Replacing NaN values with average values of the nearest numbers?
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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 Comment
Answers (2)
Michael Haderlein
on 29 Oct 2014
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
1 Comment
Joshua Larkin
on 14 Dec 2018
Edited: Joshua Larkin
on 14 Dec 2018
Finally I found somthing that worked for what I needed. Thanks for the great answer!
Chad Greene
on 4 Nov 2014
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 Comment
Ismet Handzic
on 10 Apr 2020
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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