In my understanding if FIRST.STDNT_KEY = 1 & LAST.STDNT_KEY = 1  then he would not have any duplicates, so he appears to only look for any row that is a duplicate.  True? That is true he is keeping STDNT's data who have more than one entries. I don't know what CDUP_TRANS output is supposed to represent and don't know  if the developer was only looking for dups STDNT_KEY = 1 for both first and last.  If so why did they exclude (FIRST.STDNT_KEY = 1 & LAST.STDNT_KEY = 0) from the CDUP_TRANS code? In this code he summarize each from those who have more than one entries, representing them by the first entry only. For summarize or reporting... But your code's first part is not complete. So it could be that he is report all the STDNT in the database table without duplicate. So the STDNT with more than one entries will be represented by the first record and those who has one entries will also be represented. Is (FIRST.STDNT_KEY = 1 & LAST.STDNT_KEY = 0) OR (FIRST.STDNT_KEY = 0 & LAST.STDNT_KEY = 1) redundant? No, (FIRST.STDNT_KEY = 1 & LAST.STDNT_KEY = 0) ... mean the first entry in the group (FIRST.STDNT_KEY = 0 & LAST.STDNT_KEY = 1) ... mean the last entry in the group ***** For example: STDNT_KEY 1 2 3 1 1 3 After the data sorted by STDNT_KEY: STDNT_KEY 1 1 1 2 3 3 By understanding the first and last temporary SAS variables in the sorted data, and understanding that each unique student entries is a group in it's self: STDNT_KEY 1       (FIRST.STDNT_KEY = 1 & LAST.STDNT_KEY = 0) ... the first entry in the 1's group 1       (FIRST.STDNT_KEY = 0 & LAST.STDNT_KEY = 0) ... not the first or the last 1       (FIRST.STDNT_KEY = 0 & LAST.STDNT_KEY = 1) ... the last entry in the 1's group 2       (FIRST.STDNT_KEY = 1 & LAST.STDNT_KEY = 1) ... the first entry in the 2's group and the last 3       (FIRST.STDNT_KEY = 1 & LAST.STDNT_KEY = 0) ... the first entry in the 3's group 3       (FIRST.STDNT_KEY = 0 & LAST.STDNT_KEY = 1) ... the last entry in the 3's group ********* So IF (FIRST.STDNT_KEY = 1 & LAST.STDNT_KEY = 0) OR    (FIRST.STDNT_KEY = 0 & LAST.STDNT_KEY = 0) OR    (FIRST.STDNT_KEY = 0 & LAST.STDNT_KEY = 1) THEN OUTPUT; gives you STDNT_KEY 1       (FIRST.STDNT_KEY = 1 & LAST.STDNT_KEY = 0) ... the first entry in the 1's group 1       (FIRST.STDNT_KEY = 0 & LAST.STDNT_KEY = 0) 1       (FIRST.STDNT_KEY = 0 & LAST.STDNT_KEY = 1) ... the last entry in the 1's group 2       (FIRST.STDNT_KEY = 1 & LAST.STDNT_KEY = 1) ... the first entry in the 2's group and the last 3       (FIRST.STDNT_KEY = 1 & LAST.STDNT_KEY = 0) ... the first entry in the 3's group 3       (FIRST.STDNT_KEY = 0 & LAST.STDNT_KEY = 1) ... the last entry in the 3's group ******* And IF (FIRST.STDNT_KEY = 0 & LAST.STDNT_KEY = 0) OR    (FIRST.STDNT_KEY = 0 & LAST.STDNT_KEY = 1) THEN DELETE;; gives you: STDNT_KEY 1       (FIRST.STDNT_KEY = 1 & LAST.STDNT_KEY = 0) ... the first entry in the 1's group 1       (FIRST.STDNT_KEY = 0 & LAST.STDNT_KEY = 0) 1       (FIRST.STDNT_KEY = 0 & LAST.STDNT_KEY = 1) ... the last entry in the 1's group 2       (FIRST.STDNT_KEY = 1 & LAST.STDNT_KEY = 1) ... the first entry in the 2's group and the last 3       (FIRST.STDNT_KEY = 1 & LAST.STDNT_KEY = 0) ... the first entry in the 3's group 3       (FIRST.STDNT_KEY = 0 & LAST.STDNT_KEY = 1) ... the last entry in the 3's group ... View more

Do you mean this one, Peter Song's QIF SAS Package ... View more

PROC CLUSTER requires more work to do. Why do not you use PROC FASTCLUSTER and use the maxc= option in it. Also you can added it to a macro to try different number of cluster. But if you still need to use PROC CLUSTER you can use PROC TREE after it and use nclusters= option to choose the number of cluster. Here is a good example of using PROC CLUSTER and PROC TREE. Example 91.1 Mammals’ Teeth ... View more