Interpolating non-independent values

1 次查看（过去 30 天）
Bob Thompson
Bob Thompson 2021-3-29
编辑： Matt J 2021-3-30
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 个评论
显示 6更早的评论
Matt J
Matt J 2021-3-29
编辑：Matt J 2021-3-29
If you had moderately tight upper and lower bounds on the time increments t(j+1)-t(j) then it might be possible to do something. Otherwise, I agree with John that it is a highly ill-posed problem.
Matt J
Matt J 2021-3-29
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...

请先登录，再进行评论。

请先登录，再回答此问题。

回答（2 个）

John D'Errico
John D'Errico 2021-3-29
Ran in:
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 个评论
Bob Thompson
Bob Thompson 2021-3-29
Clearly I did not communicate some of the subtext of the problem properly, please see the updated OP.
John D'Errico
John D'Errico 2021-3-29
编辑：John D'Errico 2021-3-29
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.

请先登录，再进行评论。

Matt J
Matt J 2021-3-29
编辑：Matt J 2021-3-30
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');

类别

在 Help Center 和 File Exchange 中查找有关 Arithmetic Operations 的更多信息

标签

产品


版本

R2016b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by