BASE SAS

[molla]

Hi below is the interview question which I faced,we can solve it by hardcoding ,can we solve it without hard coding. pls help me:

An exposure is linked to a limit. And a limit can have multiple exposures associated to it. Similarly a limit may have multiple collateral associated to it. Collateral can be shared between multiple limits. Please see the below sample input dataset,

ExposureId	LimitId	CollateralId
1	L1	C1
2	L1	C2
3	L2	C1
4	L2	C3
5	L3	C4
6	L3	C5
7	L4	C6
8	L5	C5

You are given the task to write code and generate below output, to cluster the related exposure, limit and collateral together with a ClusterId.

ExposureId	LimitId	CollateralId	ClusterId
1	L1	C1	1
2	L1	C2	1
3	L2	C1	1
4	L2	C3	1
5	L3	C4	2
6	L3	C5	2
7	L4	C6	3
8	L5	C5	2

[Jagadishkatam]

Could you please provide more details of the exposure limits. How the exposure limit is linked with the collateralid

[Reeza]

This isn't an easy problem to solve by the way, so it's an interesting problem from that perspective. It looks like a recursive chain look up and I think you can use the macro here. From a graph theory lens, it's asking to identify each branch. 

https://gist.github.com/statgeek/14e3aa2a9f718f551cd98134e9ceed30

Or search 'recursive lookup' on here and you'll find some hash solutions. 

[Ksharp]

I have answered this question before. Isn't it you ? 
Here  could give you a start. Once you got WANT table. It is easy to  get what you want.







data have;
infile cards ;
input from $  to $ ;
cards;
1     2
1     3
4     5
5     2
9     4
6     7
8     7
;
run;
data full;
  set have end=last;
  if _n_ eq 1 then do;
   declare hash h();
    h.definekey('node');
     h.definedata('node');
     h.definedone();
  end;
  output;
  node=from; h.replace();
  from=to; to=node;
  output;
  node=from; h.replace();
  if last then h.output(dataset:'node');
  drop node;
run;


data want(keep=node household);
declare hash ha(ordered:'a');
declare hiter hi('ha');
ha.definekey('count');
ha.definedata('last');
ha.definedone();
declare hash _ha(hashexp: 20);
_ha.definekey('key');
_ha.definedone();

if 0 then set full;
declare hash from_to(dataset:'full(where=(from is not missing and to is not missing))',hashexp:20,multidata:'y');
 from_to.definekey('from');
 from_to.definedata('to');
 from_to.definedone();

if 0 then set node;
declare hash no(dataset:'node');
declare hiter hi_no('no');
 no.definekey('node');
 no.definedata('node');
 no.definedone();
 

do while(hi_no.next()=0);
 household+1; output;
 count=1;
 key=node;_ha.add();
 last=node;ha.add();
 rc=hi.first();
 do while(rc=0);
   from=last;rx=from_to.find();
   do while(rx=0);
     key=to;ry=_ha.check();
      if ry ne 0 then do;
       node=to;output;rr=no.remove(key:node);
       key=to;_ha.add();
       count+1;
       last=to;ha.add();
      end;
      rx=from_to.find_next();
   end;
   rc=hi.next();
end;
ha.clear();_ha.clear();
end;
stop;
run;

[molla]

I dont have knowledge of hash tables,its abit difficult to understand,cud u pls brief me what is happening there actually

[Ksharp · Accepted solution]

NO. I am not going to talk about Hash Table here. It is a very big topic.

If you don't know anything about it ,check its documention at support.sas.com  .