BerndSE   You said you are new to the Community. Your problem is not for the beginner. I try to give a data step solution which is not as good as the hash solution. There is an inefficiency to search for the RANGE all over the records. If we sort the data set by age, then we can control the range and will not look records that has an age greater than the greater of the range. I am leaving it to you. The steps are: [1] Take each observation. Get the range of age. [2] Using POINT= option, examine each observation whether its AGE falls in the Range. If so, accumulate height and count. [3] At the ending of reading all data, output mean height. data sample; input Name $ Age Height; datalines; Max 10 120 Paul 12 140 Rick 15 130 Morti 8 90 Carl 9 100 Maria 14 150 Claus 13 160 Mario 6 50 Tina 11 100 ; run; %let range = 1; data want; set sample nobs = num ; loage = age - &range; hiage = age + &range; sumheight = 0; count = 0; do p = 1 to num; set sample(keep = age height rename=(height = ht age = age1)) point = p; flag = (loage <= age1 <= hiage); if flag > 0 then do; sumheight + ht; count + 1; end; end; averageheight = sumheight / count; drop age1 ht flag; run; Name Age Height loage hiage sumheight count averageheight Max 10 120 9 11 320 3 106.667 Paul 12 140 11 13 400 3 133.333 Rick 15 130 14 16 280 2 140.000 Morti 8 90 7 9 190 2 95.000 Carl 9 100 8 10 310 3 103.333 Maria 14 150 13 15 440 3 146.667 Claus 13 160 12 14 450 3 150.000 Mario 6 50 5 7 50 1 50.000 Tina 11 100 10 12 360 3 120.000 Another hash solution for you.   %let range = 1; data want; if _n_ = 1 then do; if 0 then set sample; declare hash h(dataset:"sample", ordered: 'Y'); h.definekey('Age'); h.definedata('Height'); h.definedone(); end; set sample; lo_age = age - &range; hi_age = age + &range; sumheight = 0; count = 0; do nage = lo_age to hi_age; if h.find(key:nage) = 0 then do; sumheight + height; count + 1; end; end; averageHeight = sumheight / count; run;  Hope these answers are acceptable to you. As you are new, please see there is a wheel on the top right corner. Click the Arrow next to it. Choose to tick what you like. ... View more

Better understood now data have; input id year sex a1 a2 a3 a4 a5; datalines; 1 1994 1 1 1 2 2 2 2 1992 1 1 4 4 4 4 3 1992 2 1 2 3 3 3 4 1991 1 1 2 4 4 4 5 1992 2 1 2 2 4 4 6 1985 2 1 3 4 4 4 7 1988 2 1 3 3 3 3 ; run; data need; array k[5] _temporary_; do until(eof); set have end = eof; array a a1--a5; do i = 1 to dim(k); k[i] = a[i]; end; /** Process the conditions **/ do i = 1 to dim(k); if k[i] = 4 then do; if k[i-1] = 2 then do; do j = i to dim(k); k[j] = 3; end; end; else if k[i-1] = 3 then do; do j = i to dim(k); k[j] = 2; end; end; else if k[i-1] = 1 then do; k[i] = 2; do j = i+1 to dim(k); k[j] = 3; end; end; end; end; do i = 1 to dim(k); a[i] = k[i]; end; output; end; drop i j; run; Obs id year sex a1 a2 a3 a4 a5 1 1 1994 1 1 1 2 2 2 2 2 1992 1 1 2 3 3 3 3 3 1992 2 1 2 3 3 3 4 4 1991 1 1 2 3 3 3 5 5 1992 2 1 2 2 3 3 6 6 1985 2 1 3 2 2 2 7 7 1988 2 1 3 3 3 3 . It is easier to copy a1 - a5 into a _temporary_ array first. Then process the array for 4, and apply the conditions. Then copy the elements of array to a1 - a5 and write the output.   ... View more