找出每组中最大值并保留数据

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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！！

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关键词：最大值 Thank you proc sql datetime Select 最大值

加个 DENSE_ROW,然后用PARTITON BY 就好了~

Tony-CQNU 发表于 2015-1-7 11:11
加个 DENSE_ROW,然后用PARTITON BY 就好了~

您好！可否详细点？我的程序是可以直接运行的。谢谢啦！

1. data raw;
   
2. input time datetime16. n a i j;
   
3. cards;
   
4. 13JUL14:00:00:01 1 0 0 1
   
5. 13JUL14:00:00:02 2 0 0 1
   
6. 13JUL14:00:00:03 3 0 1 1
   
7. 13JUL14:00:00:04 4 1 3 1
   
8. 13JUL14:00:00:05 5 1 0 1
   
9. 13JUL14:00:00:06 6 3 0 1
   
10. 13JUL14:00:00:07 7 1 0 1
    
11. 13JUL14:00:00:08 8 0 1 1
    
12. 13JUL14:00:00:09 9 0 0 1
    
13. 13JUL14:00:00:10 10 0 1 1
    
14. 13JUL14:00:00:11 1 1 0 2
    
15. 13JUL14:00:00:12 2 0 0 2
    
16. 13JUL14:00:00:13 3 1 0 2
    
17. 13JUL14:00:00:14 4 1 5 2
    
18. 13JUL14:00:00:15 5 0 0 2
    
19. 13JUL14:00:00:16 6 0 0 2
    
20. 13JUL14:00:00:17 7 5 1 2
    
21. 13JUL14:00:00:18 8 1 0 2
    
22. 13JUL14:00:00:19 9 0 0 2
    
23. 13JUL14:00:00:20 10 1 0 2
    
24. ;
    
25. run;
    
26. proc sql;
    
27. create table b as select * from raw by group j having max(a)=n or max(a)=i ;
    
28. quit;
    
29. data b(keep=n i diff);set b;diff=time-lag(time);run;

复制代码

sushe1527 发表于 2015-1-7 12:15

您好！特别感谢您的解答！非常好！请问如果做成快照式的该如何写呢？我现在是用j分了两组，如果是每个a与前10条观测中n的值相等的数据保留，该如何做呢？劳驾啦！