HIERARCHIZE Function

The first method demonstrates how HIERARCHIZE works for a set whose dimensionality is 1. This method will be generalized to sets with arbitrary dimensionality.

Sets with Dimensionality = 1

Consider HIERARCHIZE(S), where S = {Kansas, USA, Canada, Buffalo, Topeka}. As detailed in the section Literal Sets, you create two tables, S1 and S2, that contain the fully qualified member names in the sets S1 and S2, respectively.

Gather information as follows:

1. Use the following query to verify that all members belong to the same dimension. If the returned count is greater than 1, the set is invalid.

   SELECT COUNT DISTINCT COMPONENT(S.Name, -1)
      FROM S
   

2. Get the dimension name by using the following query:

   SELECT DISTINCT COMPONENT(S.Name, -1)
      FROM S
   

3. Get the level number of each level represented in the input set:

   SELECT DISTINCT LEVEL(S.Name)
      FROM S
   

The objective is to create a table T that looks like the following table.

For each member of set S, there should be a RANKX column that contains the rank of each of its ancestors. To get the hierarchized set from table T, use the following statement:

SELECT Name, Rank
FROM
SELECT * FROM T
   RANK ROWS AS Rank RANKORDER BY Rank1, Rank2, Rank3
ORDER BY Rank

To create the table

1. Define a table that has elements of set S and its levels:

   CREATE LOCAL TEMPORARY VIEW S1(Name)
   AS
   SELECT S.Name
   FROM S JOIN members AS M ON(S.Name = M.Level_Name)
   

2. Use a UNION statement to combine the results of three SELECT operations. (Three is the level of the lowermost member in S.)

   Note

   The WHERE clause on each UNION iterates from Level = 0 through the level of the lowermost member in S1. Also, the ANCESTOR function in the FROM iterates from 1 through 1 + the level of the lowermost member in S1.

   SELECT S1.*, M1.Natural_Sort_Rank AS Rank1,
      NULL AS Rank2, NULL AS Rank3
   FROM S1 JOIN Members AS M1 ON S1.Name = M1.Member_Name
   WHERE S1.Level = 0
   UNION
   SELECT S1.*, M1.Natural_Sort_Rank AS Rank1,
      M2.Natural_Sort_Rank AS Rank2, NULL AS Rank3
   FROM (S1 JOIN Members AS M2 ON S1.Name = M2.Member_Name)
   JOIN Members AS M1 ON ANCESTOR(S1.Name, 1)= M1.Member_Name
   WHERE S1.Level = 1
   UNION
   SELECT S1.*, M1.Natural_Sort_Rank AS Rank1,
      M2.Natural_Sort_Rank AS Rank2, M3.Natural_Sort_Rank AS M3
   FROM (((S1 JOIN Members AS M3 ON S1.Name = M3.Name)
   JOIN Members AS M2 ON ANCESTOR(S1.Name, 2) = M2.Name)
   JOIN Members AS M1 ON ANCESTOR(S1.Name, 3) = M1.Name)
   WHERE S1.Level = 2
   

Generalizing for Sets with Arbitrary Dimensionality

Consider HIERARCHIZE(S), where:

S = {(Kansas, 1996), (Buffalo, 1995.Q4), (USA, 1995.Mar), (Buffalo, 1995), (USA, 1995), (Kansas, 1996.Q4), (Kansas, 1996.Q1), (USA, 1995.Q1)}

The steps in hierarchizing this set are as follows:

1. Let S(Name1, Name2, Rank) be the table associated with the set S.

2. Let S1, S2 be tables:

   • First step:

     CREATE LOCAL TEMPORARY VIEW S1(Name, Rank)
     AS
     SELECT DISTINCT Name1, Rank FROM S
     

   • Second step:

     CREATE LOCAL TEMPORARY VIEW S2(Name, Rank)
     AS
     SELECT DISTINCT Name2, Rank FROM S
     

3. Perform HIERARCHIZE on S1 and S2 as explained in Sets with Dimensionality = 1, above. Name the resulting hierarchized sets as D1 and D2.

4. Use the following query to create the hierarchized set:

   SELECT S.Name1, S.Name2, NewRank as Rank
   FROM
      (SELECT S.Name1, S.Name2
      FROM (S JOIN D1 ON S.Name1 = D1.Name)
         JOIN D2 ON S.Name2 = D2.Name)
      RANK ROWS AS NewRank RANKORDER BY D1.Rank, D2.Rank
   

Generalizing this to sets with dimensionality greater than two should be straightforward.