topic Re: Identify long-term stays in SAS Programming

Identify long-term stays

lichee — Tue, 26 Jul 2022 18:36:19 GMT

I wanted to identify a long-term stay through discharge status given patient ID and service begin and end dates. As the example data shows below, when discharge status is 3 or missing, it means one claim/record is a part of a stay. For example, the first two records for pat_ID is actually one stay as discharge_status is 3 in the first record. When discharge status is not 3 or missing, it means one patient was discharged somewhere else.

Any suggestion is greatly appreciated!

pat_ID	service_beg_dt	service_end_dt	discharge_status
1	1/1/2018	3/31/2018	3
1	1/1/2018	4/30/2018	1
1	5/16/2018	5/31/2018	3
1	5/16/2018	6/30/2018	3
1	5/16/2018	7/31/2018	3
1	5/16/2018	8/31/2018	3
1	5/16/2018	9/30/2018	3
1	5/16/2018	10/31/2018	3
1	5/16/2018	11/30/2018	3
1	5/16/2018	12/31/2018	3
2	11/1/2017	1/31/2018	1
2	2/12/2018	5/1/2018	3
2	5/1/2018	5/31/2018	2

Re: Identify long-term stays

PaigeMiller — Tue, 26 Jul 2022 18:38:33 GMT

Can you please show us the desired output from this data?

I have to admit, I don't think the explanation is very clear of what a long-term stay is, can you explain in more detail?

Re: Identify long-term stays

lichee — Tue, 26 Jul 2022 18:47:38 GMT

Thanks for the response!

The desired output would be:

pat_ID	service_beg_dt	service_end_dt	discharge_status
1	1/1/2018	4/30/2018	1
1	5/16/2018	12/31/2018	3
2	11/1/2017	1/31/2018	1
2	2/12/2018	5/31/2018	2

where the begin and end date of a stay are captured, and the final discharge status is also recorded. The time frame is Jan. 1, 2018 to Dec. 31, 2018, which is why for pat_ID=1 had the discharge status=3 in the second stay.

Re: Identify long-term stays

andreas_lds — Wed, 27 Jul 2022 05:54:05 GMT

I am sure that i didn't fully understand the logic, but this step seems to produce the expected result:

data want;
   set have;
   
   if discharge_status ^= 3 or service_end_dt = '31Dec2018'd;
run;

Re: Identify long-term stays

lichee — Wed, 27 Jul 2022 11:25:57 GMT

I hope the side-by-side tables of the original data and the desired output can make my intent a bit more clear. I should have used title "Construct long-term stays". In the desired output, the begin and end dates of a stay are identified from the original data.

For example, for pat_ID=1, the first record (1/1/2018 - 3/31/2018) has discharge status=3 (still patient), so the stay continues to the next record and the end date of the stay is 4/30/2018 as the patient is discharged to home.

The rest of records for pat_ID=1 go from 5/16/2018 to 12/31/2018 with the same discharge status 3-still patient, so the patient remains in the same stay during the period.

The first record for pat_ID=2 has discharge status 1 - discharged to home, so the single record is a complete stay. The rest of records are for another stay as the record with service dates 2/12/2018-5/1/2018 has discharge status 3-still patient, and it continued to the next record with service end date of 5/31/2018.

Length of stay is the number of days from stay begin date to stay end date.

Thank you very much!

Original Data Desired Output

pat_ID	service_beg_dt	service_end_dt	discharge_status	Note	pat_ID	stay_beg_dt	stay_end_dt	discharge_status	Length_of_stay
1	1/1/2018	3/31/2018	3	3 - still patient	1	1/1/2018	4/30/2018	1	119
1	1/1/2018	4/30/2018	1	1 - discharge to home	1	5/16/2018	12/31/2018	3	229
1	5/16/2018	5/31/2018	3		2	11/1/2017	1/31/2018	1	91
1	5/16/2018	6/30/2018	3		2	2/12/2018	5/31/2018	2	108
1	5/16/2018	7/31/2018	3
1	5/16/2018	8/31/2018	3
1	5/16/2018	9/30/2018	3
1	5/16/2018	10/31/2018	3
1	5/16/2018	11/30/2018	3
1	5/16/2018	12/31/2018	3	end of study timeframe, still patient
2	11/1/2017	1/31/2018	1	1 - discharge to home
2	2/12/2018	5/1/2018	3	3 - still patient
2	5/1/2018	5/31/2018	2	2 - discharge to inpatient care

