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It's with SAS data steps and especially SAS proc's often easier to work with narrow data structures. That's what below code creates. data have; infile datalines dsd truncover; input member (start1 end1 start2 end2) (:date11.) /*days2019 days2020 days2021 days2022 days2023*/ ; format start1 end1 start2 end2 date11.; datalines; 123,1-Mar-20,1-Apr-20,1-Feb-22,30-Jun-23,0,31,0,333,180 456,1-Jan-19,1-Jan-20,,,365,1,0,0,0 789,13-Apr-23,15-Apr-23,,,0,0,0,0,3 ; data want; set have; array dates{2,2} start1 end1 start2 end2; format start_dt end_dt from_dt to_dt date11.; do k=1 to dim1(dates); if nmiss(dates[k,1],dates[k,2]) then continue; start_dt =dates[k,1]; end_dt =dates[k,2]; from_dt=start_dt; do i=0 by 1; to_dt=min(intnx('year',start_dt,i+1,'b'),end_dt); year=year(from_dt); days=to_dt-from_dt; output; if to_dt=end_dt then leave; from_dt=intnx('year',start_dt,i+1,'b'); end; end; drop start1 end1 start2 end2 k i; run; proc print data=want; run;   ... View more

I think you have chosen a poor layout of the data to do this. A better layout would be   Want: ID_1                  DOB                    year                                                   age A2543              08OCT1993         2018                                                   26                   A2543              08OCT1993         2019                                                   27 A3667              12JUL2003          2018                                                   15 A3667              12JUL2003          2019                                                   16   @gabagotati wrote: Thanks for you response. The reason I am interested in extracting the specific month and day is that is I have another variable (date of enrollment), and I'd like to find their age at this date.   For A2543 in my example, they are 18 after their birthday of 08OCT2018 and before their next birthday 08OCT2019. If their enrollment date was on 01JUN2019, they would be 18 on this day. Your solution would yield age_2019 of 19, which would not be what I want.   Apologies for not including this piece in my original post.       However, to address the problem you now state, I make up some data and address the solution.   data have; input ID $ DOB :date9. date_of_enrollment :date9.; cards; A2543 08OCT1993 01JUN2019 A3667 12JUL2003 12JUL2020 A3624 30MAR2000 19JAN2021 ; data want; set have; age_at_enrollment = intck('year',dob,date_of_enrollment,'c'); run;   ... View more

data have; input (ID_1 ID_2 ID_3 ID_4)($); datalines; A2543 A3700 C3404 A4314 A3667 C1000 F3123 A3414 A3624 D1434 F1334 A0394 A6000 A4303 A3802 A4137 B7000 B6205 B5310 B1394 ; data relevant_ID; length salesID $3; input salesID; datalines; A36 A37 A38 A39 A40 A41 A42 A43 A44 A45 A46 A47 A48 A49 A50 ; proc sql noprint; select distinct salesID into :salesID separated by '|' from relevant_ID; quit; options noquotelenmax; data want; set have; length all $ 4000; all=catx('|',of ID_:); relevant_match=0; if prxmatch("/&salesID./",all) then relevant_match=1; drop all; run; ... View more

data have; input subject $ year disease1 disease2 disease3; datalines; a 2019 1 1 1 a 2020 0 0 0 a 2021 0 0 0 a 2022 0 0 0 b 2019 0 1 1 b 2020 1 0 0 b 2021 1 0 0 b 2022 0 0 1 ; data d1; set have(where=(disease1=1)); year=year+1;output; year=year+1;output; keep subject year; run; data d2; set have(where=(disease2=1)); year=year+1;output; year=year+1;output; keep subject year; run; data d3; set have(where=(disease3=1)); year=year+1;output; year=year+1;output; keep subject year; run; data want; if _n_=1 then do; declare hash d1(dataset:'d1'); d1.definekey('subject','year'); d1.definedone(); declare hash d2(dataset:'d2'); d2.definekey('subject','year'); d2.definedone(); declare hash d3(dataset:'d3'); d3.definekey('subject','year'); d3.definedone(); end; set have; if d1.check()=0 then disease1=1; if d2.check()=0 then disease2=1; if d3.check()=0 then disease3=1; run;   ... View more