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:

  1. 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

解释计划:

  1. SQL> select avg(a.salary) from
  2. 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;
  3. 已用时间: 00: 00: 00.00
  4.  
  5. 执行计划
  6. ----------------------------------------------------------
  7. Plan hash value: 9035349
  8.  
  9. ---------------------------------------------------------------------------------------------
  10. | Id | Operation | Name | Rows | Bytes |TempSpc| Cost (%CPU)| Time |
  11. ---------------------------------------------------------------------------------------------
  12. | 0 | SELECT STATEMENT | | 1 | 39 | | 8850 (2)| 00:01:47 |
  13. | 1 | SORT AGGREGATE | | 1 | 39 | | | |
  14. |* 2 | VIEW | | 2000K| 74M| | 8850 (2)| 00:01:47 |
  15. | 3 | WINDOW SORT | | 2000K| 7812K| 22M| 8850 (2)| 00:01:47 |
  16. | 4 | WINDOW SORT | | 2000K| 7812K| 22M| 8850 (2)| 00:01:47 |
  17. | 5 | TABLE ACCESS FULL| TB_EMPLOYEE | 2000K| 7812K| | 3045 (2)| 00:00:37 |
  18. ---------------------------------------------------------------------------------------------
  19.  
  20. Predicate Information (identified by operation id):
  21. ---------------------------------------------------
  22.  
  23. 2 - filter(ABS("A"."SEQ"-"A"."REVSEQ")<=1)

从Cost看是目前最快的。

执行时间:

  1. SQL> select avg(a.salary) from
  2. 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;
  3.  
  4. AVG(A.SALARY)
  5. -------------
  6. 5500
  7.  
  8. 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:

  1. 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)

解释计划:

  1. SQL> select avg(b.salary) from
  2. 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. 3 where b.diff=(select min(c.diff) from
  4. 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);
  5. 已用时间: 00: 00: 00.00
  6.  
  7. 执行计划
  8. ----------------------------------------------------------
  9. Plan hash value: 3874635296
  10.  
  11. -----------------------------------------------------------------------------------------------
  12. | Id | Operation | Name | Rows | Bytes |TempSpc| Cost (%CPU)| Time |
  13. -----------------------------------------------------------------------------------------------
  14. | 0 | SELECT STATEMENT | | 1 | 39 | | 19969 (2)| 00:04:00 |
  15. | 1 | SORT AGGREGATE | | 1 | 39 | | | |
  16. |* 2 | VIEW | | 2000K| 74M| | 11119 (2)| 00:02:14 |
  17. | 3 | WINDOW SORT | | 2000K| 19M| 38M| 11119 (2)| 00:02:14 |
  18. | 4 | WINDOW SORT | | 2000K| 19M| 38M| 11119 (2)| 00:02:14 |
  19. | 5 | TABLE ACCESS FULL | TB_EMPLOYEE | 2000K| 19M| | 3045 (2)| 00:00:37 |
  20. | 6 | SORT AGGREGATE | | 1 | 26 | | | |
  21. | 7 | VIEW | | 2000K| 49M| | 8850 (2)| 00:01:47 |
  22. | 8 | WINDOW SORT | | 2000K| 7812K| 22M| 8850 (2)| 00:01:47 |
  23. | 9 | WINDOW SORT | | 2000K| 7812K| 22M| 8850 (2)| 00:01:47 |
  24. | 10 | TABLE ACCESS FULL| TB_EMPLOYEE | 2000K| 7812K| | 3045 (2)| 00:00:37 |
  25. -----------------------------------------------------------------------------------------------
  26.  
  27. Predicate Information (identified by operation id):
  28. ---------------------------------------------------
  29.  
  30. 2 - filter(ABS("A"."SEQ"-"A"."REVSEQ")= (SELECT MIN(ABS("A"."SEQ"-"A"."REVSEQ"))
  31. FROM (SELECT DENSE_RANK() OVER ( ORDER BY "SALARY") "SEQ",DENSE_RANK() OVER ( ORDER
  32. BY INTERNAL_FUNCTION("SALARY") DESC ) "REVSEQ" FROM "TB_EMPLOYEE" "TB_EMPLOYEE" ORDER
  33. 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:

  1. 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

解释计划:

  1. SQL> select avg(a.salary) from
  2. 2 (select
  3. 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. 4 from tb_employee) a
  5. 5 where a.seq=a.ceil or a.revseq=a.ceil;
  6. 已用时间: 00: 00: 00.00
  7.  
  8. 执行计划
  9. ----------------------------------------------------------
  10. Plan hash value: 3541675306
  11.  
  12. -----------------------------------------------------------------------------------------------
  13. | Id | Operation | Name | Rows | Bytes |TempSpc| Cost (%CPU)| Time |
  14. -----------------------------------------------------------------------------------------------
  15. | 0 | SELECT STATEMENT | | 1 | 52 | | 14656 (3)| 00:02:56 |
  16. | 1 | SORT AGGREGATE | | 1 | 52 | | | |
  17. | 2 | SORT AGGREGATE | | 1 | | | | |
  18. | 3 | INDEX FAST FULL SCAN| SYS_C0012264 | 2000K| | | 1474 (2)| 00:00:18 |
  19. |* 4 | VIEW | | 2000K| 99M| | 14656 (3)| 00:02:56 |
  20. | 5 | WINDOW SORT | | 2000K| 7812K| 22M| 14656 (3)| 00:02:56 |
  21. | 6 | WINDOW SORT | | 2000K| 7812K| 22M| 14656 (3)| 00:02:56 |
  22. | 7 | TABLE ACCESS FULL | TB_EMPLOYEE | 2000K| 7812K| | 3045 (2)| 00:00:37 |
  23. -----------------------------------------------------------------------------------------------
  24.  
  25. Predicate Information (identified by operation id):
  26. ---------------------------------------------------
  27.  
  28. 4 - filter("A"."SEQ"="A"."CEIL" OR "A"."REVSEQ"="A"."CEIL")

消耗时间:

  1. SQL> ;
  2. 1 select avg(b.salary) from
  3. 2 (select
  4. 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
  5. 4 - filter("A"."SEQ"="A"."CEIL" OR "A"."REVSEQ"="A"."CEIL");
  6. 5* where a.seq=a.ceil or a.revseq=a.ceil
  7. SQL> select avg(a.salary) from
  8. 2 (select
  9. 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
  10. 4 from tb_employee) a
  11. 5 where a.seq=a.ceil or a.revseq=a.ceil;
  12.  
  13. AVG(A.SALARY)
  14. -------------
  15. 5498
  16.  
  17. 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:

  1. SQL> select count(*) from tb_employee;
  2. 已用时间: 00: 00: 00.01
  3.  
  4. 执行计划
  5. ----------------------------------------------------------
  6. Plan hash value: 2866911338
  7.  
  8. ------------------------------------------------------------------------------
  9. | Id | Operation | Name | Rows | Cost (%CPU)| Time |
  10. ------------------------------------------------------------------------------
  11. | 0 | SELECT STATEMENT | | 1 | 1474 (2)| 00:00:18 |
  12. | 1 | SORT AGGREGATE | | 1 | | |
  13. | 2 | INDEX FAST FULL SCAN| SYS_C0012264 | 2000K| 1474 (2)| 00:00:18 |
  14. ------------------------------------------------------------------------------

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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