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Replacing NaN values with average values of the nearest numbers?

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Isaac
Isaac 2014-10-29
评论： Dyuman Joshi 2023-12-13
Hi, I have a data set with blocks of NaN Values. Is there a command that can fill in missing values by using the average of the values on either side?
For example, in the data set:
2
3
5
7
NaN
NaN
12
Is there a way to get something like 8.66 and 10.33 for the NaN numbers?
Also, would this work if there were multiple columns?
Thanks, Isaac
  1 个评论
Dyuman Joshi
Dyuman Joshi 2023-12-13
Ran in:
For newer versions, fillmissing (Introduced in R2016b) is a robust option -
nandata=[2;3;5;7;nan;nan;12];
fillmissing(nandata, 'linear')
ans = 7×1
2.0000 3.0000 5.0000 7.0000 8.6667 10.3333 12.0000

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回答（2 个）

Michael Haderlein
Michael Haderlein 2014-10-29
Hi Isaac,
you can use interp1:
nandata=[2;3;5;7;nan;nan;12];
xdata=(1:length(data))';
data=interp1(xdata(~isnan(nandata)),nandata(~isnan(nandata)),xdata)
data =
2.0000
3.0000
5.0000
7.0000
8.6667
10.3333
12.0000
For multiple columns, you'll need bsxfun:
nandata2=[2 1;3 nan;5 2;7 nan;nan nan;nan 1;12 10];
xdata2=(1:size(nandata,1))';
data2=bsxfun(@(x,y) interp1(y(~isnan(x)),x(~isnan(x)),y),nandata2,xdata2)
data2 =
2.0000 1.0000
3.0000 1.5000
5.0000 2.0000
7.0000 1.6667
8.6667 1.3333
10.3333 1.0000
12.0000 10.0000
Chad Greene
Chad Greene 2014-11-4
With repnan,
nandata=[2;3;5;7;nan;nan;12];
data = repnan(nandata)
2.0000
3.0000
5.0000
7.0000
8.6667
10.3333
12.0000
  1 个评论
Ismet Handzic
Ismet Handzic 2020-4-10
hmmm.... only seems to work when there are bounded values, like that last 12 at the end there. But good suggestion!

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