Overview and Introduction
How Analytic Functions Work
The Syntax
Calculate a running Total
Top-N Queries
    Example 1
    Example 2
Windows
    Range Windows
    Compute average salary for defined range
    Row Windows
    Accessing Rows Around Your Current Row
LAG
LEAD
Determine the First Value / Last Value of a Group
Crosstab or Pivot Queries
ROLLUP and RANK Examples

Overview

Analytic Functions are designed to address such problems as "Calculate a running total", "Find percentages within a group", "Top-N queries", "Compute a moving average" and many more. Most of these problems can be solved using standard PL/SQL, however the performance is often not what it should be. Analytic Functions add extensions to the SQL language that not only make these operations easier to code; they make them faster than could be achieved with pure SQL or PL/SQL.

How Analytic Functions Work ?

Analytic functions compute an aggregate value based on a group of rows. They differ from aggregate functions in that they return multiple rows for each group. The group of rows is called a window and is defined by the analytic clause. For each row, a "sliding" window of rows is defined. The window determines the range of rows used to perform the calculations for the "current row". Window sizes can be based on either a physical number of rows or a logical interval such as time.
Analytic functions are the last set of operations performed in a query except for the final ORDER BY clause. All joins and all WHERE, GROUP BY, and HAVING clauses are completed before the analytic functions are processed. Therefore, analytic functions can appear only in the select list or ORDER BY clause.

The Syntax

The Syntax of analytic functions is straightforward in appearance:

Analytic-Function(<Argument>,<Argument>,...)
OVER (
  <Query-Partition-Clause>
  <Order-By-Clause>
  <Windowing-Clause>
)

• Analytic-Function
  Specify the name of an analytic function, Oracle actually provides many analytic functions such as AVG, CORR, COVAR_POP, COVAR_SAMP, COUNT, CUME_DIST, DENSE_RANK, FIRST, FIRST_VALUE, LAG, LAST, LAST_VALUE, LEAD, MAX, MIN, NTILE, PERCENT_RANK, PERCENTILE_CONT, PERCENTILE_DISC, RANK, RATIO_TO_REPORT, STDDEV, STDDEV_POP, STDDEV_SAMP, SUM, VAR_POP, VAR_SAMP, VARIANCE.

• Arguments
  Analytic functions take 0 to 3 arguments.

• Query-Partition-Clause
  The PARTITION BY clause logically breaks a single result set into N groups, according to the criteria set by the partition expressions. The words "partition" and "group" are used synonymously here. The analytic functions are applied to each group independently, they are reset for each group.

• Order-By-Clause
  The ORDER BY clause specifies how the data is sorted within each group (partition). This will definitely affect the outcome of any analytic function.

• Windowing-Clause
  The windowing clause gives us a way to define a sliding or anchored window of data, on which the analytic function will operate, within a group. This clause can be used to have the analytic function compute its value based on any arbitrary sliding or anchored window within a group. More information on windows can be found here.

The GROUPING_ID function returns a 0 or 1 when given a single column. (In this case, it works just like the GROUPING function.) If the return value is 0, indicating a detail record value for that particular column, then the column in question was not aggregated over (was not collapsed). If the function returns 1, then the column in question was aggregated over—any aggregates in the SELECT list would have been computed over that entire column's set of values. GROUPING_ID differs from GROUPING, in that you can send a list of columns and the GROUPING_ID function will treat the list as bits and return a decimal number. That means that the call to GROUPING_ID(a,b,c) might return any number between 0 and 7, because different 0/1 combinations are returned. Given that fact, we can use a CASE statement in the query to see if the row is a detail row for DEPTNO, for JOB, for neither, or for both.

Using GROUPING SETS in Listing 1, we asked for GROUP BY only on DEPTNO and then only on JOB. So, that one query was like running the following query

select deptno, null, sum(sal) 
from emp group by deptno
union all
select null, job, sum(sal) 
from emp group by job;

Code Listing 1: Using GROUPING_ID with GROUPING SETS

SQL> select deptno,
  2         job,
  3         sum(sal),
  4         grouping_id(deptno) gid_d,
  5         grouping_id(job) gid_j,
  6         grouping_id(deptno,job) gid_dj,
  7             bin_to_num(grouping_id(deptno),grouping_id(job)) b2n,
  8         case when grouping_id(deptno,job) = 0
  9              then 'Dtl both'
 10              when grouping_id(deptno,job) = 1
 11              then 'Agg over job'
 12              when grouping_id(deptno,job) = 2
 13              then 'Agg over deptno'
 14              when grouping_id(deptno,job) = 3
 15              then 'Agg over both'
 16              end what
 17    from emp
 18   group by grouping sets( (deptno), (job) )
 19  /
 
DEPTNO	    JOB	            SUM(SAL)	  GID_D	  GID_J	   GID_DJ     B2N  WHAT
________    ____________    ___________   _______ ________  ________  ____ __________________
10		            8750	  0	  1	    1	      1	   Agg over job
20		            0875	  0	  1	    1	      1	   Agg over job
30		            9400	  0	  1	    1	      1	   Agg over job
	    ANALYST	    6000	  1	  0	    2	      2	   Agg over deptno
            CLERK	    4150	  1	  0	    2	      2	   Agg over deptno
            MANAGER	    8275	  1	  0	    2         2	   Agg over deptno
            PRESIDENT	    5000	  1	  0	    2	      2	   Agg over deptno
            SALESMAN	    5600	  1	  0	    2	      2	   Agg over deptno
 
8 rows selected.

