Applies To: SQL Server 2014, SQL Server 2016 Preview
Computes a specific percentile for sorted values in an entire rowset or within distinct partitions of a rowset in SQL Server. 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. For example, PERCENTILE_DISC (0.5) will compute the 50th percentile (that is, the median) of an expression. PERCENTILE_DISC calculates the percentile based on a discrete distribution of the column values; the result is equal to a specific value in the column.
Applies to: SQL Server (SQL Server 2012 through current version), Azure SQL Data Warehouse Public Preview.
The percentile to compute. The value must range between 0.0 and 1.0.
- WITHIN GROUP ( ORDER BY order_by_expression [ ASC | DESC ])
Specifies a list of numeric values to sort and compute the percentile over. Only one order_by_expression is allowed. The default sort order is ascending.
- OVER ( <partition_by_clause> )
Divides the result set produced by the FROM clause into partitions to which the percentile function is applied. For more information, see OVER Clause (Transact-SQL). The <ORDER BY clause> and <rows or range clause>cannot be specified in a PERCENTILE_DISC function.
The return type is determined by the order_by_expression type.
Under compatibility level 110 and higher, WITHIN GROUP is a reserved keyword. For more information, see ALTER DATABASE Compatibility Level (Transact-SQL).
Any nulls in the data set are ignored.
PERCENTILE_DISC is nondeterministic. For more information, see Deterministic and Nondeterministic Functions.
The following example uses PERCENTILE_CONT and PERCENTILE_DISC to find the median employee salary in each department. Note that these functions may not return the same value. This is because PERCENTILE_CONT interpolates the appropriate value, whether or not it exists in the data set, while PERCENTILE_DISC always returns an actual value from the set.
USE AdventureWorks2012; SELECT DISTINCT Name AS DepartmentName ,PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY ph.Rate) OVER (PARTITION BY Name) AS MedianCont ,PERCENTILE_DISC(0.5) WITHIN GROUP (ORDER BY ph.Rate) OVER (PARTITION BY Name) AS MedianDisc FROM HumanResources.Department AS d INNER JOIN HumanResources.EmployeeDepartmentHistory AS dh ON dh.DepartmentID = d.DepartmentID INNER JOIN HumanResources.EmployeePayHistory AS ph ON ph.BusinessEntityID = dh.BusinessEntityID WHERE dh.EndDate IS NULL;
Here is a partial result set.
DepartmentName MedianCont MedianDisc
-------------------- ---------- ----------
Document Control 16.8269 16.8269
Engineering 34.375 32.6923
Executive 54.32695 48.5577
Human Resources 17.427850 16.5865