How to Interpolate an entire data set

6 vues (au cours des 30 derniers jours)
Madison
Madison le 1 Août 2023
Commenté : Madison le 1 Août 2023
How to Interpolate an entire data set.I have two datasets and I am trying to determine tidal energy for the lat and lon values of my elasmobranch sightings. Since teh specific sighting lat and lon dont exist in my tidal data I am trying to interpolate them.
I have tried interpolting each value seperatly, but I recieve a value of 0. I've tried both interp1 and interp2
Tides=readtable('Tide_Data_Final.csv')
C = Tides.Longitude
M= Tides.Latitude
A=Tides.Tide
F=scatteredInterpolant(C,M,A);
Anew= F(52.00,-6.00);
figure(1)
stem3(C,M, A)
hold on
stem3(52.000,-6.00, Anew, 'r')
hold off
grid on
xlabel('\sigma')
ylabel('\alpha')
interp2(C,M,A,52.000,-6.000)
  4 commentaires
Madison
Madison le 1 Août 2023
The data ius now linked, thank you
Madison
Madison le 1 Août 2023
When I use scatteredInterpolant the value only returns with an interpolated value of 0 and I know the value is not 0 for every lat and long camobination.

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dpb
dpb le 1 Août 2023
Déplacé(e) : dpb le 1 Août 2023
unzip Tide_Data_Final
Tides=readtable('Tide_Data_Final.csv');
head(Tides)
Longitude Latitude Tide _________ ________ ____ -7 50.5 0 -7 50.505 0 -7 50.511 0 -7 50.516 0 -7 50.521 0 -7 50.526 0 -7 50.532 0 -7 50.537 0
nnz(Tides.Tide)
ans = 283644
height(Tides)
ans = 562856
F=scatteredInterpolant(Tides.Latitude,Tides.Longitude,Tides.Tide);
Tnew= F(52.00,-6.00)
Tnew = 0.7983
You have a perfectly good table with real variable names; use it/them.
You created another set of variables C,M,A with no relation at all to the actual data and turned the reference to lat, lon around by doing so and couldn't see the obvious since the new variable names bore no relationship to the actual data.
  1 commentaire
Madison
Madison le 1 Août 2023
Thank you so much! This worked!

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Plus de réponses (1)

Torsten
Torsten le 1 Août 2023
Modifié(e) : Torsten le 1 Août 2023
I think you interchanged latitude and longitude in your call. It should be
Anew= F(-6.00,52.00);
instead of
Anew= F(52.00,-6.00);
which should give something around
Anew = 0.7983

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