. . . but without having to make two passes on the EMP table, as would be the case with the UNION ALL.

In looking at the columns involved in the query in Listing 1, we can see that the function GROUPING(column_name) shows us when a column is aggregated over or preserved as a detail record. When GROUPING(deptno) = 0, DEPTNO is preserved in the output. When it is 1, it is aggregated over. However, we have two columns in this set we are aggregating by, for a total of four possible 0/1 combinations. (In this query, only two are possible.) Using the GROUPING_ID function on this vector of columns, we can easily see what each row represents. I've also included the alternative, more verbose way to accomplish this—the BIN_TO_NUM() function, to which we can send a list of 0s and 1s and get back a decimal number as well. I'm pretty sure you'll agree that GROUPING_ID(c1,c2,c3) is easier than the corresponding BIN_TO_NUM call with three GROUPING calls.

In Listing 2, we take a look at GROUP BY ROLLUP. A rollup by the two columns DEPTNO and JOB will produce

1. Detail records by DEPTNO, JOB (sum of SAL for each DEPTNO/JOB combination).
2. A summary record for each DEPTNO over JOB (like a subtotal).
3. A summary record over DEPTNO and JOB—a single aggregate for the entire result. Listing 2 shows the query and the results.

Code Listing 2: Using GROUPING_ID with GROUP BY ROLLUP

SQL> select deptno, job, sum(sal),
  2         grouping_id(deptno) gid_d,
  3         grouping_id(job) gid_j,
  4         case when grouping_id(deptno,job) = 0
  5              then 'Dtl both'
  6              when grouping_id(deptno,job) = 1
  7              then 'Agg over job'
  8              when grouping_id(deptno,job) = 2
  9              then 'Agg over deptno'
 10              when grouping_id(deptno,job) = 3
 11              then 'Agg over both'
 12              end what
 13    from emp
 14   group by rollup( deptno, job )
 15  /
 
DEPTNO	  JOB	        SUM(SAL)       GID_D	GID_J	  WHAT
________  _____________ ___________    _______  _______   ________________
10	  CLERK	         1300	       0	0	  Dtl both
10	  MANAGER	 2450	       0	0	  Dtl both
10	  PRESIDENT	 5000	       0	0	  Dtl both
10		         8750	       0	1	  Agg over job
20	  CLERK	         1900	       0	0	  Dtl both
20	  ANALYST	 6000	       0	0	  Dtl both
20	  MANAGER	 2975	       0	0	  Dtl both
20		        10875	       0	1	  Agg over job
30	  CLERK	          950	       0	0	  Dtl both
30	  MANAGER	 2850	       0	0	  Dtl both
30	  SALESMAN	 5600	       0	0	  Dtl both
30 		         9400	       0	1	  Agg over job
			29025	       1	1	  Agg over both
 
13 rows selected.

A rollup is sort of like a running total report, and GROUPING_ID tells us when the rollups happened. So the data is sorted by DEPTNO, JOB, and we have subtotals by DEPTNO (aggregated over JOB) and by DEPTNO, JOB (aggregated over both) along with the details by DEPTNO/JOB.

As you can see in Listing 2, the GROUPING_ID function was useful in telling us when we were dealing with a rolled-up record and the level of detail we could expect in that record.

Last, we'll look at GROUP BY CUBE. CUBE is similar to ROLLUP, in that you get the same three record types as shown in Listing 2 but also get all possible aggregations. CUBE grouping by DEPTNO and JOB will give you records by all of the following:

1. DEPTNO and JOB
2. DEPTNO over JOB
3. JOB over DEPTNO
4. A single total aggregate

You get every possible aggregate. Listing 3 shows the syntax and output and how to use the GROUPING_ID function to see what the level of detail is for each row. It is interesting to note that GROUP BY CUBE produces a superset of the rows we observed in the first query (in Listing 1). You could use GROUPING_ID with CUBE to generate the same result set as the original grouping sets query. That is, adding

having (grouping_id(deptno,job)=2  
    or (grouping_id(deptno,job)=1 

. . . to the GROUP BY CUBE query would cause it to be the logical equivalent of the GROUPING SETS query. But you shouldn't do that! If you need only some of the aggregates, use GROUPING SETS to get just the ones you need computed and avoid computing the others altogether. It would be fair to say that GROUPING_ID doesn't avoid multiple grouping functions but GROUPING SETS does. However, GROUPING_ID plays an important role in seeing what data is what.

