Visit Numbers for Unscheduled Visits

[Almighty]

Hi everyone,

How do I derive vist numbers for unsheduled labs. If an unsheduled lab happens between visit 1 and visit 2 then visit number for that unscheduled visit shoud be 1.1 and if there are 2 unscheduled visits between visit 1 and visit 2, then first unscheduled visit should be 1.1 and second unscheduled visit should be 1.2. and so on.. The following in the sample data. Column VISITNUM should be derived.

Thanks in advance.

Subject	LBCAT	LBTEST	LBDTC	VISIT	VISITNUM
100	hematology	RBC	10-Jan-11	Screening	-1
100	hematology	RBC	10-Mar-11	Baseline	1
100	hematology	RBC	15-Mar-11	Unscheduled	1.1
100	hematology	RBC	1-Apr-11	Unscheduled	1.2
100	hematology	RBC	10-Apr-11	Month 1	2
100	hematology	RBC	10-May-11	Month 2	3
101	hematology	RBC	11-Jan-11	Screening	-1
101	hematology	RBC	10-Feb-11	Baseline	1
101	hematology	RBC	15-Mar-11	Unscheduled	2
101	hematology	RBC	1-Apr-11	Unscheduled	2.1
101	hematology	RBC	10-Apr-11	Month 1	2.2
101	hematology	RBC	10-May-11	Month 2	3

[RW9]

Well, that depends on what you specs/team tell you.  If it is just sort by date (and what if there are multiple on the same date?), then sort the dataset, retain lstvisit and lstunsch, if unscheduled then visit=lstvisit + lstunsch.  It could however be far more complicated than that with applied logic defining visit windows, or other known patterns.

Code for the above:

data want;

     set have;

     by subjid;

     retain lstvisit lstunsch;

     if first.subjid then do;

          lstvisit=visit;

          lstunsch=0;

     end;

     else if type="UNSCHEDULED" then do;    

          lstunsch=lstunsch+0.1;          /* note I would recommend using 0.01 as there may be more than 10 */

          visit=lstvisit + lstunsch;

     end;

     else lstvisit=visit;

run;

[k6adams]

Hi RW9, I am working on a HW exercise similar to the OP. The solution you posted only partialy worked for me, SAS produces the vector lstvist such that if visit is (1, 2, ., .,3, ., 4) then lstvisit is (1, 2, 2, 2, 3, 3, 4), which is a step in the right direction. However, the lstunsch produces a vector of missing values. I have been trying to debug for a while but have had little success,  any chance you might know what is going on there? Also, I am assuming that your use of "visit" refers to the OP's use of  "visitnum", and "type" refers to  "visit".

Message was edited by: Kevin Adams

[Doc_Duke]

Your narrative and your example don't match.  Please restate more precisely.

Subject 101 on 15Mar and 10Apr do not follow the pattern in the narrative. 

[k6adams]

I think this solution should work:

DATA new;

  Set old;

  By subjid;

  Retain visit_r 0;

  If first.subjid Then visit_r = visitnum;

  If visit = "Unscheduled" Then visitnum = visit_r + 0.01;

  visit_r = visitnum;

  Drop visit_r;

RUN;

For a similar example see http://www.mwsug.org/proceedings/2009/stats/MWSUG-2009-D14.pdf

[PGStats]

Keep it simple:

data want;

set have;

by subject;

if first.subject then visitnum = -1;

visitnum + 0.1;

if VISIT ne "Unscheduled" then visitnum = ceil(visitnum);

run;

PG

[Emily_Zhang]

Really helpful! But could you please explain more detailed of the code? I can hardly understand the logic very clearly. Thanks. 🙂 

[Emily_Zhang]

Really helpful. Thanks.

[teja5959]

Hi i have one more qus for you. in case Unscheduled visit have  in same date 15-mar-11  in your data set present only one Unscheduled visit on that day  

EX;

LBDTC                        visit 

15-Mar-11                  Unscheduled

15-Mar-11                  Unscheduled

15-Mar-11                  Unscheduled

above the data is given i want visitno add 1.1 all three subjects how to do remaining  after the date as taken 1.2 

[mkeintz]

If you want all unscheduled visits on a single day  to have the same visit number,then change 2 lines of @PGStats's code:

change

    BY SUBJECT

to

    BY SUBJECT LBDTC

and change

    VISITNUM+1
to

    IF FIRST.LBDTC THEN VISITNUM+1;