在  https://www.cnblogs.com/xiandedanteng/p/12677688.html 中我列举了三种求中值方案,其中日本人MICK的做法因为不适用于二百万结果集而放弃,取而代之是新方案一。

新方案一:

经过思考后我又得出了一种中值的新解法,那就是利用排序后正向序列和反向序列交叉点为中值区的原理,如果两个序列相减小于等于一则求所在区域的均值即可。

SQL:

select avg(a.salary) from
(select salary,dense_rank() over (order by salary asc) as seq,dense_rank() over (order by salary desc) as revseq from tb_employee order by salary) a where abs(a.seq-a.revseq)<=1

解释计划:

SQL> select avg(a.salary) from
2 (select salary,dense_rank() over (order by salary asc) as seq,dense_rank() over (order by salary desc) as revseq from tb_employee order by salary) a where abs(a.seq-a.revseq)<=1;
已用时间: 00: 00: 00.00 执行计划
----------------------------------------------------------
Plan hash value: 9035349 ---------------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes |TempSpc| Cost (%CPU)| Time |
---------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 39 | | 8850 (2)| 00:01:47 |
| 1 | SORT AGGREGATE | | 1 | 39 | | | |
|* 2 | VIEW | | 2000K| 74M| | 8850 (2)| 00:01:47 |
| 3 | WINDOW SORT | | 2000K| 7812K| 22M| 8850 (2)| 00:01:47 |
| 4 | WINDOW SORT | | 2000K| 7812K| 22M| 8850 (2)| 00:01:47 |
| 5 | TABLE ACCESS FULL| TB_EMPLOYEE | 2000K| 7812K| | 3045 (2)| 00:00:37 |
--------------------------------------------------------------------------------------------- Predicate Information (identified by operation id):
--------------------------------------------------- 2 - filter(ABS("A"."SEQ"-"A"."REVSEQ")<=1)

从Cost看是目前最快的。

执行时间:

SQL> select avg(a.salary) from
2 (select salary,dense_rank() over (order by salary asc) as seq,dense_rank() over (order by salary desc) as revseq from tb_employee order by salary) a where abs(a.seq-a.revseq)<=1; AVG(A.SALARY)
-------------
5500 已用时间: 00: 00: 02.46

从运行时间上看排第二。

经局部优化后两种方案SQL如下:

原有方案二:

select avg(b.salary) from
(select a.salary,abs(a.seq-a.revseq) as diff from (select id,salary,dense_rank() over (order by salary asc) as seq,dense_rank() over (order by salary desc) as revseq from tb_employee order by salary) a ) b
where b.diff=(select min(c.diff) from
(select abs(a.seq-a.revseq) as diff from (select dense_rank() over (order by salary asc) as seq,dense_rank() over (order by salary desc) as revseq from tb_employee order by salary) a ) c)

解释计划:

SQL> select avg(b.salary) from
2 (select a.salary,abs(a.seq-a.revseq) as diff from (select id,salary,dense_rank() over (order by salary asc) as seq,dense_rank() over (order by salary desc) as revseq from tb_employee order by salary) a ) b
3 where b.diff=(select min(c.diff) from
4 (select abs(a.seq-a.revseq) as diff from (select dense_rank() over (order by salary asc) as seq,dense_rank() over (order by salary desc) as revseq from tb_employee order by salary) a ) c);
已用时间: 00: 00: 00.00 执行计划
----------------------------------------------------------
Plan hash value: 3874635296 -----------------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes |TempSpc| Cost (%CPU)| Time |
-----------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 39 | | 19969 (2)| 00:04:00 |
| 1 | SORT AGGREGATE | | 1 | 39 | | | |
|* 2 | VIEW | | 2000K| 74M| | 11119 (2)| 00:02:14 |
| 3 | WINDOW SORT | | 2000K| 19M| 38M| 11119 (2)| 00:02:14 |
| 4 | WINDOW SORT | | 2000K| 19M| 38M| 11119 (2)| 00:02:14 |
| 5 | TABLE ACCESS FULL | TB_EMPLOYEE | 2000K| 19M| | 3045 (2)| 00:00:37 |
| 6 | SORT AGGREGATE | | 1 | 26 | | | |
| 7 | VIEW | | 2000K| 49M| | 8850 (2)| 00:01:47 |
| 8 | WINDOW SORT | | 2000K| 7812K| 22M| 8850 (2)| 00:01:47 |
| 9 | WINDOW SORT | | 2000K| 7812K| 22M| 8850 (2)| 00:01:47 |
| 10 | TABLE ACCESS FULL| TB_EMPLOYEE | 2000K| 7812K| | 3045 (2)| 00:00:37 |
----------------------------------------------------------------------------------------------- Predicate Information (identified by operation id):
--------------------------------------------------- 2 - filter(ABS("A"."SEQ"-"A"."REVSEQ")= (SELECT MIN(ABS("A"."SEQ"-"A"."REVSEQ"))
FROM (SELECT DENSE_RANK() OVER ( ORDER BY "SALARY") "SEQ",DENSE_RANK() OVER ( ORDER
BY INTERNAL_FUNCTION("SALARY") DESC ) "REVSEQ" FROM "TB_EMPLOYEE" "TB_EMPLOYEE" ORDER
BY "SALARY") "A"))

