Programming

[Deep_Impact]

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

[ballardw · Accepted solution]

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.

[Deep_Impact]

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.