I need to fill the nulls of a column with the mean sum of the division of two columns multiplied by one column and rest the previous An example would be A B_01 B_02 ... B_60 5 . . 5 2 3 7 3 1,2 9 3 0,3 4 . . Well, I would like the missing value for column B_01 to be (2/5 + 3/7 + 3/9) / 3 * its corresponding column A For column B_02(3/5 + 1,2/7 + 0,3/9)/3 * its corresponding column A - his new value in B_01 I have thought about doing this, but it turns out that I have 60 columns with which to do it and the only way it comes to mi mind is to do this 60 times. thanks Proc sql; create table new as Select * , sum(B_01/A)/sum(case when B_01 is missimg then . else 1)*A end as new_B_01 , sum(B_02/A)/sum(case when B_02 is missimg then . else 1)*A-B_01 end as new_B_02 from table_one ; ... View more

I need to fill the nulls of a column with the mean sum of the division of two columns An example would be A B C ... B_01 C_01 5 . 5 2 7 3 9 3 4 .   Well, I would like the missing value for column B to be (2/5 + 3/7 + 3/9) / 3 * its corresponding column A, which for the first row would be 2.07   I have thought about doing this, but it turns out that I have 60 columns with which to do it and the only way it comes to mi mind is to do this 60 times.   Proc sql créate table new as Select *, sum(B/A)/sum(case when B is missimg then . else 1) end as new_B, sum(B/A)/sum(case when B is missimg then . else 1) end as new_B_01 from table_one; Quit;   Thanks ... View more