USING retain statement - sas

Question

Assume you have a temporary SAS data set called EPISODES that contains information about hospital episodes. The data set contains the variables ID_NO (patient ID), ADMIT_DATE (date of admission), DISC_DATE (date of discharge), and TOTAL_COST.
Using this data set, create a new data set in which you will create a separate observation for each day of each hospital episode. If a patient had a hospital episode that was 3 days long, they would have three views in the new data set from that episode -- one for each day.
Each observation in the new data set should have only three variables: the patient identifier ID_NO, the date for that particular day of hospitalization XDATE, and the cost for that day of hospitalization DAILY_COST = TOTAL_COST divided by the number of days in the episode.

Answer 1

My thought is to do this as a loop. Something like the following.
data new_data;
set input_data ;
do xdate = admit_date to disc_data;
daily_cost = .... ;
output new_data ( keep = xdate daily_cost id_no );
end;
run;

Answer 2

*This program block sets up our data set;
data episodes;
INPUT ID_NO $ ADMIT_DATE mmddyy10. TOTAL_COST DISC_DATE mmddyy10.;
DATALINES;
1 01/01/2017 3000 01/03/2017
2 01/01/2017 14000 01/14/2017
;
run;
data new_episodes (keep= ID_NO XDATE DAILY_COST);
set episodes;
NUM_DAYS= DISC_DATE-ADMIT_DATE;
DAILY_COST= TOTAL_COST/(DISC_DATE-ADMIT_DATE);
*Using the Do While loop to create a matrix of date observations;
XDATE=ADMIT_DATE;*initializing our variable;
do while(XDATE<DISC_DATE);
put XDATE=;
XDATE+1;
output;*outputting the date variable;
end;
format XDATE mmddyy10.;
run;
proc print data=new_episodes;
run;

Answer 3

Since SAS stores dates as number of days you can just use a DO loop to increment XDATE from ADMIT_DATE to DISC_DATE.
But you need to decide how to count dates. If you are admitted on Monday and discharged on Tuesday is that one day or two days? If it is one day then do you want XDATE records for Monday or Tuesday? Or both?
Let's make some test data:
data have;
input id_no $ admit_date :yymmdd. total_cost disc_date :yymmdd.;
format admit_date disc_date yymmdd10.;
put (_all_) (+0);
datalines;
1 2017-01-01 3000 2017-01-04
2 2017-01-01 5000 2017-01-06
3 2020-02-23 500 2020-02-23
;
Here is code that treats Monday to Tuesday as one day. So it doesn't output the discharge date (unless it is the same as the admission date).
data want;
set have ;
if admit_date=disc_date then daily_cost=total_cost;
else daily_cost = total_cost / (disc_date - admit_date);
do xdate=admit_date to max(admit_date,disc_date-1) ;
output;
end;
keep id_no xdate daily_cost;
format xdate yymmdd10.;
run;
Results:
daily_
Obs id_no cost xdate
1 1 1000 2017-01-01
2 1 1000 2017-01-02
3 1 1000 2017-01-03
4 2 1000 2017-01-01
5 2 1000 2017-01-02
6 2 1000 2017-01-03
7 2 1000 2017-01-04
8 2 1000 2017-01-05
9 3 500 2020-02-23
If you want to treat a stay from Monday to Tuesday as 2 days then the code is easier.
data want;
set have ;
daily_cost = total_cost / (disc_date - admit_date + 1);
do xdate=admit_date to disc_date ;
output;
end;
keep id_no xdate daily_cost;
format xdate yymmdd10.;
run;
Results:
daily_
Obs id_no cost xdate
1 1 750.000 2017-01-01
2 1 750.000 2017-01-02
3 1 750.000 2017-01-03
4 1 750.000 2017-01-04
5 2 833.333 2017-01-01
6 2 833.333 2017-01-02
7 2 833.333 2017-01-03
8 2 833.333 2017-01-04
9 2 833.333 2017-01-05
10 2 833.333 2017-01-06
11 3 500.000 2020-02-23