另一种中位数求法

我在 https://www.cnblogs.com/xiandedanteng/p/12637767.html 里谈到过两种中位数求法，我又找出了一种新的中位数求法，核心解法仍是正序和逆序序列。

如果存在奇数个元素，观察其seq，revseq和ceil三值，你将发现中值就是seq等于ceil那条，或是revseq等于ceil那条：

SQL> select
  2  id,salary,rank() over (order by salary asc) as seq,rank() over (order by salary desc) as revseq,(select ceil(count(*)/2) from tb_coder1) as ceil
  3  from tb_coder1 order by salary;

        ID     SALARY        SEQ     REVSEQ       CEIL
---------- ---------- ---------- ---------- ----------
         5       3083          1          9          5
         6       6548          2          8          5
         4       9057          3          7          5
         8       9521          4          6          5
         3      11458          5          5          5
         9      14775          6          4          5
         1      16292          7          3          5
         2      17009          8          2          5
         7      18825          9          1          5

如果存在偶数个元素，再观察seq，revseq和ceil三值，你将发现中值就是seq或是revseq等于ceil那两条：

SQL> select
  2  id,salary,rank() over (order by salary asc) as seq,rank() over (order by salary desc) as revseq,(select ceil(count(*)/2) from tb_coder2) as ceil
  3  from tb_coder2 order by salary;

        ID     SALARY        SEQ     REVSEQ       CEIL
---------- ---------- ---------- ---------- ----------
         3       3396          1         10          5
         1       3682          2          9          5
         5       5029          3          8          5
         9       5859          4          7          5
         7       9428          5          6          5
         4       9691          6          5          5
         6      13613          7          4          5
        10      14366          8          3          5
         2      16375          9          2          5
         8      19127         10          1          5

已选择10行。

于是，最终的sql可以出来了：

select avg(a.salary) from
(select 
id,salary,rank() over (order by salary asc) as seq,rank() over (order by salary desc) as revseq,(select ceil(count(*)/2) from tb_coder1) as ceil 
from tb_coder2 order by salary) a
where a.seq=a.ceil or a.revseq=a.ceil

注意，有时需要把rank函数替换成row_number函数：

select avg(a.salary) from
(select 
id,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

对于这两个函数的细微差别我目前还有欠缺的地方。

对于奇数行，结果是：

SQL> select
  2  id,salary,rank() over (order by salary asc) as seq,rank() over (order by salary desc) as revseq,(select ceil(count(*)/2) from tb_coder1) as ceil
  3  from tb_coder1 order by salary;

        ID     SALARY        SEQ     REVSEQ       CEIL
---------- ---------- ---------- ---------- ----------
         5       3083          1          9          5
         6       6548          2          8          5
         4       9057          3          7          5
         8       9521          4          6          5
         3      11458          5          5          5
         9      14775          6          4          5
         1      16292          7          3          5
         2      17009          8          2          5
         7      18825          9          1          5

已选择9行。

SQL> select avg(a.salary) from
  2  (select
  3  id,salary,rank() over (order by salary asc) as seq,rank() over (order by salary desc) as revseq,(select ceil(count(*)/2) from tb_coder1) as ceil
  4  from tb_coder1 order by salary) a
  5  where a.seq=a.ceil or a.revseq=a.ceil;

AVG(A.SALARY)
-------------
        11458

对于偶数行，结果是：

SQL> select
  2  id,salary,rank() over (order by salary asc) as seq,rank() over (order by salary desc) as revseq,(select ceil(count(*)/2) from tb_coder2) as ceil
  3  from tb_coder2 order by salary;

        ID     SALARY        SEQ     REVSEQ       CEIL
---------- ---------- ---------- ---------- ----------
         3       3396          1         10          5
         1       3682          2          9          5
         5       5029          3          8          5
         9       5859          4          7          5
         7       9428          5          6          5
         4       9691          6          5          5
         6      13613          7          4          5
        10      14366          8          3          5
         2      16375          9          2          5
         8      19127         10          1          5

已选择10行。

SQL> select avg(a.salary) from
  2  (select
  3  id,salary,rank() over (order by salary asc) as seq,rank() over (order by salary desc) as revseq,(select ceil(count(*)/2) from tb_coder1) as ceil
  4  from tb_coder2 order by salary) a
  5  where a.seq=a.ceil or a.revseq=a.ceil;

AVG(A.SALARY)
-------------
       9559.5

--2020年4月11日--

千夫诺诺，不如一士谔谔。 能正面指出你的不足的人，才是真正对你有帮助的人。