topic Re: Programming in SAS Programming

Programming

Deep_Impact — Tue, 08 Dec 2020 22:31:27 GMT

I have 2 dates (start date end date) for an encounter by patient. I have multiple encounters. How do I subtract the last date of one encounter from the first date of the next encounter for each patient?

HAVE DATASET

PATIENT ENCOUNTER START DATE END DATE

1 1 ST_DATE_1 END_DATE_1

1 2 ST_DATE_2 END_DATE_2

1 3 ST_DATE_3 END_DATE_3

1 4 ST_DATE_4 END_DATE_4

WANT DATASET

PATIENT ENCOUNTER DIFF_END DATE_START DATE_FOR CONSECUTIVE RECORDS

1 1 .

1 2 DIFF START_DATE_2 & END_DATE_1

1 3 DIFF START_DATE_3 & END_DATE_2

1 4 DIFF START_DATE_4 & END_DATE_3

Re: Programming

ballardw — Tue, 08 Dec 2020 22:50:06 GMT

First requirement: Have the dates as numeric values, and best a SAS format that displays them as a usable date. Note: a number like 20201102 is not a SAS date.

data want; 
   set have;
   by patient;
   lend = lag(end_date);
   if not first.patient then daysdiff = Start_date - Lend;
   drop lend;
run;

The LAG function maintains a queue of prior values so you can access them.

When using BY group in a data set SAS creates automatic variables First. and Last. for each variable on the By statement that you can test for true/false.

Re: Programming

Deep_Impact — Thu, 10 Dec 2020 04:31:51 GMT

Hi ballardw,

Thank you for your prompt reply. It did work and got what I wanted. Only thing in the program log was that first.patient is uninitialized.