Assuming I understood
data have;
input id date clinic ;
cards;
1 20020408 12135 1 1 1
1 20020715 12135 2 4 4
1 20030124 15109 1 1 5 /*4+1*/
1 20030308 75423 1 1 6 /*4+1+1*/
1 20030505 75423 2 4 9 /*4+1+4*/
1 20030607 75423 3 9 14 /*4+1+9*/
1 20030815 12135 3 9 19 /*9+1+9*/
2 20020615 25875 1 1 1
2 20020717 25875 2 4 4
2 20020824 55415 1 1 5 /*4+1*/
2 20020907 25875 3 9 10 /*9+1*/
2 20021215 25875 4 16 17 /*16+1*/
2 20030108 25875
;
run;
data want;
if _n_=1 then do;
if 0 then set have;
declare hash h();
declare hiter hi('h');
h.definekey('clinic');
h.definedata('clinic','count','countsquare');
h.definedone();
end;
set have;
by id;
if first.id then h.clear();
if h.find()=0 then do;count=count+1;countsquare=count**2;h.replace();end;
else do;count=1;countsquare=1;h.replace();end;
_clinic=clinic;_countsquare=countsquare;sum=0;
do while(hi.next()=0);
if _clinic ne clinic then sum+countsquare;
end;
sum+_countsquare ; clinic=_clinic;
drop _:;
run;
Hi @km0927 Sorry for the delay as I slept like a baby. Nevertheless, what better help than genius Ksharp who has inspired me a lot. I personally would choose his solution as it is very capable of scaling higher and of course just for the reason I committed last night , I am sharing mine. I hope your weekend is going well.
data have;
input id date clinic ;
cards;
1 20020408 12135 1 1 1
1 20020715 12135 2 4 4
1 20030124 15109 1 1 5 /*4+1*/
1 20030308 75423 1 1 6 /*4+1+1*/
1 20030505 75423 2 4 9 /*4+1+4*/
1 20030607 75423 3 9 14 /*4+1+9*/
1 20030815 12135 3 9 19 /*9+1+9*/
2 20020615 25875 1 1 1
2 20020717 25875 2 4 4
2 20020824 55415 1 1 5 /*4+1*/
2 20020907 25875 3 9 10 /*9+1*/
2 20021215 25875 4 16 17 /*16+1*/
2 20030108 25875
;
run;
proc sql noprint;
select max(c) into : n
from (
select count(id) as c
from have
group by id);
quit;
data want;
array t(&n,2);
array j(*) t:;
array m(%eval(&n*2)) ;
call missing(of t(*),of j(*),of m(*));
do until(last.id);
set have;
by id;
if not missing(clinic) and clinic not in t then do;
n+1;
t(n,1)=clinic;
count=1;
count_square=count**2;
t(n,2)=count;
end;
else if clinic in t then do;
k=whichn(clinic, of t(*));
count=j(K+1)+1;
count_square=count**2;
j(K+1)=count;
end;
m(whichn(clinic,of j(*)))=count_square;
sum=sum(of m(*));
output;
drop m: t: k n ;
end;
run;
thank you for all your help!
thanks to your help, i can complete my work safely.
Are you ready for the spotlight? We're accepting content ideas for SAS Innovate 2025 to be held May 6-9 in Orlando, FL. The call is open until September 25. Read more here about why you should contribute and what is in it for you!
Learn how use the CAT functions in SAS to join values from multiple variables into a single value.
Find more tutorials on the SAS Users YouTube channel.
Select SAS Training centers are offering in-person courses. View upcoming courses for: