From where do you get the list of possible values for "Item"? You need that to generate records when there is nothing in data1. ... View more

proc sql; select * from data1 a, data2 b where a.id = b.id; quit; If an ID appears more than once in data2, you will get the cartesian product (eg 3 records for '1' in data1, 5 records for '1' in data2 = 15 records for '1' in want). If data2 has only one record per ID, you can do it in a data step. (assuming both datasets are sorted by ID) data want; merge   data1 (in=a)   data2 (in=b) ; by id; if a and b; run; ... View more

This is a more complete approach: %let start=1.6.0.0; %let end=1.7.255.255; data ip_ad (keep=ip_address); start_ad = 0; do i = 1 to 4;   i1 = input(scan("&start",i,'.'),3.);   start_ad = start_ad + i1 * 256**(4-i); end; end_ad = 0; do i = 1 to 4;   i1 = input(scan("&end",i,'.'),3.);   end_ad = end_ad + i1 * 256**(4-i); end; length ip_address $ 15; do address = start_ad to end_ad;   ip_address = ' ';   char_ad = put(address,hex8.);   do i = 0 to 3;     ip_address = compress(ip_address !! put(input(substr(char_ad,i*2+1,2),hex2.),best3.));     if i ne 3 then ip_address = trim(ip_address) !! '.';   end;   output; end; run; RW9's example would have problems with start=1.7.0.0 end=2.6.255.255 ... View more

Or, in a code node, use (x being your merge variable(s)) proc sort data=A; by x; run; proc sort data=B; by x; run; data C; merge   A (in=in_a)   B (in=in_b) ; by x; if in_a and in_b; run; ... View more