how to check days overlap

[Jim_G]

Michelle    Good job of checking the output data.  I see the problem and am busy today but I will keep working on it.    Jim

[michellel]

Thank Jim!

[Jim_G · Accepted solution]

Michelle       Try this code.   It doesn't concentrate on how many drug overlaps per day but which type of overlap occurred.  It there were 5 overlaps of drug type A It shows an overlap of type A.  If there was 5 drug A overlaps on a certain day and 1 B drug on the same day, it would report a type A overlap and a type A&B overlap.  Jim

** expand each date range to one record per day;

data one;

input id  type $ drug $  (begin_date   end_date) (: mmddyy12.);

format begin_date  end_date ondate mmddyy10.;

*   put out a record for each day on the drug by type;

ondate=begin_date;

do until (ondate=end_date);

  output;  ondate+1;

end;

output;  drop begin_date  end_date;

cards;

1 a drug1  03/01/14 03/10/14

1 a drug1  03/11/14 03/20/14

1 a drug1  03/25/14 04/05/14

1 a drug2  03/02/14 03/09/14

1 a drug2  03/17/14 03/24/14

1 a drug2  03/27/14 04/04/14

1 a drug3  09/02/14 09/09/14

1 b drug4  03/01/14 03/10/14

1 b drug4  03/11/14 03/20/14

1 b drug4  03/25/14 04/05/14

1 b drug5  09/02/14 09/09/14

2 a drug1  03/01/14 03/10/14

2 a drug1  03/11/14 03/20/14

2 a drug1  03/25/14 04/05/14

2 a drug2  03/02/14 03/09/14

2 a drug2  03/17/14 03/24/14

2 a drug2  03/27/14 04/04/14

2 a drug3  09/02/14 09/09/14

;

**  sort by day to see how many drugs taken each day ;

proc sort;  by id ondate type drug ;

proc print; by id ondate; id id ondate ;

title 'how many drugs taken each day';  run;

* count number of drugs taken each day by id type;

*  collapse to one record per string of days;

data;  set;  by id ondate; * type drug ; retain typa typb;

if first.ondate then do; count=0; typa=0; typb=0; end;

if type='a' then typa+1;

if type='b' then typb+1;

count+1;

if last.ondate and count gt 1 then do;

   if typa and     typb then do;  olap='A&B'; output; end;

   if typa gt 1         then do;  olap='A';   output; end;

   if typb gt 1         then do;  olap='B';   output; end;

end;  

drop drug type typa typb;

proc sort;    by id olap ondate;

proc print;   id id olap  ondate; 

title 'Collapse by date and type where multiple drugs taken'; run;

  ** step 3 identify the string of contiguous days (episode);

data;   set;   by id olap;

dif=ondate-lag(ondate);

retain flag  strtdate; 

format strtdate mmddyy10.;

if first.olap then do;

   contdys=0; flag=0; dif=1; strtdate=ondate;

end;

if dif=1 then contdys+1;

else do;   ** end of continuous string;

    dif=1; strtdate=ondate;

    contdys=1;

end;   

drop   dif flag count;

proc print; id id olap strtdate ondate contdys olap ;

    by id olap strtdate;

title 'See episode data  overlap of drugs and type'; run;

***step 4  see if the continuous string of days was gt 5;

data;   set;   by id olap strtdate;

     drop ondate ;

  if last.strtdate then do;    

        if contdys gt 5 then output;

  end;   

proc print split='*';

  label id='id*'  strtdate='episode*start date'

    contdys='continous* days' olap='type'; 

     id id strtdate contdys olap;  by id; run;  

*/;

[michellel]

Hi Jim, Thanks so much!!!!!!!!!!!!!!!!!! You are the super!