执行时间:

SQL> select avg(b.salary) from
2 (select a.salary,abs(a.seq-a.revseq) as diff from (select id,salary,dense_rank() over (order by salary asc) as seq,dense_rank() over (order by salary desc) as revseq from tb_employee order by salary) a ) b
3 where b.diff=(select min(c.diff) from
4 (select abs(a.seq-a.revseq) as diff from (select dense_rank() over (order by salary asc) as seq,dense_rank() over (order by salary desc) as revseq from tb_employee order by salary) a ) c);

AVG(B.SALARY)
-------------
5500

已用时间: 00: 00: 04.56

原有方案三:

select avg(a.salary) from
(select
salary,row_number() over (order by salary asc) as seq,row_number() over (order by salary desc) as revseq,(select ceil(count(*)/2) from tb_employee) as ceil
from tb_employee) a
where a.seq=a.ceil or a.revseq=a.ceil

解释计划:

SQL> select avg(a.salary) from
2 (select
3 salary,row_number() over (order by salary asc) as seq,row_number() over (order by salary desc) as revseq,(select ceil(count(*)/2) from tb_employee) as ceil
4 from tb_employee) a
5 where a.seq=a.ceil or a.revseq=a.ceil;
已用时间: 00: 00: 00.00 执行计划
----------------------------------------------------------
Plan hash value: 3541675306 -----------------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes |TempSpc| Cost (%CPU)| Time |
-----------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 52 | | 14656 (3)| 00:02:56 |
| 1 | SORT AGGREGATE | | 1 | 52 | | | |
| 2 | SORT AGGREGATE | | 1 | | | | |
| 3 | INDEX FAST FULL SCAN| SYS_C0012264 | 2000K| | | 1474 (2)| 00:00:18 |
|* 4 | VIEW | | 2000K| 99M| | 14656 (3)| 00:02:56 |
| 5 | WINDOW SORT | | 2000K| 7812K| 22M| 14656 (3)| 00:02:56 |
| 6 | WINDOW SORT | | 2000K| 7812K| 22M| 14656 (3)| 00:02:56 |
| 7 | TABLE ACCESS FULL | TB_EMPLOYEE | 2000K| 7812K| | 3045 (2)| 00:00:37 |
----------------------------------------------------------------------------------------------- Predicate Information (identified by operation id):
--------------------------------------------------- 4 - filter("A"."SEQ"="A"."CEIL" OR "A"."REVSEQ"="A"."CEIL")

消耗时间:

SQL> ;
1 select avg(b.salary) from
2 (select
3 salary,row_number() over (order by salary asc) as seq,row_number() over (order by salary desc) as revseq,(select ceil(count(*)/2) from tb_employee) as ceil
4 - filter("A"."SEQ"="A"."CEIL" OR "A"."REVSEQ"="A"."CEIL");
5* where a.seq=a.ceil or a.revseq=a.ceil
SQL> select avg(a.salary) from
2 (select
3 salary,row_number() over (order by salary asc) as seq,row_number() over (order by salary desc) as revseq,(select ceil(count(*)/2) from tb_employee) as ceil
4 from tb_employee) a
5 where a.seq=a.ceil or a.revseq=a.ceil; AVG(A.SALARY)
-------------
5498 已用时间: 00: 00: 01.98

而单查表数量的解释计划大家也参考看一下:

SQL> select count(*) from tb_employee;
已用时间: 00: 00: 00.01 执行计划
----------------------------------------------------------
Plan hash value: 2866911338 ------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Cost (%CPU)| Time |
------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 1474 (2)| 00:00:18 |
| 1 | SORT AGGREGATE | | 1 | | |
| 2 | INDEX FAST FULL SCAN| SYS_C0012264 | 2000K| 1474 (2)| 00:00:18 |
------------------------------------------------------------------------------

最后的大比拼表格:

方案 Cost 耗时
新方案一 8850 2.46秒
原有方案二 19946 4.56秒
原有方案三 14646 1.98

截至目前,方案三以较高效率胜出,新方案一其实也不错,原有方案三排第三位。

--2020年4月12日--

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