Free-knot spline approximation (BSFK) problem

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Michal 2022-9-22

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1810010-free-knot-spline-approximation-bsfk-problem

编辑： Bruno Luong 2023-1-12

@BrunoLuong

My data acquisition system produce periodically 1-D measured noised data with the fixed time window length W. I want to produce smoothed data for each window W separately, with specific constraints on continuous (k=2) or smooth (k = 3 or 4) processed signal connections between consecutive time measurement windows. So, for each window W I get finally separate "pp" structure. How to set proper BSFK options setting to fulfil these constraints?

The second question is: Is there any method how to merge separate "pp" structures to one "pp" structure for several processing windows at one?

Add note: May by some processing windows overlap could be required. Do you have any experiance with using BSFK in streaming regime?

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Bruno Luong 2022-9-23

编辑：Bruno Luong 2022-9-23

You could try to recursively enforce the continuity for function/derivative when you call BSFK on the next interval using the pp of the previous interval, just tell BSFK to have function/derivative of the most left knot (current) = previous pp function/derivative at the right knot (previous).

Michal 2022-9-23

Yes, something like this I would like to try, but I don't know how exactly implement this type of fix by "shape" and "pntcon" options. This is the main reason why I need your help.

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采纳的回答

Bruno Luong 2022-9-23

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1810010-free-knot-spline-approximation-bsfk-problem#answer_1059440

编辑：Bruno Luong 2023-1-12

在 MATLAB Online 中打开

Reference to BSFK function https://fr.mathworks.com/matlabcentral/fileexchange/25872-free-knot-spline-approximation

Here is the recursive pointwise constraint. You'll see it does the job (zoom in) the transition is not nice

data=load('result_4_8.mat')
data=data.result;
[m,n] = size(data);
x = cellfun(@(data) data.x, data, 'unif', 0);
y = cellfun(@(data) data.y, data, 'unif', 0);
j = 1; % n
close all
figure
hold on
for i = 1:m
    % Normalize data so that dy/dx is comparable to y
    xij = x{i,j}/10000;
    yij = y{i,j}/10000;
    options = struct('lambda', 1e-8);
    if i >= 2
        xleft = pp.breaks(end);
        yleft = ppval(pp,xleft);
        ydleft = ppval(ppder(pp),xleft);
        xij = [xleft; xij];
        yij = [yleft; yij];
        pntcon = struct('p', {0 1}, 'x', {xleft,xleft}, 'v', {yleft ydleft});
        options.pntcon = pntcon;
    end
    pp =BSFK(xij, yij, 4, [], [], options);
    xi = linspace(min(xij),max(xij),1000);
    yi = ppval(pp, xi);
    plot(xij, yij,'c.');
    plot(xi, yi, 'r', 'Linewidth', 2);
    drawnow
end
function ppd = ppder(pp)
ppd = pp;
coefs = ppd.coefs;
n = size(coefs,2);
ppd.coefs = coefs(:,1:n-1).*(n-1:-1:1);
ppd.order = ppd.order-1;
end

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Bruno Luong 2022-9-29

I haven't not studied the complexity of BSFK.

But it seems to me the linear dependency to number of knots is not quite true, I would say it is more like quadratic.

Michal 2022-9-29

Of course, this is my mistake ... you are right! CPU time dependency to number of knots is ~ quadratic!!!

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