MySQL 求中位數 兩個案例

簡單案例

簡單案例資料準備工作見此

使用者變數法：

SET @rownum := -1;

SELECT
AVG(t.mark) as median_num
FROM
    (SELECT @rownum:=@rownum + 1 AS rowindex,
    marks AS mark
    FROM median_even
    # (或medain_odd)
    ORDER BY marks) AS t
WHERE t.rowindex IN (FLOOR(@rownum / 2) , CEIL(@rownum / 2));

原理：

1. 排序，預設asc
2. 定義使用者變數，為資料進行標號（從0開始，初始取-1）
3. 無論求中位數的欄位的個數是奇數還是偶數，都取最大編號的一半的floor()和CEIL()，就是向下和向上取整，然後求平均，結果總是中位數

關乎使用者變數（SET @xxx:=x）的拓展：https://blog.csdn.net/JesseYoung/article/details/40779631

複雜案例

題目：https://leetcode-cn.com/problems/median-employee-salary/

SELECT Id, Company, Salary
FROM Employee
WHERE Id in (
    SELECT e1.Id
    FROM Employee e1
    JOIN Employee e2
    ON e1.Company = e2.Company
    GROUP BY e1.Id
    HAVING SUM(CASE WHEN e1.Salary >= e2.Salary THEN 1 ELSE 0 END) >= COUNT(*)/2 
    AND SUM(CASE WHEN e1.Salary <= e2.Salary THEN 1 ELSE 0 END) >= COUNT(*)/2
    )
GROUP BY Company, Salary
ORDER BY Company
 

-- 然後將工資>=的數量與COUNT(*)/2，進行對比，將工資<=的數量與COUNT(*)/2 進行對比
-- 還不理解的是    HAVING SUM(CASE WHEN e1.Salary >= e2.Salary THEN 1 ELSE 0 END) >= COUNT(*)/2 AND SUM(CASE WHEN e1.Salary <= e2.Salary THEN 1 ELSE 0 END) >= COUNT(*)/2的結果是什麼

• 首先對錶進行自連線，主鍵為id，但這裡on的是company，使得比如id為1的A就會對應六次A，每行兩個A且對應著兩列工資。內連線為工資的對比建立了基礎。（因為這裡是"各個公司的員工工資中位數"，因此要內連線到Company）

• 然後按照id分組，使得每個id的行數是id所在公司的員工數，每個id都對應該公司所有員工的工資

• 對於下面語句的理解是

   HAVING SUM(CASE WHEN e1.Salary >= e2.Salary THEN 1 ELSE 0 END) >= COUNT(*)/2 AND SUM(CASE WHEN e1.Salary <= e2.Salary THEN 1 ELSE 0 END) >= COUNT(*)/2
  

  也就是說明這裡的then 1其實是在選子集，而不是真的一個一個為1然後sum。這句語句的真正含義其實是選出兩個子集中的重複值

另外拓展：https://leetcode-cn.com/problems/find-median-given-frequency-of-numbers/

方法一：

SELECT 
AVG(Number)median 
FROM
(SELECT n1.Number FROM Numbers n1 JOIN Numbers n2 ON n1.Number>=n2.Number 
 GROUP BY 
 n1.Number 
 HAVING 
 SUM(n2.Frequency)>=(SELECT SUM(Frequency) FROM Numbers)/2 
 AND 
 SUM(n2.Frequency)-AVG(n1.Frequency)<=(SELECT SUM(Frequency) FROM Numbers)/2
)s

• 核心原理：如果 n1.Number 為中位數，n1.Number（包含本身）前累計的數字應大於等於總數/2 同時n1.Number（不包含本身）前累計數字應小於等於總數/2

方法二：

select avg(n) as median from 
(
    select Number as n, @c1 + 1 as 'c1', (@c1 := @c1 + Frequency) as 'c2', t2.s
    from Numbers, (select @c1 := 0) t1, (select sum(Frequency) as s
    from Numbers) t2
    order by n
) tmp
where c1 <= s/2 + 1 and c2 >= s/2

拓展題目還未完全理解。