分享一些Code，歡迎討論和賜教！

是 否 +2 論壇幣

k人 參與回答

經(jīng)管之家送您一份

應(yīng)屆畢業(yè)生專屬福利!

求職就業(yè)群

趙安豆老師微信：zhaoandou666

經(jīng)管之家聯(lián)合CDA

送您一個全額獎學(xué)金名額~ !

立即領(lǐng)取

感謝您參與論壇問題回答

經(jīng)管之家送您兩個論壇幣！

+2 論壇幣

請有高見的大蝦出來討論和賜教！

1. data test;
   
2. input id $ name $ 14. age;
   
3. cards;
   
4. 001 張三,李四,王五 25
   
5. 002 趙六,甲乙         30
   
6. ;
   
7. run;
   
8. data test1;
   
9. retain id age;
   
10. set test;
    
11. do i=1 to countc(name,',')+1;
    
12. name1=scan(name,i,',');
    
13. output;
    
14. end;
    
15. drop i name;
    
16. run;

復(fù)制代碼

1. data aa;
   
2. input ID name $;
   
3. cards;
   
4. 1001 ABC
   
5. 1002 ABCD
   
6. 1003 ABCDE
   
7. ;
   
8. run;
   
9. 
   
10. data bb;
    
11. retain id;
    
12. set aa;
    
13. do i=1 to length(name);
    
14. do j=i to length(name);
    
15. name1=substr(name,i,j-i+1);
    
16. output;
    
17. end;
    
18. end;
    
19. drop i j name;
    
20. run;
    
21. 
    
22. 
    
23. proc sql;
    
24. create table cc as
    
25. select distinct a.id as id1, b.id as id2, count(a.name1=b.name1) as com
    
26. from bb as a, bb as b
    
27. where a.id<b.id and a.name1=b.name1
    
28. group by a.id, b.id
    
29. ;
    
30. 
    
31. create table dd as
    
32. select *,length(name) as len
    
33. from aa
    
34. ;
    
35. create table ee as
    
36. select a.id1, a.id2, b.len as len1, a.com
    
37. from cc as a, dd as b
    
38. where a.id1=b.id
    
39. ;
    
40. 
    
41. select a.id1, a.id2, a.len1, b.len as len2, a.com
    
42. from ee as a, dd as b
    
43. where a.id2=b.id
    
44. ;
    
45. quit;

復(fù)制代碼

掃碼加我 拉你入群

請注明：姓名-公司-職位

以便審核進群資格，未注明則拒絕

分享0 收藏1 回帖

關(guān)鍵詞：code ODE COD proc sql Distinct

頂樓主，我也需要