Oracle9i SQL Reference Release 1 (9.0.1) Part Number A9012501 

Functions, 89 of 166
percentile_cont::=
percentile_cont
See Also:
"Analytic Functions" for information on syntax, semantics, and restrictions of the 
The PERCENTILE_CONT
function is an inverse distribution function that assumes a continuous distribution model. It takes a percentile value and a sort specification, and returns an interpolated value that would fall into that percentile value with respect to the sort specification. Nulls are ignored in the calculation.
The first expr must evaluate to a numeric value between 0 and 1, because it is a percentile value. This expr must be constant within each aggregation group. The ORDER
BY
clause takes a single expression that must be a numeric or datetime value, as these are the types over which Oracle can perform interpolation.
The result of PERCENTILE_CONT
is computed by linear interpolation between values after ordering them. Using the percentile value (P) and the number of rows (N) in the aggregation group, we compute the row number we are interested in after ordering the rows with respect to the sort specification. This row number (RN) is computed according to the formula RN = (1+ (P*(N1))
. The final result of the aggregate function is computed by linear interpolation between the values from rows at row numbers CRN = CEILING(RN)
and FRN = FLOOR(RN)
.
The final result will be:
if (CRN = FRN = RN) then (value of expression from row at RN) else (CRN  RN) * (value of expression for row at FRN) + (RN  FRN) * (value of expression for row at CRN)
You can use the PERCENTILE_CONT
function as an analytic function. You can specify only the query_partitioning_clause in its OVER clause. It returns, for each row, the value that would fall into the specified percentile among a set of values within each partition.
The following example computes the median salary in each department.
SELECT department_id, PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY salary DESC) "Median cont", PERCENTILE_DISC(0.5) WITHIN GROUP (ORDER BY salary DESC) "Median disc" FROM employees GROUP BY department_id; DEPARTMENT_ID Mediancont Mediandisc    10 4400 4400 20 9500 13000 30 2850 2900 40 6500 6500 50 3100 3100 60 4800 4800 70 10000 10000 80 8800 8800 90 17000 17000 100 8000 8200 110 10150 12000
PERCENTILE_CONT
and PERCENTILE_DISC
may return different results. PERCENTILE_CONT
returns a computed result after doing linear interpolation. PERCENTILE_DISC
simply returns a value from the set of values that are aggregated over. When the percentile value is 0.5, as in this example, PERCENTILE_CONT
returns the average of the two middle values for groups with even number of elements, whereas PERCENTILE_DISC
returns the value of the first one among the two middle values. For aggregate groups with an odd number of elements, both functions return the value of the middle element.
In the following example, the median for Department 60 is 4800, which has a corresponding percentile (Percent_Rank
) of 0.5. None of the salaries in Department 30 have a percentile of 0.5, so the median value must be interpolated between 2900 (percentile 0.4) and 2800 (percentile 0.6), which evaluates to 2850.
SELECT last_name, salary, department_id, PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY salary DESC) OVER (PARTITION BY department_id) "Percentile_Cont", PERCENT_RANK() OVER (PARTITION BY department_id ORDER BY salary DESC) "Percent_Rank" FROM employees WHERE department_id IN (30, 60); LAST_NAME SALARY DEPARTMENT_ID Percentile_Cont Percent_Rank      Raphaely 11000 30 2850 0 Khoo 3100 30 2850 .2 Baida 2900 30 2850 .4 Tobias 2800 30 2850 .6 Himuro 2600 30 2850 .8 Colmenares 2500 30 2850 1 Hunold 9000 60 4800 0 Ernst 6000 60 4800 .25 Austin 4800 60 4800 .5 Pataballa 4800 60 4800 .5 Lorentz 4200 60 4800 1

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