/*********************************GET TINS WITH TOTAL ADMISSIONS >=10 ONLY*********************/
PROC SQL;
CREATE TABLE ReAdm.TIN_Merge3 AS
(SELECT *
FROM ReAdm.TIN_Merge2
WHERE ADMIT >=10
GROUP BY TIN);
RUN;
I need to code this to grab all my severities. Right now it will only grab those that have admit >=10 and my example table looks like this
who svrty admit
1111 1 5
1111 2 4
1111 3 12
1111 4 7
If I run the code above it only returns the svrty 3 and I need it actually to return all because the admits summed together are 28. I guess I need to code it where the sum of admit >=10 so it will return all svrty items.
try:
PROC SQL;
CREATE TABLE ReAdm.TIN_Merge3 AS
SELECT *
FROM ReAdm.TIN_Merge2
where tin in (select tin from ReAdm.TIN_Merge2
GROUP BY TIN
having sum(ADMIT) >=10)
;
quit;
PROC SQL;
CREATE TABLE ReAdm.TIN_Merge3 AS
SELECT *
FROM ReAdm.TIN_Merge2
GROUP BY TIN
having sum(ADMIT) >=10
;
RUN;
That does not return anything other than
who svrty admit
1111 3 12
It leaves out the other svrty codes which I need them to be returned. A case where it would not be returned is this:
who svrty admit
2222 1 1
2222 2 3
2222 3 0
2222 4 0
None of these would be returned because the sum of the admit is not >=10 it is 4
I need all severities returned if the who has an admit sum of >=10 so in my first example all 4 severities should be returned because the sum of the TIN is 28
From your example data your group by variable should be WHO not TIN.
try:
PROC SQL;
CREATE TABLE ReAdm.TIN_Merge3 AS
SELECT *
FROM ReAdm.TIN_Merge2
where tin in (select tin from ReAdm.TIN_Merge2
GROUP BY TIN
having sum(ADMIT) >=10)
;
quit;
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 the difference between classical and Bayesian statistical approaches and see a few PROC examples to perform Bayesian analysis in this video.
Find more tutorials on the SAS Users YouTube channel.