data Session; input ID SessionTime v1 v2 v3 ; cards; 1 18.2 2 3 6 1 18.2 3 7 8 1 18.2 4 6 12 1 18.2 6 8 10 1 18.2 8 9 22 1 18.2 9 10 25 1 18.2 10 12 30 1 16.7 12 17 4 2 16.7 3 7 6 2 16.7 4 11 10 2 16.7 5 10 8 2 16.7 7 12 20 2 16.7 9 13 23 2 16.7 10 14 28 2 16.7 11 16 2 2 16.7 13 21 4 3 18.2 4 11 8 3 18.2 5 15 6 3 18.2 6 14 18 3 18.2 8 16 21 3 18.2 10 17 26 3 18.2 11 18 0 3 18.2 12 20 2 3 18.2 14 25 6 4 19.3 5 15 4 4 19.3 6 19 16 4 19.3 7 18 19 4 19.3 9 20 24 4 19.3 11 21 7 4 19.3 12 22 0 4 19.3 13 24 4 4 19.3 15 29 2 5 20.3 6 19 14 5 20.3 7 23 17 5 28.4 8 22 22 5 28.4 10 24 5 5 28.4 12 25 5 5 28.4 13 26 2 5 28.4 14 28 0 5 28.4 16 33 12 ;;; run; proc sql ; create table chk as select id, sessiontime, count(*) as frq from Session group by id, sessiontime having frq=8 order by id, sessiontime ; quit; proc sql; create table session_ok as select s.* from session s where id in (select id from chk); quit; Also ID=1 have 18.2 and 16.7 different SessionTime. ... View more

Data have1; input AccountID trasactionKey; cards; 123 222777 123 999999 456 333888 ; Run; Data have2; Input AccountID customerID; cards; 123 999 123 888 123 333 123 222 456 333 ; Run; /**** sql full join ****/ proc sql; create table Want as select * from have1 full join have2 on have1.AccountID=have2.AccountID; quit; ... View more

data rawcode; input code $3.; cards; 001 026 099 126 523 525 ; data ab_code; input acode :$3. bcode :$7. desc :$10.; cards; 001 001 chairs 027 026 cups 099 027 plates 126 098 pencils 399 112 books 400 113 pens 522 523 clips 525 525 pillows ; run; proc sql; create table want as select c.code, ab.desc from rawcode C inner join ab_code ab on (c.code =ab.acode or c.code =ab.bcode); quit; ... View more

perfect ! ... View more

very elegant program ! ... View more

why calculate count(*) as Accounts_Count... ? proc sql; create table want as select A.* from A as A inner join (select distinct party from A where balance > 50 ) as B on A.party = B.party; quit; ... View more

data try; input name $ yesno $ price; datalines; chris yes 10 chris no 10 chris yes 30 chris yes 40 chris no 50 chris yes 60 ken yes 60 ken yes 40 ken no 30 ken yes 10 tom no 10 tom no 20 tom no 30 tom yes 70 ; run; proc sql; select name , sum(yesno='yes') as yes, sum(yesno='no') as no , sum( price * (yesno='yes')) as sum_of_yes, sum( price * (yesno='no')) as sum_of_no from try group by name ; quit; ... View more