Code Listing 3: Using GROUPING_ID with GROUP BY CUBE

 SQL> select deptno, job, sum(sal),
  2         grouping_id(deptno) gid_d,
  3         grouping_id(job) gid_j,
  4         case when grouping_id(deptno,job) = 0
  5              then 'Dtl both'
  6              when grouping_id(deptno,job) = 1
  7              then 'Agg over job'
  8              when grouping_id(deptno,job) = 2
  9              then 'Agg over deptno'
 10              when grouping_id(deptno,job) = 3
 11              then 'Agg over both'
 12              end what
 13    from emp
 14   group by cube( deptno, job )
 15   order by grouping_id(deptno,job)
 16  /
 
DEPTNO	    JOB	            SUM(SAL)	 GID_D	   GID_J    WHAT
_________   _____________   ___________  _______   _______  ___________________
10	    CLERK	     1300	 0	   0	    Dtl both
10	    MANAGER	     2450	 0	   0	    Dtl both
10	    PRESIDENT	     5000	 0	   0	    Dtl both
20	    CLERK	     1900	 0	   0	    Dtl both
30	    CLERK	      950	 0	   0	    Dtl both
30	    SALESMAN	     5600	 0	   0	    Dtl both
30	    MANAGER	     2850	 0	   0	    Dtl both
20	    MANAGER	     2975	 0	   0	    Dtl both
20	    ANALYST	     6000	 0	   0	    Dtl both
10		             8750	 0	   1	    Agg over job
20		            10875	 0	   1	    Agg over job
30		             9400	 0	   1	    Agg over job
	    CLERK	     4150	 1	   0	    Agg over deptno
	    ANALYST	     6000	 1	   0	    Agg over deptno
	    MANAGER	     8275	 1	   0	    Agg over deptno
	    PRESIDENT	     5000	 1	   0	    Agg over deptno
	    SALESMAN	     5600	 1	   0	    Agg over deptno
			    29025	 1	   1	    Agg over both
 
18 rows selected.

Example: Calculate a running Total

This example shows the cumulative salary within a departement row by row, with each row including a summation of the prior rows salary.

set autotrace traceonly explain
break on deptno skip 1
column ename format A6
column deptno format 999
column sal format 99999
column seq format 999

SELECT ename "Ename", deptno "Deptno", sal "Sal",
  SUM(sal) OVER (ORDER BY deptno, ename) "Running Total",
  SUM(SAL) OVER (PARTITION BY deptno
                 ORDER BY ename) "Dept Total",
  ROW_NUMBER()
    OVER (PARTITION BY deptno ORDER BY ENAME) "Seq"
FROM emp
ORDER BY deptno, ename
/

Ename  Deptno    Sal Running Total Dept Total  Seq
------ ------ ------ ------------- ---------- ----
CLARK      10   2450          2450       2450    1
KING            5000          7450       7450    2
MILLER          1300          8750       8750    3

ADAMS      20   1100          9850       1100    1
FORD            3000         12850       4100    2
JONES           2975         15825       7075    3
SCOTT           3000         18825      10075    4
SMITH            800         19625      10875    5

ALLEN      30   1600         21225       1600    1
BLAKE           2850         24075       4450    2
JAMES            950         25025       5400    3
MARTIN          1250         26275       6650    4
TURNER          1500         27775       8150    5
WARD            1250         29025       9400    6

Execution Plan
---------------------------------------------------
   0      SELECT STATEMENT Optimizer=CHOOSE
   1    0   WINDOW (SORT)
   2    1     TABLE ACCESS (FULL) OF 'EMP'
Statistics
---------------------------------------------------
          0  recursive calls
          0  db block gets
          3  consistent gets
          0  physical reads
          0  redo size
       1658  bytes sent via SQL*Net to client
        503  bytes received via SQL*Net from client
          2  SQL*Net roundtrips to/from client
          1  sorts (memory)
          0  sorts (disk)
         14  rows processed

The example shows how to calculate a "Running Total" for the entire query. This is done using the entire ordered result set, via SUM(sal) OVER (ORDER BY deptno, ename).
Further, we were able to compute a running total within each department, a total that would be reset at the beginning of the next department. The PARTITION BY deptno in that SUM(sal) caused this to happen, a partitioning clause was specified in the query in order to break the data up into groups.
The ROW_NUMBER() function is used to sequentially number the rows returned in each group, according to our ordering criteria (a "Seq" column was added to in order to display this position).
The execution plan shows, that the whole query is very well performed with only 3 consistent gets, this can never be accomplished with standard SQL or even PL/SQL.

Top-N Queries

How can we get the Top-N records by some set of fields ?
Prior to having access to these analytic functions, questions of this nature were extremely difficult to answer. There are some problems with Top-N queries however; mostly in the way people phrase them. It is something to be careful about when designing reports. Consider this seemingly sensible request:
I would like the top three paid sales reps by department

Using the "traditional approach you can perform:
select *  from
(your_query)
 where rownum <= 10;

The problem with this question is that it is ambiguous because of repeated values, there might be four people who all make the same salary, what should we do then ?
Let's look at three examples, all use the well known table EMP.

