Interpolating NaN-s
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    Karlito
 el 3 de Abr. de 2012
  
    
    
    
    
    Comentada: Matteo Soldini
 el 13 de Feb. de 2020
            I have a time series, where there are some missing values. I've marked them as NaN. How would it be possible for me to interpolate them from the rest of the data. So my data to interpolate looks like that (just example numbers):
x=
0.482230405436799
0.0140930751890233
0.622880344434796
NaN
0.527433962776634
0.724991961357239
0.607415791144935
0.588366445795251
0.433434840033058
0.244172893507025
0.428960353778007
0.0101774555217771
0.608821449044360
0.957975188197358
NaN
0.0355905633801477
0.886235111325751
0.246941365057497
0.00891505625333700
0.814920271448039
0.140499436548767
0.879866440284003
0.0953767860905247
0.352560101248640
How to get a value for those NaN-s? I've tried Interp1 but i can't get it working properly.
1 comentario
  Jan
      
      
 el 3 de Abr. de 2012
				It is a good idea to post your trials with INTERP1, such that the problem can be fixed. This is usually easier that creating a solution from the scratch and you could learn, what went wrong.
Respuesta aceptada
  Andrei Bobrov
      
      
 el 3 de Abr. de 2012
        inn = ~isnan(x);
i1 = (1:numel(x)).';
pp = interp1(i1(inn),x(inn),'linear','pp')
out = fnval(pp,linspace(i1(1),i1(end),1000));
plot(i1,x,'ko',linspace(i1(1),i1(end),1000),out,'b-')
grid on
0 comentarios
Más respuestas (2)
  Jan
      
      
 el 3 de Abr. de 2012
        nanx = isnan(x);
t    = 1:numel(x);
x(nanx) = interp1(t(~nanx), x(~nanx), t(nanx));
This does not catch NaNs at the margins, such that you have to care for them by your own, e.g. by repeating the first or last non-NaN value.
8 comentarios
  Matteo Soldini
 el 13 de Feb. de 2020
				If I have 50 variables and I use Jan's code to interpolate each nan in each variable (variables have different ranges of values, for example var1 = [1,2], var2 = [300,400]), will it work or should I use a for loop?
  John D'Errico
      
      
 el 3 de Abr. de 2012
        My inpaint_nans will fill them in with interpolated values, even if there are multiple consecutive nans, and it will extrapolate nicely at the ends. It works on 1-d problems.
x = inpaint_nans(x);
1 comentario
  Jan
      
      
 el 3 de Abr. de 2012
				+1. And it works on >1 D problems, where our direct INTERP1 approachs fail.
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