How to replace NaN with column mean if less than b NaN in a column?

Hello, Say I have the martix:

MA=[1 2 3 NaN; 6 NaN NaN 9; NaN NaN NaN 9;NaN 45 NaN 9;NaN NaN NaN 19;1 12 3 34] I would like to replace the NaNs in each column with the average of the column if the number of NaNs in the column is less than 4. Any easy way to do this please? I know that to find the average of the column I can use the nanmean function. Note that the actual matrices that I have are much larger, but I know the total number of rows and columns.

thanks, K

Answer 1

[Andrei Bobrov]

n = nanmean(MA);
nn = isnan(MA);
ii = sum(nn) < 4;
z = MA(:,ii);
z(nn(:,ii)) =  nonzeros(bsxfun(@times,nn(:,ii),n(ii)));
MA(:,ii) = z;

or

n = nanmean(MA);
nn = isnan(MA);
ii = bsxfun(@and,nn,sum(nn) < 4);
MA(ii) = n(nonzeros(bsxfun(@times,ii,1:numel(n))));

or

n = nanmean(MA);
nn = isnan(MA);
ii = bsxfun(@and,nn,sum(nn) < 4);
[~,idx] = find(ii);
MA(ii) = n(idx);

[Katerina F]

Thanks, this seems nice, because there is no loop, but in a slightly different case it gives me a strange result: Say: MA=[1 2 NaN NaN; 6 NaN NaN 9;4 NaN NaN 9;NaN 45 NaN 9;5 NaN NaN 19;NaN 12 NaN 34], which is: MA =

     1     2   NaN   NaN
     6   NaN   NaN     9
     4   NaN   NaN     9
   NaN    45   NaN     9
     5   NaN   NaN    19
   NaN    12   NaN    34

then, the result is: MA =

    1.0000    2.0000   16.0000       NaN
    6.0000   19.6667       NaN    9.0000
    4.0000   19.6667       NaN    9.0000
    4.0000   45.0000       NaN    9.0000
    5.0000   19.6667       NaN   19.0000
    4.0000   12.0000       NaN   34.0000
_

[Andrei Bobrov]

corrected

[Katerina F]

Thank you very much Andrei, it works fine. K.