Example 1

Let's look at what ROW_NUMBER can do. Here is an example query using ROW_NUMBER to assign an increasing number to each row in the EMP table after sorting by SAL DESC:

select ename, sal,
      row_number() over (order by sal desc) rn
  from emp
  order by sal desc;

ENAME    SAL    RN
-----   ----    --  
 KING   5000     1
 FORD   3000     2
SCOTT   3000     3
JONES   2975     4
 .
 .
 .
 JAMES    950   13
 SMITH    800   14

I can apply a predicate to ROW_NUMBER after it is assigned. For example

select * from (select ename, sal,
                      row_number() over (order by sal desc) rn
                from emp)
  where rn <= 3
  order by sal desc;

ENAME    SAL    RN
-----   ----    --  
 KING   5000     1
SCOTT   3000     2
 FORD   3000     3

So, that demonstrates how to perform a top-n query by using ROW_NUMBER and also points out a general issue with top-n queries. If you look at that result, you see two rows with the value 3000. What if, in the EMP table, three people, instead of just two, had a salary of 3000? The result obtained by the above query would be ambiguous—I would get three records, but the records I retrieved would be somewhat random. We will analyze that on the example 2:

Another Example:
Sort the sales people by salary from greatest to least. Give the first three rows. If there are less then three people in a department, this will return less than three records.

set autotrace on explain
break on deptno skip 1

SELECT * FROM (SELECT deptno, ename, sal, ROW_NUMBER()
                 OVER (PARTITION BY deptno
                          ORDER BY sal DESC)
                 Top3 FROM emp)
  WHERE Top3 <= 3;

    DEPTNO ENAME             SAL       TOP3
---------- ---------- ---------- ----------
        10 KING             5000          1
           CLARK            2450          2
           MILLER           1300          3

        20 SCOTT            3000          1
           FORD             3000          2
           JONES            2975          3

        30 BLAKE            2850          1
           ALLEN            1600          2
           TURNER           1500          3

This query works by sorting each partition (or group, which is the deptno), in a descending order, based on the salary column and then assigning a sequential row number to each row in the group as it is processed. The use of a WHERE clause after doing this to get just the first three rows in each partition.

Example 2

Bearing this in mind, I can use other analytic functions to remove the ambiguity from example 1. They will do so, but the analytic functions might return more than n rows. In my opinion, when the attribute I order by is not unique, I want my query to return all of the relevant records—not just the first n arbitrary ones. To that end, I can use the RANK and DENSE_RANK analytic functions. Let's take a look at what they do:

select ename,sal,
   row_number() over (order by sal desc)rn,
   rank() over (order by sal desc)rnk,
   dense_rank() over (order by sal desc)drnk
   from emp
  order by sal desc;

ENAME    SAL   RN   RNK   DRNK
-----   ----   --   ---   ----
 KING   5000    1     1      1
 FORD   3000    2     2      2
SCOTT   3000    3     2      2
JONES   2975    4     4      3
BLAKE   2850    5     5      4
CLARK   2450    6     6      5
 .
 .
 .

The main things to note here are the following:

• ROW_NUMBER assigns contiguous, unique numbers from 1..N to a result set.
• RANK does not assign unique numbers—FORD and SCOTT tied for second place—nor does it assign contiguous numbers. No record was assigned the value of 3, because two people tied for second place, and no one came in third, according to RANK.
• DENSE_RANK, like RANK, does not assign unique numbers, but it does assign contiguous numbers. Even though two records tied for second place, there is a third-place record.

You can use RANK and DENSE_RANK in the same way you would use ROW_NUMBER to restrict the number of rows returned, but obviously you'll get subtly different results. For example

select * from (select ename,sal,
                      dense_rank() over (order by sal desc)drnk
                 from emp) 
  where drnk <= 3
  order by sal desc;

ENAME    SAL   DRNK   
-----   ----   ---- 
 KING   5000      1
SCOTT   3000      2
 FORD   3000      2
JONES   2975      3

That query returns "the set of people who make the top three salaries," which is likely the desired result. Getting the first three records from EMP after sorting by SAL is rather arbitrary, because using exactly the same set of data, simply inserted in different orders, you could observe different result sets with ROW_NUMBER (because SAL is not unique). Using DENSE_RANK, however, I don't get precisely three records but, instead, a repeatable (deterministic) result set. And I suspect that I retrieve the set the end user really meant to retrieve—the set of people making the top three salaries.

Another Example:
Give me the set of sales people who make the top 3 salaries - that is, find the set of distinct salary amounts, sort them, take the largest three, and give me everyone who makes one of those values.

