How to fix error about interp1?

Hello,

I have some data and i want to run this code:

 xi = [85];
 yi = interp1(y,x,xi)

but I have this error:

Error using griddedInterpolant
The grid vectors are not strictly monotonic increasing.

Error in interp1 (line 183)
        F = griddedInterpolant(X,V,method);

Error in Q4 (line 19)
yi = interp1(y,x,xi)

How can I fix it?

Answer 1

[Star Strider]

One way (without seeing your data, so this is a guess on my part):

y85 = find(y <= 85, 1, 'last');
yi = interp1(y(y85-1:y85+1),x(y85-1:y85+1),xi, 'linear','extrap');

That is how I would do it. This assumes there is only 1 where ‘y’ is approximately 85.

Note — This is obviously UNTESTED CODE. It should work, if there is one ‘y’ near 85.

EDIT — Added the 'extrap' option to make my approach more robust.

[Ghazal Hnr]

There is only one where y is 85.

I use your code but i still have error:

Index exceeds matrix dimensions.
Error in Q4 (line 20)
yi = interp1(y(y85-1:y85+1),x(y85-1:y85+1),xi, 'linear','extrap')

[Star Strider]

Without your data, I cannot tell.

Try this:

yi = interp1(y(y85-1:y85),x(y85-1:y85),xi, 'linear','extrap')

or this:

yi = interp1(y(y85:y85+1),x(y85:y85+1),xi, 'linear','extrap')

One of them should work.

Also, be certain you are interpolating the correct value. If you want the value of ‘y’ at ‘x=85’, do this instead:

yi = interp1(x, y, xi, 'linear','extrap')

You may have misinterpreted the documation for interp1.

[Ghazal Hnr]

thanks my problem solved

[Star Strider]

My pleasure.

Answer 2

[John D'Errico]

The fix is easy. Don't use interp1 to interpolate data that is not strictly monotone in the independent variable.

After all, how can interp1 know which value to predict on a function that is apparently multi-valued? For example, consider points around the perimeter of acircle. For any given x, there are TWO solutions. Interp1 requires a single valued function for any given x.