Skip Headers
Oracle® Database SQL Language Reference
11g Release 2 (11.2)

Part Number E26088-02
Go to Documentation Home
Home
Go to Book List
Book List
Go to Table of Contents
Contents
Go to Index
Index
Go to Master Index
Master Index
Go to Feedback page
Contact Us

Go to previous page
Previous
Go to next page
Next
PDF · Mobi · ePub

PERCENTILE_DISC

Syntax

Description of percentile_disc.gif follows
Description of the illustration percentile_disc.gif

See Also:

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

Purpose

PERCENTILE_DISC is an inverse distribution function that assumes a discrete distribution model. It takes a percentile value and a sort specification and returns an element from the set. 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 conversion

The first expr must evaluate to a numeric value between 0 and 1, because it is a percentile value. This expression must be constant within each aggregate group. The ORDER BY clause takes a single expression that can be of any type that can be sorted.

For a given percentile value P, PERCENTILE_DISC sorts the values of the expression in the ORDER BY clause and returns the value with the smallest CUME_DIST value (with respect to the same sort specification) that is greater than or equal to P.

Aggregate Example

See aggregate example for PERCENTILE_CONT.

Analytic Example

The following example calculates the median discrete percentile of the salary of each employee in the sample table hr.employees:

SELECT last_name, salary, department_id,
       PERCENTILE_DISC(0.5) WITHIN GROUP (ORDER BY salary DESC)
         OVER (PARTITION BY department_id) "Percentile_Disc",
       CUME_DIST() OVER (PARTITION BY department_id 
         ORDER BY salary DESC) "Cume_Dist"
  FROM employees
  WHERE department_id in (30, 60)
  ORDER BY last_name, salary, department_id;

LAST_NAME                     SALARY DEPARTMENT_ID Percentile_Disc  Cume_Dist
------------------------- ---------- ------------- --------------- ----------
Austin                          4800            60            4800         .8
Baida                           2900            30            2900         .5
Colmenares                      2500            30            2900          1
Ernst                           6000            60            4800         .4
Himuro                          2600            30            2900 .833333333
Hunold                          9000            60            4800         .2
Khoo                            3100            30            2900 .333333333
Lorentz                         4200            60            4800          1
Pataballa                       4800            60            4800         .8
Raphaely                       11000            30            2900 .166666667
Tobias                          2800            30            2900 .666666667

The median value for Department 30 is 2900, which is the value whose corresponding percentile (Cume_Dist) is the smallest value greater than or equal to 0.5. The median value for Department 60 is 4800, which is the value whose corresponding percentile is the smallest value greater than or equal to 0.5.