SELECT * FROM (SELECT deptno, ename, sal,
                 DENSE_RANK() OVER (PARTITION BY deptno ORDER BY sal desc)
                 TopN FROM emp)
   WHERE TopN <= 3
   ORDER BY deptno, sal DESC;

    DEPTNO ENAME             SAL       TOPN
---------- ---------- ---------- ----------
        10 KING             5000          1
           CLARK            2450          2
           MILLER           1300          3

        20 SCOTT            3000          1  <--- !
           FORD             3000          1  <--- !
           JONES            2975          2
           ADAMS            1100          3

        30 BLAKE            2850          1
           ALLEN            1600          2
        30 TURNER           1500          3

Here the DENSE_RANK function was used to get the top three salaries. We assigned the dense rank to the salary column and sorted it in a descending order.
The DENSE_RANK function computes the rank of a row in an ordered group of rows. The ranks are consecutive integers beginning with 1. The largest rank value is the number of unique values returned by the query. Rank values are not skipped in the event of ties. Rows with equal values for the ranking criteria receive the same rank.
The DENSE_RANK function does not skip numbers and will assign the same number to those rows with the same value. Hence, after the result set is built in the inline view, we can simply select all of the rows with a dense rank of three or less, this gives us everyone who makes the top three salaries by department number.

Windows

The windowing clause gives us a way to define a sliding or anchored window of data, on which the analytic function will operate, within a group. The default window is an anchored window that simply starts at the first row of a group an continues to the current row.
We can set up windows based on two criteria: RANGES of data values or ROWS offset from the current row. It can be said, that the existance of an ORDER BY in an analytic function will add a default window clause of RANGE UNBOUNDED PRECEDING. That says to get all rows in our partition that came before us as specified by the ORDER BY clause.
Let's look at an example with a sliding window within a group and compute the sum of the current row's SAL column plus the previous 2 rows in that group. If we need a report that shows the sum of the current employee's salary with the preceding two salaries within a departement, it would look like this.

break on deptno skip 1
column ename format A6
column deptno format 999
column sal format 99999

SELECT deptno "Deptno", ename "Ename", sal "Sal",
  SUM(SAL) OVER (PARTITION BY deptno
                 ORDER BY ename
                 ROWS 2 PRECEDING) "Sliding Total"
FROM emp
ORDER BY deptno, ename;

Deptno Ename     Sal Sliding Total
------ ------ ------ -------------
    10 CLARK    2450          2450
       KING     5000          7450
       MILLER   1300          8750

    20 ADAMS    1100          1100
       FORD     3000          4100
       JONES    2975          7075  ^
       SCOTT    3000          8975  |
       SMITH     800          6775  \-- Sliding Window

    30 ALLEN    1600          1600
       BLAKE    2850          4450
       JAMES     950          5400
       MARTIN   1250          5050
       TURNER   1500          3700
       WARD     1250          4000

The partition clause makes the SUM (sal) be computed within each department, independent of the other groups. Tthe SUM (sal) is ' reset ' as the department changes. The ORDER BY ENAME clause sorts the data within each department by ENAME; this allows the window clause: ROWS 2 PRECEDING, to access the 2 rows prior to the current row in a group in order to sum the salaries.
For example, if you note the SLIDING TOTAL value for SMITH is 6 7 7 5, which is the sum of 800, 3000, and 2975. That was simply SMITH's row plus the salary from the preceding two rows in the window.

Range Windows

Range windows collect rows together based on a WHERE clause. If I say ' range 5 preceding ' for example, this will generate a sliding window that has the set of all preceding rows in the group such that they are within 5 units of the current row. These units may either be numeric comparisons or date comparisons and it is not valid to use RANGE with datatypes other than numbers and dates.

Example
Count the employees which where hired within the last 100 days preceding the own hiredate. The range window goes back 100 days from the current row's hiredate and then counts the rows within this range. The solution ist to use the following window specification:

COUNT(*) OVER (ORDER BY hiredate ASC RANGE 100 PRECEDING)

column ename heading "Name" format a8
column hiredate heading "Hired" format a10
column hiredate_pre heading "Hired-100" format a10
column cnt heading "Cnt" format 99

SELECT ename, hiredate, hiredate-100 hiredate_pre,
       COUNT(*)  OVER (ORDER BY hiredate ASC
                       RANGE 100 PRECEDING) cnt
  FROM emp
 ORDER BY hiredate ASC;

Name     Hired      Hired-100  Cnt
-------- ---------- ---------- ---
SMITH    17-DEC-80  08-SEP-80    1
ALLEN    20-FEB-81  12-NOV-80    2
WARD     22-FEB-81  14-NOV-80    3
JONES    02-APR-81  23-DEC-80    3
BLAKE    01-MAY-81  21-JAN-81    4
CLARK    09-JUN-81  01-MAR-81    3
TURNER   08-SEP-81  31-MAY-81    2
MARTIN   28-SEP-81  20-JUN-81    2
KING     17-NOV-81  09-AUG-81    3
JAMES    03-DEC-81  25-AUG-81    5
FORD     03-DEC-81  25-AUG-81    5
MILLER   23-JAN-82  15-OCT-81    4
SCOTT    09-DEC-82  31-AUG-82    1
ADAMS    12-JAN-83  04-OCT-82    2

We ordered the single partition by hiredate ASC. If we look for example at the row for CLARK we can see that his hiredate was 09-JUN-81, and 100 days prior to that is the date 01-MAR-81. If we look who was hired between 01-MAR-81 and 09-JUN-81, we find JONES (hired: 02-APR-81) and BLAKE (hired: 01-MAY-81). This are 3 rows including the current row, this is what we see in the column "Cnt" of CLARK's row.