Re: Identify long-term stays

lichee — Wed, 27 Jul 2022 13:13:20 GMT

The tables below may help. I should have used the title "Construct long-term stays". When the discharge status of a record is 3-still patient, it means the stay continues to the next record. When the discharge status is not 3 (1-discharge to home, 2-discharge to inpatient care), it means a stay ends.

For example, the first record for pat_ID=1 has discharge status 3 (still patient), which means the stay continues to the next record till 4/30/2018 when the patient discharged to home (discharge_status=1). The next stay for pat_ID=1 went from 5/16/2018 to 12/31/2018 as the rest of records all have discharge_status=3.

For pat_ID=2, the record with service dates 1/1/2018-1/31/2018 has discharge_status=1, so the single record is a complete stay. The second and third records are for one stay because the second record has discharge_status=3 (still patient).

Assuming that the study timeframe is 1/1/2018 - 12/31/2018.

Thank you!!

Original data

pat_ID	service_beg_dt	service_end_dt	discharge_status	Note
1	1/1/2018	3/31/2018	3	3 - still patient
1	1/1/2018	4/30/2018	1	1 - discharge to home
1	5/16/2018	5/31/2018	3
1	5/16/2018	6/30/2018	3
1	5/16/2018	7/31/2018	3
1	5/16/2018	8/31/2018	3
1	5/16/2018	9/30/2018	3
1	5/16/2018	10/31/2018	3
1	5/16/2018	11/30/2018	3
1	5/16/2018	12/31/2018	3	end of study timeframe, still patient
2	1/1/2018	1/31/2018	1	1 - discharge to home
2	2/12/2018	5/1/2018	3	3 - still patient
2	5/1/2018	5/31/2018	2	2 - discharge to inpatient care

Desired output:

pat_ID	stay_beg_dt	stay_end_dt	discharge_status	Length_of_stay
1	1/1/2018	4/30/2018	1	119
1	5/16/2018	12/31/2018	3	229
2	11/1/2017	1/31/2018	1	91
2	2/12/2018	5/31/2018	2	108

Re: Identify long-term stays

andreas_lds — Thu, 28 Jul 2022 04:54:18 GMT

And now please post the original data as working data step using datalines or cards.

Re: Identify long-term stays

lichee — Thu, 28 Jul 2022 11:08:26 GMT

data have;
infile datalines truncover dsd;
input pat_ID (service_beg_dt service_end_dt) (:mmddyy10.) discharge_status;
format service_beg_dt service_end_dt mmddyy10.;
datalines;
1,1/1/2018,3/31/2018,3
1,1/1/2018,4/30/2018,1
1,5/16/2018,5/31/2018,3
1,5/16/2018,6/30/2018,3
1,5/16/2018,7/31/2018,3
1,5/16/2018,8/31/2018,3
1,5/16/2018,9/30/2018,3
1,5/16/2018,10/31/2018,3
1,5/16/2018,11/30/2018,3
1,5/16/2018,12/31/2018,3
2,1/1/2018,1/31/2018,1
2,2/12/2018,5/1/2018,3
2,5/1/2018,5/31/2018,2
;

Re: Identify long-term stays

mkeintz — Thu, 28 Jul 2022 11:17:26 GMT

Given we don't have sample data in the form of a working data step, you'll need to test the code below:

data want (drop=service_beg_dt    rename=(service_end_dt=stay_end_dt) );
  do until (discharge_status^=3 or last.pat_id=1);
    set have;
    by pat_id;
    stay_beg_dt=min(stay_beg_dt,service_beg_dt);
  end;
  length_of_stay=service_end_dt-stay_beg_dt;
  format stay_beg_dt mmddyy10. ;
run;

This assumes that the only continuation indicator is DISCHARGE_STATUS=3, and that the data are sorted by PAT_ID/SERVICE_END_DT.