If this is more than an exercise, I suggest that you refine your definition of "pregnancy start date" and "end of first/second/third trimester". A good place to start is the Wikipedia page on human Pregnancy. ... View more

Adapted from the SAS code provided with this publication, you could start with:   Data have; input year $ month $ count; monthDate = input(cats("01", substr(month, 1, 3), year), date9.); drop year month; format monthDate yymmdd10.; cards; 2017 Jan 0 2017 Feb 0 2017 Mar 2 2017 Apr 3 2017 May 1 2017 Jun 6 2017 Jul 12 2017 Aug 23 2017 Sept 25 2017 Oct 39 2017 Nov 40 2017 Dec 45 2018 Jan 44 2018 Feb 40 2018 Mar 38 2018 Apr 37 2018 May 35 2018 Jun 34 2018 Jul 24 2018 Aug 23 2018 Sept 25 2018 Oct 39 2018 Nov 40 2018 Dec 45 2019 Jan 0 2019 Feb 0 2019 Mar 2 2019 Apr 3 2019 May 1 2019 Jun 6 2019 Jul 12 2019 Aug 23 2019 Sept 25 2019 Oct 39 2019 Nov 40 2019 Dec 45 ; proc glimmix data=have; effect splDate = spline(monthDate); model count = splDate / dist=Poisson covb ddfm = residual; random _residual_; output out=glmixout pred(noblup ilink) = p lcl(noblup ilink)= l ucl(noblup ilink)= u; run; proc sgplot data=glmixout; band x=monthDate lower=l upper=u; scatter y=count x=monthDate; series y=p x=monthDate; xaxis type=time; run; (Not sure the confidence interval makes much sense at the beginning of the series... ) ... View more

With SQL you could do:   proc sql; select * from mydata group by id having abs(days) = min(abs(days)); quit; ... View more

To avoid replicated characters within a value use the permutation (k out of n) routine:   data want; array a $1 a_1-a_26; do i = 1 to dim(a); a{i} = byte(i - 1 + rank("A")); end; seed = 87896; length final $7; do i = 1 to 50; call ranperk(seed, 7, of a{*}); final = cats(of a_1-a_7); output; end; stop; keep i final; run;   ... View more

The function call removes leading zeros if starting at the first character. Is equivalent to prxchange("s/^0+//", 1, NBR). Note that prxchange("s/^\s*0+//", 1, NBR) would account for leading spaces. ... View more

No need to create copies of table A. Simply name it twice in SQL. To eliminate duplicate combinations you can arbitrarly pick the one where the first part is greater than the second.   proc sql; select * from A(rename=part1=part1a), A(rename=part1=part1b), B, C, D where part1a gt part1b; quit; ... View more

You don't say what's wrong with either result.   Maybe... data crs(where=(anl04fl='Y')); merge adbmk(in=bmk) adae(in=ae where=(crsfl='Y') rename=(anlfl=anl04fl)); by usubjid; if bmk; if not ae or (astdt-1 <= adt <= aeendt2 +3) then output; run; This will keep the usubjid that are absent from table adae. ... View more

Is usubjid a unique key in table adae?   Is usubjid a unique key in table adbmk?   Why is this a left join? The condition anl02fl="Y" will essentially transform it into an inner join. ... View more

Select the makes, then extract the sample:   proc sql; create table makes as select unique make from sashelp.cars; quit; proc surveyselect data=makes n=5 out=makes_5 seed=123; run; proc sql; create table cars_samp as select * from sashelp.cars where make in (select make from makes_5); drop table makes_5; quit; ... View more

If ID is indeed a primary key in my_small_in_table, then this should be an equivalent query:   proc sql; create table filtered_table as select * from my_huge_in_table where ID in (select ID from my_small_in_table where my_filter_variable = 1 ); quit; and might shed some light (or confusion!) on what's going on here. ... View more

At first sight, the only thing suspicious to me in your query is   select ...., PID, count(distinct PID) .... from .... group by PID   if you group by PID, then count(distinct PID) will be one, by definition.   Also note that the first subquery will require remerging because you do not group by INSERTDATETIME. This will likely generate more records than you expect. Either remove INSERTDATETIME from the select list or add INSERTDATETIME to the group by clause. ... View more