If I remember my number theory correctly, that's not the technical definition of significant digits which has to do with numeric precision.  There are rules around how to count them.  For example, 0.000052 has two significant digits while 0.520000 has 6. but...I may not be remember the theory correctly...and it may be more information than you actually need. ... View more

%let colour=blue; data _null_; if "&colour"="blue" then do;      call symput('fname','FilePath/FileA');      end; else do;      call symput('fname','FilePath/FileB');      end; run; %include "&fname"; ... View more

perhaps by sorting by chartno admdate descending and then processing similarly? ... View more

data have; input @1  chartno $6.    @7  acctno  $3.    @10 admdate yymmdd8.    @19 sepdate yymmdd8.; format admdate yymmdd10. sepdate yymmdd10.; cards; 1111110012012032920120331 2222220022012050120120515 3333330032012060120120607 1111110042012040220120410 4444440052012060320120615 5555550062012053120120603 2222220072012070120120713 2222220082012072020120725 4444440092012081320120817 3333330102012061520120621 1111110112012110120121130 run; proc sort in=have; by chartno acctno; run; data want (drop=acctno_lag sepdate_lag); set have; by chartno; format     sepdate_lag date9.     index_acctno $3.     ; index_acctno=''; acctno_lag=lag(acctno); sepdate_lag=lag(sepdate); if first.chartno then dif=.; else do;     dif = admdate - sepdate_lag;     if coalesce(dif,0) <=30 then index_acctno=acctno_lag;     end; run; ... View more

sorry...my background is SQL, not SAS.  Even after using SAS for 10 years, I still find new SAS ways of doing things that are a surprise. ... View more

How about this..although I didn't see any case where the indicator variable was anything other than 0 DATA HAVE; INPUT firmid year rating $ notch $; CARDS; 1001 1990 AAA middle 1001 1991 AA+ plus 1001 1992 AA+ plus 1001 1993 AA middle 1001 1994 AA+ plus 1002 1990 BBB+ plus 1002 1991 BBB+ plus 1002 1992 A- minus 1002 1993 A+ plus 1002 1994 A middle 1002 1995 A- minus 1003 1990 BBB middle 1003 1991 BBB- minus 1003 1992 BB+ plus 1003 1993 BB+ plus 1003 1994 BB- minus 1003 1995 BB- minus ; RUN; PROC SQL; CREATE TABLE WANT AS SELECT     Curr.*     ,case when lastyear.notch='middle' and curr.notch='plus' and year1.notch='plus' and year2.notch='plus' then 1 else 0 end as Indicator_Var FROM        HAVE as curr     left join have as LastYear         on curr.firmid=lastyear.firmid             and curr.year-1=lastyear.year     left join have as year1         on curr.firmid=year1.firmid             and curr.year+1=year1.year     left join have as year2         on curr.firmid=year2.firmid             and curr.year+2=year2.year; quit; ... View more

Presuming that you want all of the records associated with 80% of unique IDs to be identified: data have; input id score forum; cards; 12     89          98    12     87          67    13     56          87    13     45          98    14     78          98 15    23           87    16    54          23 ; proc sql; create table ids as select distinct id, 0 as id_rand_val from work.have order by id; update ids set id_rand_val=rand('uniform'); create table want as select     t1.*,     case when t2.id_rand_val <= .8 then 'Group1' else 'Group2' end as ID_Group from     work.have t1     inner join work.ids t2         on t1.id=t2.id; quit; ... View more

how about CALL SLEEP(rand(uniform)*120000, .001); ... View more

That was IT's suggested approach...use EZTrieve to create a snap text extract and download it....sorta old school, isn't it. ... View more