Replace NaN values with median for each row

Hi guys,

I have this table:

Date var_1 var_2 var_3 var_4 var_5 var_6

01-01-90 0.2 0.34 0.43 0.12 0.18 0.1

01-01-91 0.3 0.54 0.53 0.62 0.38 0.2

01-01-92 0.7 0.40 0.77 0.42 NaN 0.2

01-01-93 0.7 0.40 0.77 0.42 NaN 0.2

How do I replace the NaN values in the table with the median of the row for each row?

The result for this table should be:

Date var_1 var_2 var_3 var_4 var_5 var_6

01-01-90 0.2 0.34 0.43 0.12 0.18 0.1

01-01-91 0.3 0.54 0.53 0.62 0.38 0.2

01-01-92 0.7 0.40 0.77 0.42 Med_Row 0.2

01-01-93 0.7 0.40 0.77 0.42 Med_Row 0.2

[KSSV]

Read about fillmissing

Answer 1

[madhan ravi]

m = TablE{:,2:end};
ix = any(isnan(m),2);
mix = m(ix,:);
m1 = median(m(ix,:),2,'omitnan') .* isnan(m(ix,:))
m1(m1==0) = mix(~isnan(mix));
m(ix,:) = m1
TablE{:,2:end} = m

[gcicceri]

what if there are more Nan values in the row?

[madhan ravi]

See edited answer.

[gcicceri]

Thanks, could you write the comments for each line code to understand all the steps?

[madhan ravi]

1) converting table to a matrix

2) detecting which rows of matrix has nans in it

3) saving those twos of matrix in a variable called mix

4) calculating median for those rows and assigning it to the matrix where nana occurs

5) copying the values other than nan

6) replacing those rows to the original matrix

7) converting it back to a table

[gcicceri]

Perfect. thanks