如何找重复的id号

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data tem;
input id1 id2;
cards;
1 2
2 1
4 7
5 8
6 9
8 5
;

要求结果是
1 2
4 7
5 86 9

也就是说：
如果id1的号码在id2的某个观测存在的话，就把对应的id2的那条观测删掉，只保留当前观测，如果id1的号码在id2中不存在的话，就保留下来
比如id1=1的号码，在第二条观测id2也等于1，就把第二条观测删掉，只保留第一条观测

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关键词：100论坛币 悬赏100 Input cards Data 如何

1. /* the code works for the given data, not try for other data */
   
2. 
   
3. data tem;
   
4. input id1 id2;
   
5. cards;
   
6. 1 2
   
7. 2 1
   
8. 4 7
   
9. 5 8
   
10. 6 9
    
11. 8 5
    
12. ; run;
    
13. proc print data=tem; title 'tem'; run;
    
14. 
    
15. data repeated_id;
    
16.      set tem (rename=(id2=id3));
    
17.        n = _n_;
    
18.      do i = n+1 to nobs;
    
19.        set tem nobs=nobs point=i;
    
20.         if id1 = id3 then output;
    
21.      end;
    
22.     keep id1 id2;
    
23. run;
    
24. 
    
25. proc sql;
    
26.   create table want as
    
27.       select *
    
28.       from tem
    
29.       except select * from repeated_id;
    
30. quit;
    
31. proc print data=want; title 'want'; run;

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Come on, your rule is incomplete. How do you break cycles, for example, delete which one from (a,b), (b,c), (c,a)?

邓贵大 发表于 2013-10-29 12:43
Come on, your rule is incomplete. How do you break cycles, for example, delete which one from (a,b), ...

sorry, missing a THEN after id1=id3.
It was add.

data wanted(keep=id1 id2);
   set tem;
   length _id1 $100;
   retain _id1 ;
   if index(_id1,quote(cats(id2))) then flag=0;
   else do;
        flag=1;
        _id1=catx(",",_id1,quote(cats(id1)));
        end;
        if flag;
run;

好厉害！

1. data tem;
   
2. input id1 id2;
   
3. cards;
   
4. 1 2
   
5. 2 1
   
6. 4 7
   
7. 5 8
   
8. 6 9
   
9. 8 5
   
10. ;
    
11. 
    
12. proc sql noprint;
    
13.         select id1 into :id1 separated by "" from tem;
    
14. quit;
    
15. 
    
16. %put &id1.;
    
17. 
    
18. data test;
    
19.         set tem;;
    
20.         if index("&id1.",scan(compress(put(id2,best12.)),1))>_n_ or index("&id1.",scan(compress(put(id2,best12.)),1))=0;
    
21.         keep id1 id2;
    
22. run;

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yongyitian 发表于 2013-10-29 12:28

哥们data 步的应用太娴熟了，你学多久SAS了

yongyitian 发表于 2013-10-29 12:28

请问两个set是几个指针，几个pdv

yongyitian 发表于 2013-10-29 12:47
sorry, missing a THEN after id1=id3.
It was add.

No, your code is OK.
I was just saying the solution may not be unique based on Imasasor's description. The solution you get will depend on the order of data.
For example, your code returns (1,2) when the order is

1. 1 2
   
2. 2 3
   
3. 3 1

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but it yields (1,2), (2,3) if the order is

1. 2 3
   
2. 1 2
   
3. 3 1

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Another example, given a path like (1,2), (2,3), (3,4). If you delete (2,3) first, you'll end up with two records; or if you erase (3,4) first, you'll still get a chance to delete (2,3) and return only 1 record. Both solutions are correct but they are different.
I'm just not sure in which order one should break a loop, a cycle, or a path.
Imasasor has to define a complete set of rules for others to work on a program.