topic Re: Difference from Baseline Calculation in SAS Programming

Difference from Baseline Calculation

kryden — Sat, 24 Feb 2018 05:39:46 GMT

Hey all,

So I have some output from PROC MEANS where I have some basic values spread across four visits. The first visit is considered the baseline visit.

I am trying to crate a dataset that contains the difference from baseline for each variable at each visit:

Anyone have any thoughts on how I can do this?

Re: Difference from Baseline Calculation

art297 — Sat, 24 Feb 2018 23:40:48 GMT

Here is one way:

data have;
 informat visit $25.;
 input visit & n Mean SD Med Q1 Q3 Min Max;
 cards;
Baseline Visit   48 33.8 9.6 32.6 26.2 38.1 18.6 59.7
6 Week Follow-up   35 37.8 7.1 38.1 31.2 42.2 22.8 52.5
3 Month Follow-up    25 42.3 8 42.4 37.3 47.8 18.4 56.4
6 Month Follow-up    2 54.5 1.5 54.5 53.5 55.6 53.5 55.6
;

data want (drop=_:);
 set have;
 retain _Mean _SD _Med _Q1 _Q3 _Min _Max;
 if _n_ eq 1 then do;
 _Mean=Mean;
 _SD=SD;
 _Med=Med;
 _Q1=Q1;
 _Q3=Q3;
 _Min=Min;
 _Max=Max;
 end;
 else do;
 Mean+-_Mean;
 SD+-_SD;
 Med+-_Med;
 Q1+-_Q1;
 Q3+-_Q3;
 Min+-_Min;
 Max+-_Max;
 output;
 end;
run;

Art, CEO, AnalystFinder.com

Note: Corrected to drop retained variables

Re: Difference from Baseline Calculation

mkeintz — Sun, 25 Feb 2018 17:09:56 GMT

This is an alternative to @art297's suggestion. It takes advantage of the DIF function (where dif(x) is defined as x-lag(x)), and interspersing a baseline observation between each of the other visits:

data have;
 informat visit $25.;
 input visit & n Mean SD Med Q1 Q3 Min Max;
 cards;
Baseline Visit           48 33.8 9.6 32.6 26.2 38.1 18.6 59.7
6 Week Follow-up         35 37.8 7.1 38.1 31.2 42.2 22.8 52.5
3 Month Follow-up        25 42.3 8 42.4 37.3 47.8 18.4 56.4
6 Month Follow-up        2 54.5 1.5 54.5 53.5 55.6 53.5 55.6
run;


data want (drop=i);
  set have (where=(visit=:'B'))
      have (where=(visit=:'6 Week') in=in6w)
      have (where=(visit=:'B'))
      have (where=(visit=:'3 M') in=in3m)
      have (where=(visit=:'B'))
      have (where=(visit=:'6 M') in=in6m);
  array x {*} mean--max;
  do i=1 to dim(x);
    x{i}=dif(x{i}); 
  end;
  if in6w or in3m or in6m;
run;

The SET statement interleaves 6 subsets of HAVE, where one subset (visit=:'B') is read 3 times, once preceding the visit=:'6 W', once before visit=:'3 M' and once preceding '6 M'. This means the DIF function, which calculates x-lag(X), will be generating

X at 6 weeks minus X at baseline

X at 3 months minus X at baseline
X at 6 months minus X at baseline

The "where" parameters use the operator "=:", which says to take the shorter string and compare it only to the start of the longer string.