Skip Headers

Oracle® Database SQL Reference
10g Release 1 (10.1)

Part Number B10759-01
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
Feedback

Go to previous page
Previous
Go to next page
Next
View PDF

PERCENTILE_DISC


Syntax

percentile_disc::=
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 datatype or any nonnumeric datatype that can be implicitly converted to a numeric datatype. The function returns the same datatype as the numeric datatype of the argument.


See Also:

Table 2-11, "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);

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

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.