Trend analysis for arrays: fitting a trend line to find the slope as trend

[Primavera]

Hi all,

I have a dataset of patients using a drug. Each person has used the drug for Total_days and the dosage of each day is shown in dose_day[i]. Most patients have interruptions (days with dose_day[i]=0) when they received the drug. A simplified data set looks like this:

ID	Total_days	dose_day1	dose_day2	dose_day3	dose_day4	dose_day5	dose_day6	dose_day7	dose_day8	dose_day9	dose_day10	dose_day11	dose_day12	dose_day13	dose_day14	dose_day15	dose_day16	dose_day17	dose_day18	dose_day19	dose_day20	dose_day21	dose_day22	dose_day23	dose_day24	dose_day25
1	18	30	30	30	15	0	0	45	45	45	60	60	0	0	0	45	45	60	60	0	0	0	0	0	0	0
2	25	15	15	15	25	25	25	25	25	40	0	0	0	0	40	40	40	40	40	40	40	40	25	25	25	25
3	21	70	70	70	70	70	50	50	50	50	50	50	50	30	30	0	0	0	20	20	20	40	0	0	0	0
4	19	40	40	40	40	60	60	60	0	0	0	50	50	50	50	50	50	65	65	65	0	0	0	0	0	0
5	15	15	15	15	15	15	25	25	25	25	25	25	10	10	10	10	0	0	0	0	0	0	0	0	0	0
6	25	80	80	80	80	80	80	80	50	50	50	50	0	0	0	0	0	50	50	50	90	90	90	90	90	90
7	22	25	25	25	40	40	40	40	40	65	65	65	65	65	65	65	0	0	0	0	40	40	40	0	0	0

I want to know what their dose trend was. Basically, if I was doing this task manually I would make a plot for each person with x being 1 to i where i=Total_days and y being the amount of drug received on that day (Total_dose[i]) and then I would fit a trendline on that plot. My goal is to have a variable "slope" that is "Pos" if the slope of the trend line is positive "Neg" if the slope is negative and "Zero" in the unlikely case that the slope is actually zero.

Does anyone know how can this be done in SAS 9.4 TS?

Many thanks,

[PaigeMiller]

First, you would run PROC TRANSPOSE to create a long data set, rather than a wide data set. Then turn the zeros into missing. Lastly, run regressions or estimate trends with a BY ID; statement.

Here's the PROC TRANSPOSE part and the turning zeros into missing part

proc transpose data=have out=t;
    by id;
    var dose_day:;
run;
data t1;
    set t;
    if col1=0 then col1=.;
    day=input(substr(_name_,9),3.);
run;