Description of the illustration percentile_cont.gif

See Also:

"Analytic Functions" for information on syntax, semantics, and restrictions of the`OVER`

clause`PERCENTILE_CONT`

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.

This function takes as an argument any numeric data type or any nonnumeric data type that can be implicitly converted to a numeric data type. The function returns the same data type as the numeric data type of the argument.

See Also:

Table 3-10, "Implicit Type Conversion Matrix" for more information on implicit conversionThe 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, you can compute the row number you 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*(N-1))`

. 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 the result is (value of expression from row at RN) Otherwise the result is (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 `MEDIAN`

function is a specific case of `PERCENTILE_CONT`

where the percentile value defaults to 0.5. For more information, refer to MEDIAN.

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 ORDER BY department_id; DEPARTMENT_ID Median cont Median disc ------------- ----------- ----------- 10 4400 4400 20 9500 13000 30 2850 2900 40 6500 6500 50 3100 3100 60 4800 4800 70 10000 10000 80 8900 9000 90 17000 17000 100 8000 8200 110 10154 12008 7000 7000

`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) ORDER BY last_name, salary, department_id; LAST_NAME SALARY DEPARTMENT_ID Percentile_Cont Percent_Rank ------------------------- ---------- ------------- --------------- ------------ Austin 4800 60 4800 .5 Baida 2900 30 2850 .4 Colmenares 2500 30 2850 1 Ernst 6000 60 4800 .25 Himuro 2600 30 2850 .8 Hunold 9000 60 4800 0 Khoo 3100 30 2850 .2 Lorentz 4200 60 4800 1 Pataballa 4800 60 4800 .5 Raphaely 11000 30 2850 0 Tobias 2800 30 2850 .6