Compute average salary for defined range

As an example, compute the average salary of people hired within 100 days before for each employee. The query looks like this:

column ename heading "Name" format a8
column hiredate heading "Hired" format a10
column hiredate_pre heading "Hired-100" format a10
column avg_sal heading "Avg-100" format 999999

SELECT ename, hiredate, sal,
       AVG(sal) OVER (ORDER BY hiredate ASC
                      RANGE 100 PRECEDING) avg_sal
  FROM emp
ORDER BY hiredate ASC;

Name     Hired             SAL Avg-100
-------- ---------- ---------- -------
SMITH    17-DEC-80         800     800
ALLEN    20-FEB-81        1600    1200
WARD     22-FEB-81        1250    1217
JONES    02-APR-81        2975    1942
BLAKE    01-MAY-81        2850    2169
CLARK    09-JUN-81        2450    2758
TURNER   08-SEP-81        1500    1975
MARTIN   28-SEP-81        1250    1375
KING     17-NOV-81        5000    2583
JAMES    03-DEC-81         950    2340
FORD     03-DEC-81        3000    2340
MILLER   23-JAN-82        1300    2563
SCOTT    09-DEC-82        3000    3000
ADAMS    12-JAN-83        1100    2050

Look at CLARK again, since we understand his range window within the group. We can see that the average salary of 2758 is equal to (2975+2850+2450)/3. This is the average of the salaries for CLARK and the rows preceding CLARK, those of JONES and BLAKE. The data must be sorted in ascending order.

Row Windows

Row Windows are physical units; physical number of rows, to include in the window. For example you can calculate the average salary of a given record with the (up to 5) employees hired before them or after them as follows:

set numformat 9999
SELECT ename, hiredate, sal,
AVG(sal)
  OVER (ORDER BY hiredate ASC ROWS 5 PRECEDING) AvgAsc,
COUNT(*)
  OVER (ORDER BY hiredate ASC ROWS 5 PRECEDING) CntAsc,
AVG(sal)
  OVER (ORDER BY hiredate DESC ROWS 5 PRECEDING) AvgDes,
COUNT(*)
  OVER (ORDER BY hiredate DESC ROWS 5 PRECEDING) CntDes
FROM emp
ORDER BY hiredate;

ENAME      HIREDATE    SAL AVGASC CNTASC AVGDES CNTDES
---------- --------- ----- ------ ------ ------ ------
SMITH      17-DEC-80   800    800      1   1988      6
ALLEN      20-FEB-81  1600   1200      2   2104      6
WARD       22-FEB-81  1250   1217      3   2046      6
JONES      02-APR-81  2975   1656      4   2671      6
BLAKE      01-MAY-81  2850   1895      5   2675      6
CLARK      09-JUN-81  2450   1988      6   2358      6
TURNER     08-SEP-81  1500   2104      6   2167      6
MARTIN     28-SEP-81  1250   2046      6   2417      6
KING       17-NOV-81  5000   2671      6   2392      6
JAMES      03-DEC-81   950   2333      6   1588      4
FORD       03-DEC-81  3000   2358      6   1870      5
MILLER     23-JAN-82  1300   2167      6   1800      3
SCOTT      09-DEC-82  3000   2417      6   2050      2
ADAMS      12-JAN-83  1100   2392      6   1100      1

The window consist of up to 6 rows, the current row and five rows " in front of " this row, where " in front of " is defined by the ORDER BY clause. With ROW partitions, we do not have the limitation of RANGE partition - the data may be of any type and the order by may include many columns. Notice, that we selected out a COUNT(*) as well. This is useful just to demonstrate how many rows went into making up a given average. We can see clearly that for ALLEN's record, the average salary computation for people hired before him used only 2 records whereas the computation for salaries of people hired after him used 6.

Accessing Rows Around Your Current Row

Frequently you want to access data not only from the current row but the current row " in front of " or " behind " them. For example, let's say you need a report that shows, by department all of the employees; their hire date; how many days before was the last hire; how many days after was the next hire.
Using straight SQL this query would be difficult to write. Not only that but its performance would once again definitely be questionable. The approach I typically took in the past was either to " select a select " or write a PL/SQL function that would take some data from the current row and " find " the previous and next rows data. This worked, but introduce large overhead into both the development of the query and the run-time execution of the query.
Using analytic functions, this is easy and efficient to do.

set echo on
column deptno format 99 heading Dep
column ename format a6 heading Ename
column hiredate heading Hired
column last_hire heading LastHired
column days_last heading DaysLast
column next_hire heading NextHire
column days_next heading NextDays
break on deptno skip 1

