How to find interpolation of x in a function of y by using intrep1 ?

So I'm trying to understand how interp1 works in Matlab. I have two variables x and y. I have NaN values in x and I have y values. So I want to find NaN values in x by using y.

I found this code for interp1.

nanx = isnan(x);
t    = 1:numel(x);
x(nanx) = interp1(t(~nanx), x(~nanx), t(nanx));

from this link : Interpolating NaN-s

However in this case from what I understand it interpolate based on the previous value i x list not based on y values.

I really hope someone can explain this to me.

Answer 1

[Jan]

You are right: In this example the NaNs are treated as gaps, which are filled by a linear interpolation of the surrounding points. An y does not appear in the code and therefore it does not solve your problem.

 I have NaN values in x and I have y values.
 So I want to find NaN values in x by using y.

This is not clear yet. To find NaN values in x use: isnan(x). I do not see a method to use another variable y to find NaNs in x. Please explain this again. Post an example:

x = [1, 2, 3, 4, NaN, NaN, 8, 10]
y = [0, 1, 4, 5, 7,   11,  3, 2]

So what is the wanted result?

[Ss]

The answers for NaN values in x supposed to be 3 and -2 because all point x are dependent on y and calculated by using the linear interpolation method. The NaN values won't be regulated on x only but it also on y.

I'm sorry I'm not really familiar with MATLAB's code yet, mathematically the formula somehow like this:

xnan = x1 + (y-y1)*(x2-x1)/(y2-y1)

But I don't know how to apply it in intero1 and isnan(x). Do you get what I mean?

[Stephen23]

@Ss: please take a look at the example x and y vectors in Jan Simon's answer. Please shows the exact vector you would expect the output to be, given those two input vectors.

Answer 2

[Walter Roberson]

Is y monotonic, either constantly increasing or constantly decreasing? If so then you can reverse the order of interpolation.

x_nan_mask = isnan(x);
y_at_nan = y(x_nan_mask);
x_no_nan = x(~x_nan_mask);
y_no_nan = y(~x_nan_mask);
x_reconstructed_at_nan = interp1(y_no_nan, x_no_nan, y_at_nan);
x_reconstructed = x;
x_reconstructed(x_nan_mask) = x_reconstructed_at_nan;

If y is not monotonically increasing then there is no unique solution unless you assume that it is piecewise continuous and that the x that are nan are in the interior (not boundary) of pieces. In such a case the solution is to break the problem up into sections that are individually piecewise and then reverse the order of interpolation.

[Ss]

yes it is piecewise continuous and x are the nan.

Can't I just try to find NaN values of x and interpolate based on y using interp1 directly?

[Walter Roberson]

interp1() can only work for monotonic increasing or decreasing data.

You could certainly write a loop that stepped through a point at a time, detected nan, and interpolated from the neighbours.