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

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Katerina F
Katerina F 2014-10-20
评论： Katerina F 2014-10-22
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

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Andrei Bobrov
Andrei Bobrov 2014-10-20
编辑：Andrei Bobrov 2014-10-21
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);
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