If you consider that a unique ID is one that occurs once only, it is possible to find by using Hash objects. But how much time it will take compared to SQL can be found only by you, by running both ways. Xia has already given a hash solution. It can not tell which ID occurs once only. Here is another hash way with an example, the length of ID need not be 6 in your case. It It collects all records having one ID into one data set(Unique) and the rest to another data set(Duplicates). data have; input ID $6.; datalines; aaaaaa aaaaaa aaaaaa bbbbbb dddddd ; run; data _null_;    if _n_ = 1 then do;       declare hash h();       h.definekey('ID');       h.definedata('ID','freq');       h.definedone();    end;    set have end = eof;    if h.find() ne 0 then freq = 0;    freq + 1;    h.replace();    if eof;      h.output(dataset:'Unique (where=(freq=1)');      h.output(dataset:'Duplicates (where=(freq > 1)'); run; proc print data = Unique; run; proc print data = Duplicates; run; ... View more

PG: Your observation to replace the lower boundary of the array to 0 is correct. The program works if the array starts with 0-index for your example. On practical consideration, such a situation may not arise. Look at the subgroup: subject = 13, date = 3 and product = 1. It is a lone observation with a missing price value. In this case, my COUNT which counts the non-missing price-values in the subset, is zero. This results in imputing a missing value by another missing value -- helped by 0-index. So, either I change to 0-start or add a condition to check that there is at least one nonmissing price value in a subgroup before doing the imputation -- in that case there is no need to change the array index. For the benefit of the Original Poster, I place the revised program here: data want;    array k[10] _temporary_;    count = 0;    do until(last.product);       set have ;       by date product;       if not missing(price) then do; count + 1; k[count] = price; end;    end;    do until(last.product);       set have;       by date product;       if price = . and count then do;          i = ceil(ranuni(123) * count);          price = k;       end;       output;    end; drop i count; run; ------------- DataSP ... View more

I think that PG Stat has provided a completely correct solution. I suspect that I am having a problem that I will refer to my IT group. Thank you, Ian ... View more

Additional 4-th digit is not an issue. At present the use of $ sign is by default takes 8 characters. Even if it is longer than 8, use the appropriate width to $. No need to substring it. ... View more

Just to clarify, proc sql does not suck.  It is very useful, and has a variety of uses.  The simple fact is that the underlying concept of SAS is not a relational database and SQL was built to deal with relational databases.  Again, its using something in a way it was not intended.  In your case you are dealing with large volumes of data, so you should be looking at datawarehouses/marts and processes and tools designed for such tasks. ... View more

Dear Poster Look at the note: If you post a question and someone replies with information that you can use, it's a good idea to mark their reply as either helpful or correct. That way, others who have the same question can quickly see which replies might be most useful for them. A helpful reply is on the right track, while a correct reply answered your question. Keep in mind that you and others get more status points for posting replies that are marked as helpful or correct! When you click Correct Answer for one of the replies, the original question becomes marked answered and you see a clip of the correct answer in your original question. To mark responses to your question, go to the question in the community, look for responses, and click Helpful Answer or Correct Answer for each response that fits.  If you want to undo a correct or helpful answer, simply click Unmark as Helpful or Unmark as Correct. ... View more

For SAS UE technical support email sasanalyticsu@sas.com ... View more

Thank you Tom. ... View more

data have; input empid salary ; cards; 1 15000 2 20000 3 30000 4 45000 5 60000 6 70000 7 10000 8 20000 9 100000 10 35000 11 40000 12 7000 13 90000 14 80000 15 70000 16 65000 17 87000 18 56000 19 8000 20 7000 21 12000 22 19000 23 23000 24 34000 25 46000 26 58000 27 67000 28 78000 29 5000 30 84000 ; run; proc format; value fmt 0-20000='1' 20000<-40000='2' 40000<-60000='3' 60000<-80000='4' 80000<-100000='5'; run; data want; set have; group=put(salary,fmt.); run; ... View more

Thank you datasp for showing two diff methods to resolve. I learn new concept of Hash. Thank you Arthur very much. Nicely explained. My issue resolved. Thanks once again. ... View more