topic Re: Splitting a row into two based on date in SAS Programming

Splitting a row into two based on date

Kc2 — Wed, 24 Jun 2015 13:06:59 GMT

Hello ,

I have a dataset of patient and the dr. office dates.

Some visits are mandatory while other are just when the patient has an issue.

The visits are grouped into periods.

Mandatory visits are 6 and visit 9

Have dataset

User stdate enddate visit6dt visit9dt

A 2013-10-25 2013-11-16 2013-10-25 2014-01-25

B 2013-02-01 2013-04-21 2013-03-10 2013-06-10

C 2013-09-01 2013-09 2013-09-10 2013-12-01

Want datset

User stdate enddate visit6dt visit9dt

A 2013-10-25 2013-10-25 2013-10-25 2014-01-25

A 2013-10-26 2014-01-25 2013-10-25 2014-01-25

B 2013-02-01 2013-02-01 2013-03-10 2013-06-10

B 2013-02-02 2013-06-10 2013-03-10 2013-06-10

C 2013-09-01 2013-09-09 2013-09-10 2013-12-01

C 2013-09-10 2013-09-10 2013-09-10 2013-12-01

C 2013-09-11 2013-09 -30 2013-09-10 2013-12-01

C 2013-10-01 2013-12-01 2013-09-10 2013-12-01

becasue the en day in one case is missing i had to split it into 4 rows;

first until the 6 visit Then until the en of the month and then again until the end date.

Thank you for your help

Re: Splitting a row into two based on date

RW9 — Wed, 24 Jun 2015 13:48:08 GMT

Hi,

It would seem to be straightforward if then statements, something like:

period=0; /* I.e. before v6 */

new_start=stdate;

new_end=visit6dt;

output;

period=6;

new_start=visit6dt;

new_end=visit9dt;

output;

How you handle missing data items is up to you, maybe check length of date and assign a value or cover with some other error checking logic.

Re: Splitting a row into two based on date

ballardw — Wed, 24 Jun 2015 13:51:13 GMT

I think you need to provide some more rules. Such as how the STDATE for the second or subsequent line is determined.

Why does User C end up with 4 lines?

And are any of these variables SAS date valued? Date manipulations are simpler with those.

Re: Splitting a row into two based on date

Steelers_In_DC — Wed, 24 Jun 2015 13:52:26 GMT

Here is a solution for user C.

data have;

infile cards;

informat user $1. stdate enddate visit6dt visit9dt yymmdd10.;

format user $1. stdate enddate visit6dt visit9dt mmddyy10.;

input User$ stdate enddate visit6dt visit9dt;

cards;

A 2013-10-25 2013-11-16 2013-10-25 2014-01-25

B 2013-02-01 2013-04-21 2013-03-10 2013-06-10

C 2013-09-01 2013-09 2013-09-10 2013-12-01

;

data prep;

set have;

output;

if missing(enddate) then do;

enddate = visit6dt-1;output;

enddate = visit6dt;output;

enddate = intnx('month',stdate,0,'e');output;

enddate = visit9dt;output;

end;

run;

data want;

set prep;

if missing(enddate) then delete;

run;

Re: Splitting a row into two based on date

Kc2 — Wed, 24 Jun 2015 16:39:14 GMT

Thank you everyone for the quick response. Mark Johnson's solution is the closest one . I still need to make a few adjustemnt but I am getting there. (user C si the problem).

Re: Splitting a row into two based on date

Steelers_In_DC — Wed, 24 Jun 2015 16:47:49 GMT

I didn't see what you were doing with the duplicate A and B rows, here's another more complete solution:

data prep;

set have;

output;

if missing(enddate) then do;

enddate = visit6dt-1;output;

enddate = visit6dt;output;

enddate = intnx('month',stdate,0,'e');output;

enddate = visit9dt;output;

end;

if not missing(enddate) then do;

retain User enddate visit6dt visit9dt;

stdate = stdate+1;output;

end;

run;

data want;

set prep;

if missing(enddate) then delete;

run;

Re: Splitting a row into two based on date

Kc2 — Wed, 24 Jun 2015 19:51:52 GMT

Thank you Mark.