data a;
input time datetime16. n a i j;
cards;
13JUL14:00:00:01 1 0 0 1
13JUL14:00:00:02 2 0 0 1
13JUL14:00:00:03 3 0 1 1
13JUL14:00:00:04 4 1 3 1
13JUL14:00:00:05 5 1 0 1
13JUL14:00:00:06 6 3 0 1
13JUL14:00:00:07 7 1 0 1
13JUL14:00:00:08 8 0 1 1
13JUL14:00:00:09 9 0 0 1
13JUL14:00:00:10 10 0 1 1
13JUL14:00:00:11 1 1 0 2
13JUL14:00:00:12 2 0 0 2
13JUL14:00:00:13 3 1 0 2
13JUL14:00:00:14 4 1 5 2
13JUL14:00:00:15 5 0 0 2
13JUL14:00:00:16 6 0 0 2
13JUL14:00:00:17 7 5 1 2
13JUL14:00:00:18 8 1 0 2
13JUL14:00:00:19 9 0 0 2
13JUL14:00:00:20 10 1 0 2
;
run;
proc sql;
create table b as
select max(a) into :x from a by group j;
quit;
数据集a中观测按照n分两组，10个观测为一组，找出每组中a的最大值max(a) ，创建数据集b，使b中只保留每组中n=max(a) 和i=max(a) 的数据，然后计算时间差，如何做到？请指教！thank you！！