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Interpolating NaN-s

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Karlito
Karlito il 3 Apr 2012
Commentato: Matteo Soldini il 13 Feb 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 Commento
Jan
Jan il 3 Apr 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.

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Risposta accettata

Andrei Bobrov
Andrei Bobrov il 3 Apr 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

Più risposte (2)

Jan
Jan il 3 Apr 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 Commenti
Steven Lord
Steven Lord il 20 Ago 2017
Check if the fillmissing function does what you want.
Matteo Soldini
Matteo Soldini il 13 Feb 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?

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John D'Errico
John D'Errico il 3 Apr 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 Commento
Jan
Jan il 3 Apr 2012
+1. And it works on >1 D problems, where our direct INTERP1 approachs fail.

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