scatteredinterpolant differs from sample points

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Olaf
Olaf le 4 Août 2016
a surface created with scatteredinterpolant differs from the sample points it was made from, why ?
  1 commentaire
Adam
Adam le 4 Août 2016
You told it to interpolate the points so it introduced new points. Depending on what parameters you used these will potentially do what you have shown there when it comes to extrapolating.

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John D'Errico
John D'Errico le 2 Sep 2016
scatteredinterpolant will ALWAYS reproduce the data exactly, although it may sometimes introduce tiny noise on the order of eps, just due to floating point arithmetic. Interpolation is interpolation.
Others have suggested extrapolation. That has NOTHING to do with interpolation, and prediction of the original points in your set.
Most likely, this is due to a bug in your code. That essentially, you are not plotting the real data that was used to interpolate the surface. That happens.
So, if you want a valid answer, then you need to provide the data. Show how you tried to plot it. Show what you did. Give an example where you think scatteredinterpolant fails. When you do that, along with the data itself, I'll be able to show what you did wrong.

Plus de réponses (3)

Pelin Demir
Pelin Demir le 2 Sep 2016
Because corners must be broken.

James
James le 2 Sep 2016
Because the interpolation points are not always the scatters.

Sridhar Palagiri
Sridhar Palagiri le 2 Sep 2016
Due to Interpolation and Extrapolation

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