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Interpolating non-independent values

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Bob Thompson
Bob Thompson 2021 年 3 月 29 日
編集済み: Matt J 2021 年 3 月 30 日
I'm not really sure how to ask this in a concise enough way to find it elsewhere, so I'm just going to ask a new question.
I have two arrays of data, one with time and x values, and one with x and y values. I want to get the y values based on time, but the x values are not uniquely associated with y values.
tx = [0 0;
0.1 1;
0.2 2;
0.3 3;
0.4 2.5;
0.5 2.8;
0.6 2.6];
xy = [0 0;
0.5 0.1;
1 0.15;
1.5 0.2;
2.1 0.22;
2.7 0.3;
3 0.33;
2.6 0.37;
2.5 0.39;
2.7 0.42;
2.8 0.45;
2.6 0.5];
% I can interpolate the axial data to time easily.
x_proper = interp1(tx(:,1),tx(:,2),[0:0.05:0.6]);
% Interpolating with interp1 to get y data limits the y values to <= 0.33,
% because it doesn't understand the bounce in x values.
Here are the assumptions that can be made about the two datasets:
1) tx and xy are applicable over the same total time. Therefore tx([1,end],1) == time of xy([1,end],:)
2) The pattern of x in tx and in xy are matching, and share the same peaks and troughs
3) Due to the time nature of the dataset, each unique (x,y) set will correspond to a unique time
  8 件のコメント
Matt J
Matt J 2021 年 3 月 29 日
Here are the assumptions that can be made about the two datasets:
It's basically a 1D image registration problem. The question is, what will be your deformation model...

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回答 (2 件)

John D'Errico
John D'Errico 2021 年 3 月 29 日
x is non-monotonic as a function of time. I have no idea what you mean by bounce, but I assume it indicates the failure to be monotonic.
You can predict x as a function of time. But now you want to predict y as a function of x? How should MATLAB (or anyone, including me) know how to predict y? x does not vary in any kind of increasing order, so which value of x should be used to predict y? Sorry, but you are asking for something that is not posible.
xy = [0 0;
0.5 0.1;
1 0.15;
1.5 0.2;
2.1 0.22;
2.7 0.3;
3 0.33;
2.6 0.37;
2.5 0.39;
2.7 0.42;
2.8 0.45;
2.6 0.5];
plot(xy(:,1),xy(:,2),'o-')
xline(2.65);
Consider x = 2.65. Which of several (four possible) values of y should be predicted?Can the computer possibly know? Looking at it, I don't even know.
  2 件のコメント
John D'Errico
John D'Errico 2021 年 3 月 29 日
Yes, you did communicate that. What you did not communicate is how you can possibly know from one curve what to do with the second.

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Matt J
Matt J 2021 年 3 月 29 日
編集済み: Matt J 2021 年 3 月 30 日
Assuming you had moderately tight upper and lower bounds U and L on the time increments t(j+1)-t(j) of the xy data set, I can sort of see you attempting an inverse problem to solve for the unknown t(j) as follows.
Fx=griddedInterpolant(tx(:,1), tx(:,2),'spline');
tstart=tx(1,1);
tstop=tx(end,1);
xdata=xy(:,1);
ydata=xy(:,2);
N=numel(xdata)-2;
D=diff(speye(N),1,1); %differencing matrix
Aineq=[D;-D];
bineq=max(0, repelem([U;-L],N-1) );
lb=repelem(tstart,N-1,1);
ub=repelem(tstop,N-1,1);
t0=linspace(tstart,tstop,N+2);
t0([1,end])=[];
t=fmincon(@(t) norm(Fx(t(:))-xdata).^2, t0,Aineq,bineq,[],[],lb,ub,[]);
t=[tstart,t(:).',tstop];
Fy=griddedInterpolant(t,ydata,'spline');

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