Interpolating non-independent values

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

[Matt J]

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]

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]

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]

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]

"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?

[Bob Thompson]

Yes.

[Matt J]

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]

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...

Answer 1

[John D'Errico]

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.

[Bob Thompson]

Clearly I did not communicate some of the subtext of the problem properly, please see the updated OP.

[John D'Errico]

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.

Answer 2

[Matt J]

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');