SELECT deptno, ename, hiredate,
LAG(hiredate,1,NULL)
  OVER (PARTITION BY deptno
        ORDER BY hiredate, ename) last_hire,
hiredate - LAG(hiredate,1,NULL)
  OVER (PARTITION BY deptno
        ORDER BY hiredate, ename) days_last,
LEAD(hiredate,1,NULL)
  OVER (PARTITION BY deptno
        ORDER BY hiredate, ename) next_hire,
LEAD(hiredate,1,NULL)
  OVER (PARTITION BY deptno
        ORDER BY hiredate, ename) - hiredate days_next
FROM emp
ORDER BY deptno, hiredate;

Dep Ename  Hired     LastHired DaysLast NextHire  NextDays
--- ------ --------- --------- -------- --------- --------
 10 CLARK  09-JUN-81                    17-NOV-81      161
    KING   17-NOV-81 09-JUN-81      161 23-JAN-82       67
    MILLER 23-JAN-82 17-NOV-81       67

 20 SMITH  17-DEC-80                    02-APR-81      106
    JONES  02-APR-81 17-DEC-80      106 03-DEC-81      245
    FORD   03-DEC-81 02-APR-81      245 09-DEC-82      371
    SCOTT  09-DEC-82 03-DEC-81      371 12-JAN-83       34
    ADAMS  12-JAN-83 09-DEC-82       34

 30 ALLEN  20-FEB-81                    22-FEB-81        2
    WARD   22-FEB-81 20-FEB-81        2 01-MAY-81       68
    BLAKE  01-MAY-81 22-FEB-81       68 08-SEP-81      130
    TURNER 08-SEP-81 01-MAY-81      130 28-SEP-81       20
    MARTIN 28-SEP-81 08-SEP-81       20 03-DEC-81       66
    JAMES  03-DEC-81 28-SEP-81       66

The LEAD and LAG routines could be considered a way to " index into your partitioned group ". Using these functions you can access any individual row. Notice for example in the above printout, it shows that the record  for KING  includes the data (in bold red font) from the prior row (LAST HIRE) and the next row (NEXT-HIRE). We can access the fields in records preceding or following the current record in an ordered partition easily.

LAG

LAG ( value_expr [, offset] [, default] )
   OVER ( [query_partition_clause] order_by_clause )

LAG provides access to more than one row of a table at the same time without a self join. Given a series of rows returned from a query and a position of the cursor, LAG provides access to a row at a given physical offset prior to that position. If you do not specify offset, then its default is 1. The optional default value is returned if the offset goes beyond the scope of the window. If you do not specify default, then its default value is null. The following example provides, for each person in the EMP table, the salary of the employee hired just before:

SELECT ename,hiredate,sal,
       LAG(sal, 1, 0) OVER (ORDER BY hiredate) AS PrevSal
FROM emp
WHERE job = 'CLERK';

Ename  Hired       SAL PREVSAL
------ --------- ----- -------
SMITH  17-DEC-80   800       0
JAMES  03-DEC-81   950     800
MILLER 23-JAN-82  1300     950
ADAMS  12-JAN-83  1100    1300

LEAD

LEAD ( value_expr [, offset] [, default] )
   OVER ( [query_partition_clause] order_by_clause )

LEAD provides access to more than one row of a table at the same time without a self join. Given a series of rows returned from a query and a position of the cursor, LEAD provides access to a row at a given physical offset beyond that position. If you do not specify offset, then its default is 1. The optional default value is returned if the offset goes beyond the scope of the table. If you do not specify default, then its default value is null. The following example provides, for each employee in the EMP table, the hire date of the employee hired just after:

SELECT ename, hiredate,
    LEAD(hiredate, 1) OVER (ORDER BY hiredate) AS NextHired
FROM emp WHERE deptno = 30;

Ename  Hired     NEXTHIRED
------ --------- ---------
ALLEN  20-FEB-81 22-FEB-81
WARD   22-FEB-81 01-MAY-81
BLAKE  01-MAY-81 08-SEP-81
TURNER 08-SEP-81 28-SEP-81
MARTIN 28-SEP-81 03-DEC-81
JAMES  03-DEC-81

Determine the First Value / Last Value of a Group

The FIRST_VALUE and LAST_VALUE functions allow you to select the first and last rows from a group. These rows are especially valuable because they are often used as the baselines in calculations.
Example
The following example selects, for each employee in each department, the name of the employee with the lowest salary.

break on deptno skip 1
SELECT deptno, ename, sal,
  FIRST_VALUE(ename)
  OVER (PARTITION BY deptno
        ORDER BY sal ASC) AS MIN_SAL_HAS
FROM emp
ORDER BY deptno, ename;

    DEPTNO ENAME             SAL MIN_SAL_HAS
---------- ---------- ---------- -----------
        10 CLARK            2450 MILLER
           KING             5000 MILLER
           MILLER           1300 MILLER

        20 ADAMS            1100 SMITH
           FORD             3000 SMITH
           JONES            2975 SMITH
           SCOTT            3000 SMITH
           SMITH             800 SMITH

        30 ALLEN            1600 JAMES
           BLAKE            2850 JAMES
           JAMES             950 JAMES
           MARTIN           1250 JAMES
           TURNER           1500 JAMES
           WARD             1250 JAMES

The following example selects, for each employee in each department, the name of the employee with the highest salary.

