interp1 function error
2 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
fima v
le 7 Mai 2020
Réponse apportée : Ameer Hamza
le 8 Mai 2020
Hello i have made an interpolation of a descrete CSV file which is attached and turned it into continues function using interp1.
it went without an error , but when i tried to plot continues function using xq vector,it said
'Subscript indices must either be real positive integers or logicals.'
Where did i go wrong?
Thanks.
data = load('t2.csv');
x = data(:,1);
y = data(:,2);
%plot(x,y);
%hold on
x_data=x/(1e+6);
y_data=abs(y/(100));
plot(x_data,y_data);
coef_fun = @(lambda) interp1(x_data, y_data(x_data),lambda, 'linear', 'extrap'); %%%%FIRST FUNCTION
xq = linspace(5e-6,13.5e-6,10000);
plot(xq, coef_fun(xq))
%hold off
0 commentaires
Réponse acceptée
Ameer Hamza
le 8 Mai 2020
No need to index y_data with x_data
coef_fun = @(lambda) interp1(x_data, y_data, lambda, 'linear', 'extrap'); %%%%FIRST FUNCTION
0 commentaires
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Interpolation dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!