Replacing NaN values with average values of the nearest numbers?

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

[Dyuman Joshi]

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

Answer 1

[Michael Haderlein]

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

[Joshua Larkin]

Finally I found somthing that worked for what I needed. Thanks for the great answer!

Answer 2

[Chad Greene]

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

[Ismet Handzic]

hmmm.... only seems to work when there are bounded values, like that last 12 at the end there. But good suggestion!