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How to fix error about interp1?

11 vues (au cours des 30 derniers jours)
Ghazal Hnr
Ghazal Hnr le 12 Mar 2017
Commenté : Star Strider le 13 Mar 2017
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?

Réponse acceptée

Star Strider
Star Strider le 12 Mar 2017
Modifié(e) : Star Strider le 12 Mar 2017
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.
  4 commentaires
Ghazal Hnr
Ghazal Hnr le 13 Mar 2017
thanks my problem solved
Star Strider
Star Strider le 13 Mar 2017
My pleasure.

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Plus de réponses (1)

John D'Errico
John D'Errico le 12 Mar 2017
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.

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