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Interpolating NaN-s

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Karlito
Karlito 2012 年 4 月 3 日
コメント済み: Matteo Soldini 2020 年 2 月 13 日
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 件のコメント
Jan
Jan 2012 年 4 月 3 日
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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採用された回答

Andrei Bobrov
Andrei Bobrov 2012 年 4 月 3 日
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

その他の回答 (2 件)

Jan
Jan 2012 年 4 月 3 日
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 件のコメント
Steven Lord
Steven Lord 2017 年 8 月 20 日
Check if the fillmissing function does what you want.
Matteo Soldini
Matteo Soldini 2020 年 2 月 13 日
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 2012 年 4 月 3 日
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 件のコメント
Jan
Jan 2012 年 4 月 3 日
+1. And it works on >1 D problems, where our direct INTERP1 approachs fail.

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