SELECT deptno, ename, sal, 
  FIRST_VALUE(ename)
  OVER (PARTITION BY deptno
        ORDER BY sal DESC) AS MAX_SAL_HAS
FROM emp
ORDER BY deptno, ename;

    DEPTNO ENAME             SAL MAX_SAL_HAS
---------- ---------- ---------- -----_-----
        10 CLARK            2450 KING
           KING             5000 KING
           MILLER           1300 KING

        20 ADAMS            1100 FORD
           FORD             3000 FORD
           JONES            2975 FORD
           SCOTT            3000 FORD
           SMITH             800 FORD

        30 ALLEN            1600 BLAKE
           BLAKE            2850 BLAKE
           JAMES             950 BLAKE
           MARTIN           1250 BLAKE
           TURNER           1500 BLAKE
           WARD             1250 BLAKE

The following example selects, for each employee in department 30 the name of the employee with the lowest salary using an inline view

SELECT deptno, ename, sal,
  FIRST_VALUE(ename)
  OVER (ORDER BY sal ASC) AS MIN_SAL_HAS
FROM (SELECT * FROM emp WHERE deptno = 30)

    DEPTNO ENAME             SAL MIN_SAL_HAS
---------- ---------- ---------- -----------
        30 JAMES             950 JAMES
           MARTIN           1250 JAMES
           WARD             1250 JAMES
           TURNER           1500 JAMES
           ALLEN            1600 JAMES
           BLAKE            2850 JAMES

Crosstab or Pivot Queries

A crosstab query, sometimes known as a pivot query, groups your data in a slightly different way from those we have seen hitherto. A crosstab query can be used to get a result with three rows (one for each project), with each row having three columns (the first listing the projects and then one column for each year) -- like this:

Project        2001        2002
     ID         CHF         CHF
-------------------------------
    100      123.00      234.50
    200      543.00      230.00
    300      238.00      120.50

Example:
Let's say you want to show the top 3 salary earners in each department as columns. The query needs to return exactly 1 row per department and the row would have 4 columns. The DEPTNO, the name of the highest paid employee in the department, the name of the next highest paid, and so on. Using analytic functions this almost easy, without analytic functions this was virtually impossible.

SELECT deptno,
  MAX(DECODE(seq,1,ename,null)) first,
  MAX(DECODE(seq,2,ename,null)) second,
  MAX(DECODE(seq,3,ename,null)) third
FROM (SELECT deptno, ename,
       row_number()
       OVER (PARTITION BY deptno
             ORDER BY sal desc NULLS LAST) seq
       FROM emp)
WHERE seq <= 3
GROUP BY deptno;

    DEPTNO FIRST      SECOND     THIRD
---------- ---------- ---------- ----------
        10 KING       CLARK      MILLER
        20 SCOTT      FORD       JONES
        30 BLAKE      ALLEN      TURNER

Note the inner query, that assigned a sequence (RowNr) to each employee by department number in order of salary.

SELECT deptno, ename, sal,
 row_number()
 OVER (PARTITION BY deptno
       ORDER BY sal desc NULLS LAST) RowNr
FROM emp;

    DEPTNO ENAME             SAL      ROWNR
---------- ---------- ---------- ----------
        10 KING             5000          1
        10 CLARK            2450          2
        10 MILLER           1300          3
        20 SCOTT            3000          1
        20 FORD             3000          2
        20 JONES            2975          3
        20 ADAMS            1100          4
        20 SMITH             800          5
        30 BLAKE            2850          1
        30 ALLEN            1600          2
        30 TURNER           1500          3
        30 WARD             1250          4
        30 MARTIN           1250          5
        30 JAMES             950          6

The DECODE in the outer query keeps only rows with sequences 1, 2 or 3 and assigns them to the correct "column". The GROUP BY gets rid of the redundant rows and we are left with our collapsed result. It may be easier to understand if you see the resultset without the aggregate function MAX grouped by deptno.

SELECT deptno,
  DECODE(seq,1,ename,null) first,
  DECODE(seq,2,ename,null) second,
  DECODE(seq,3,ename,null) third
FROM (SELECT deptno, ename,
       row_number()
       OVER (PARTITION BY deptno
             ORDER BY sal desc NULLS LAST) seq
       FROM emp)
WHERE seq <= 3;

    DEPTNO FIRST      SECOND     THIRD
---------- ---------- ---------- ----------
        10 KING
        10            CLARK
        10                       MILLER
        20 SCOTT
        20            FORD
        20                       JONES
        30 BLAKE
        30            ALLEN
        30                       TURNER

The MAX aggregate function will be applied by the GROUP BY column DEPTNO. In any given DEPTNO above only one row will have a non-null value for FIRST, the remaining rows in that group will always be NULL. The MAX function will pick out the non-null row and keep that for us. Hence, the group by and MAX will collapse our resultset, removing the NULL values from it and giving us what we want.