interp1 function error

1 visualización (últimos 30 días)
fima v
fima v el 7 de Mayo de 2020
Respondida: Ameer Hamza el 8 de Mayo de 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

Respuesta aceptada

Ameer Hamza
Ameer Hamza el 8 de Mayo de 2020
No need to index y_data with x_data
coef_fun = @(lambda) interp1(x_data, y_data, lambda, 'linear', 'extrap'); %%%%FIRST FUNCTION

Más respuestas (0)

Categorías

Más información sobre Graph and Network Algorithms en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by