Interpolating non-independent values

2021 3 29
2 个回答
更新时间：2021 3 30
3 次查看（30 天）

1 个投票

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
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Matt J
Matt J 2021-3-29
Well, clearly t=0.6 maps to both y=0.37 and y=0.5. What rule should be applied to resolve the ambiguity?
Bob Thompson
Bob Thompson 2021-3-29
I think that's part of what I'm trying to figure out, syntaxically.
t = 0.6 does not map to y = 0.37 or y = 0.5 directly, but rather it maps directly to x = 2.6. The problem is that when a curve is drawn along the tx points there are four places where x can equal 2.6, but t = 0.6 and x = 2.6 only happens once, at the very end of the series. Therefore, at t = 0.6, y = 0.5, because that is also the last relationship between x and y. The other 'bounces' are much more difficult to logic.
It is assumed that both data sets cover the same total time, but not necessarily the same time intervals. I suppose I didn't make that clear before, but the xy data is technically time dependent, but the relationship is not directly known, which is why I'm trying to draw the relationships.
Matt J
Matt J 2021-3-29
编辑：Matt J 2021-3-29
Is the sampling always equi-spaced in time? t(j+1)-t(j) is a constant independent of j?
It is assumed that both data sets cover the same total time, but not necessarily the same time intervals.
Does this mean t(end)-t(1) is the same in both data sets, but t(1) may vary?
Bob Thompson
Bob Thompson 2021-3-29
It's the opposite of what you're thinking. t(end)- t(1) is constant, and t(1) is constant, but j is not a constant, so the sampling is not constant. I guess that's what I meant by 'interval'.
Matt J
Matt J 2021-3-29
"j is not constant" looks like a typo. Because j is an index ranging over the length of the data set, it is obviously never going to be constant.
Do you mean that t(j+1)-t(j) is not constant over the data set? The sampling is not equi-spaced in time over a fixed data set?
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...

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回答（2 个）

John D'Errico
John D'Errico 2021-3-29
Ran in:

1 个投票

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.

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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.

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Matt J
Matt J 2021-3-29
编辑：Matt J 2021-3-30

0 个投票

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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Matt J
Matt J
2021-3-30

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