fitting over is the first dimension? Access (and assignment) over the first dimension is faster.

Justin 2012-8-14

Walter,

I attempted reordering the matrix and performing the process as above, over the 1st dimension. There was an incremental improvement in speed, but not nearly the type of advance I was hoping I could achieve.

Check out the answer by Teja below. Its exactly what I was hoping for and works perfectly.

Thanks for the input! Justin

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Answer 1

Teja Muppirala 2012-8-10

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https://ww2.mathworks.cn/matlabcentral/answers/45680-faster-method-for-polyfit-along-3rd-dimension-of-a-large-3d-matrix#answer_55902

在 MATLAB Online 中打开

This can be accomplished in a fraction of the time with some matrix operations.

dataCube  = rand(100,100,300);
sizeCube = size(dataCube);
x = rand(300,1);
z = 3;
V = bsxfun(@power,x,0:z);
M = V*pinv(V);
polyCube = M*reshape(permute(dataCube,[3 1 2]),sizeCube(3),[]);
polyCube = reshape(polyCube,[sizeCube(3) sizeCube(1) sizeCube(2)]);
polyCube = permute(polyCube,[2 3 1]);

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Pavel Psota 2022-2-10

+1, cool, indeed! Could the fitting coefficients be obtained from your solution?

Jan 2022-2-11

在 MATLAB Online 中打开

A small improvement is to avoid the expensive power operation:

% Replace:
V = bsxfun(@power,x,0:z);
% by:
V = [ones(300, 1), cumprod(repmat(x, 1, z), 2)];

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Answer 2

Martin Offterdinger 2019-4-9

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Dear Teja,

I am having a similar problem- actually a simpler one even. I have the same array, but I always need to fit a first-order polynom (linear, z=1 in your code). Is it possible to get the coefficients of the linear fit (p1,p2) from your solution as well?

Thanks,

Martin

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Jan 2019-4-9

编辑：Jan 2019-4-9

@Martin: Please do not attach a new question in the section for answer. Open a new thread instead and remove this pseudo-answer. Including a link to this thread is a good idea. Thanks.

What's wrong with setting z=1? Which array is "the same" and why do you need to determine the fit multiple times for the same array? (Please explain this in your new question...)

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