在MySQL中，如何計算一組數據的中位數？

要得到一組數據的中位數（例如某個地區或某家公司的收入中位數），我們首先要將這一任務細分為3個小任務：

1. 將數據排序，並給每一行數據給出其在所有數據中的排名。
2. 找出中位數的排名數字。
3. 找出中間排名對應的值。

舉例說明：

建表語句：

CREATE TABLE `income` (
  `name`   VARCHAR(10) NOT NULL DEFAULT '',
  `income` INT(11)     NOT NULL DEFAULT '0'
)
  ENGINE = InnoDB
  DEFAULT CHARSET = utf8;

INSERT INTO test.income (name, income) VALUES ('麻子', 20000);
INSERT INTO test.income (name, income) VALUES ('李四', 12000);
INSERT INTO test.income (name, income) VALUES ('張三', 10000);
INSERT INTO test.income (name, income) VALUES ('王二', 16000);
INSERT INTO test.income (name, income) VALUES ('土豪', 40000);

　　

小任務1的查詢語句：

SELECT
  a1.name,
  a1.income,
  count(*) AS rank
FROM income AS a1, income AS a2
WHERE a1.income < a2.income OR (a1.income = a2.income AND a1.name <= a2.name)
GROUP BY a1.name, a1.income
ORDER BY rank;

小任務2的查詢語句：

SELECT (COUNT(*) + 1) DIV 2
FROM income;

小任務3的查詢語句：

SELECT income AS median
FROM
  (SELECT
     a1.name,
     a1.income,
     count(*) AS rank
   FROM income AS a1, income AS a2
   WHERE a1.income < a2.income OR (a1.income = a2.income AND a1.name <= a2.name)
   GROUP BY a1.name, a1.income
   ORDER BY rank) a3

WHERE rank = (SELECT (COUNT(*) + 1) DIV 2
              FROM income)

至此，我們就找到了如何從一組數據中獲得中位數的方法。

下面，來介紹另外一種優化排名語句的方法。

我們都知道如何給一組數據做排序操作，在本例中，實現方法如下：

SELECT
  name,
  income
FROM income
ORDER BY income DESC

那我們可不可以更進一步，對查詢出的結果加一列，這一列的數據為排名呢？

我們可以通過3個自定義變量的方法來實現這一目標：

• 第一個變量用來記錄當前行數據的收入
• 第二個變量用來記錄上一行數據的收入
• 第三個變量用來記錄當前行數據的排名

SET @curr_income := 0;
SET @prev_income := 0;
SET @rank := 0;

SELECT
  name,
  @curr_income := income                                      AS income,
  @rank := if(@prev_income != @curr_income, @rank + 1, @rank) AS rank,
  @prev_income := @curr_income                                AS dummy
FROM income
ORDER BY income DESC

查詢結果如下：

然后再找出中位數的排名數字，進一步找出收入的中位數：

SET @curr_income := 0;
SET @prev_income := 0;
SET @rank := 0;

SELECT income AS median
FROM
  (SELECT
     name,
     @curr_income := income                                      AS income,
     @rank := if(@prev_income != @curr_income, @rank + 1, @rank) AS rank,
     @prev_income := @curr_income                                AS dummy
   FROM income
   ORDER BY income DESC) AS a1
WHERE a1.rank = (SELECT (COUNT(*) + 1) DIV 2
                 FROM income)

至此，我們找了兩種方法來解決中位數的問題。撒花。