I did remove the SAS VM from virtual box. Then added the SAS VM from virtual box.  SAS studio then started and did the download and successful update.  I tried to do this early in the morning before the server was busy.  Thanks for the reply ... View more

data; infile cards delimiter='09'x; Input @1 name $1. @3 price: date: date12. time: time8.; format date date10. time time12. ; cards; A 3 2-Apr-08 9:44:38 A 5 2-Apr-08 9:53:02 A 7 2-Apr-08 9:55:23 A 4 2-Apr-08 9:58:31 0.333333 A 6 2-Apr-08 11:17:33 A 2 2-Apr-08 11:18:33 A 1 2-Apr-08 11:19:33 -0.83333 A 4 2-Apr-08 11:57:33 A 8 7-Apr-08 9:08:39 A 4 7-Apr-08 9:10:22 A 2 7-Apr-08 9:13:41 -0.75 A 3 7-Apr-08 9:32:14 A 7 7-Apr-08 9:41:04 1.333333 A 8 17-Apr-08 9:15:51 A 2 17-Apr-08 10:24:15 B 4 22-Jul-08 10:16:19 B 7 22-Jul-08 10:20:51 0.75 B 8 22-Jul-08 11:12:19 B 4 22-Jul-08 11:15:27 B 2 22-Jul-08 11:17:30 B 5 22-Jul-08 11:18:36 -0.375 B 1 22-Jul-08 11:31:50 B 3 14-Oct-08 9:05:07 B 1 14-Oct-08 9:37:11 B 6 14-Oct-08 9:46:16 B 7 14-Oct-08 9:50:56 B 7 14-Oct-08 9:52:52 B 8 14-Oct-08 9:59:23 7 B 5 14-Oct-08 10:05:13 B 7 14-Oct-08 11:19:44 B 2 14-Oct-08 11:20:15 -0.71429 ; * proc print; run; data one; set; by name date ; retain start; format start time12.; if first.date then start=time; else if time-start ge '1800' then halfhour+1; *proc print; run; data want; set; by name date halfhour; retain initial; if first.halfhour then initial=price; else if last.halfhour then return=(price-initial)/initial; proc print; run; ... View more

Thanks @art ... View more

Thank you @novinosrin ... View more

It could be done in one data step in SAS but there are several reasons why 2 data steps are better.   Data step 1 is not just inputting a flat file, it is flattening a hierarchical file with 2 different record types. Data step 2 uses the BY statement with last. feature that tells we are on the last record of this household.  The by statement is not valid after the infile statement. So to put the logic to detect last household   in the first data step would complicate it unnecessarily.   In programming world there is maintenance. Either you or your peer will need to update this program and keeping the functions of flattening the hierarchical file and summarizing the results in different data steps keeps the logic  straight forward and makes maintenance easier.   It could be done but would require lots more logic.   Jim       ... View more

Data household (drop=HorP_indicator) ; Infile datalines dsd ; retain Household_identity Type No_of_members; input HorP_indicator :$10. @ ; if length(HorP_indicator )>1 then do; input @1 Household_identity :$10. @12 Type $ No_of_members; *Member_no H_Indicator $ DOB :date9. Gender $ M_status $ Edu_level E_Status $ M_Income; end; else do; input @1 Member_no H_Indicator $ DOB :date9. Gender $ M_status $ Edu_level E_Status $ M_Income ; output; end; format DOB date9.; datalines; A123456789,A,3 1,Y,15Feb1960,M,M,3,FT,55000 2,N,3Jun1965,F,M,3,UE,0 3,N,24Jan1988,M,S,2,NA,3000 A135790234,B,1 1,Y,19Oct1944,F,D,0,NA,3000 B234523456,A,2 1,Y,30Jun1978,F,M,1,PT,4000 2,N,21May1975,M,M,2,FT,30000 proc print; run; /* summary */; data want; set household; by household_identity; if first.household_identity then age+intck('year',dob,today())-12; if e_status in ('FT','PT') then employed+1; income+M_income; if gender='M' then males+1; else females+1; if last.household_identity then do; income=income/no_of_members; output; employed=0; income=0; males=0; females=0; age=0; end; keep type males females age employed income household_identity; proc print label split='*'; id household_identity type males females employed age income; label males='number of* male members' females='number of*female members' age='age of *householder' income='average*household*income'; run;   Jim ... View more

If your two dates are in SAS date informat, the INTCK function returns the number of days between the two dates. duration=intck('day',effectivedate,enddate);   I think that you are going to want some sorting and logic to see if the shorter dates completely fit inside the longer periods.   Maybe sort by ID descending addrestype effectivedate duration. then delete observations that are overlapped.   jim ... View more

data new; retain want 0; input have; if have=. then do; want=.; end; else want+1; datalines; 2 1 9 . 1 2 . 5 1 2 . 9 7 5 ; proc print; run; ... View more

Not exactly your results but on the right track.   data dt1; input id sdate :date8.; format sdate date8.; cards; 1 3mar2017 1 5Mar2017 2 12Mar2017 3 1Apr2017 3 4Apr2017 3 20Apr2017 5 5May2017 ; data dt2; input id edate : date8.; format edate date8.; cards; 1 3Mar2017 2 20Apr2017 3 22Apr2017 3 1Apr2017 4 2Mar2017 6 3Jun2017 ; data want; merge dt1 (in=in1) dt2 (in=in2); by id; days=intck('day',sdate,edate); output; proc print; run;     It needs some logic to handle the case of more end dates then start dates for an id.      jim ... View more