在MySQL中，如何计算一组数据的中位数？

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要得到一组数据的中位数（例如某个地区或某家公司的收入中位数），我们首先要将这一任务细分为3个小任务：

将数据排序，并给每一行数据给出其在所有数据中的排名。
找出中位数的排名数字。
找出中间排名对应的值。

举例说明：

建表语句：

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 ('张三', 10000);
INSERT INTO test.income (name, income) VALUES ('王二', 16000);
INSERT INTO test.income (name, income) VALUES ('土豪', 40000);

小任务1的查询语句：

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;

小任务2的查询语句：

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

小任务3的查询语句：

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) a3WHERE rank = (SELECT (COUNT(*) + 1) DIV 2
FROM income)

至此，我们就找到了如何从一组数据中获得中位数的方法。

下面，来介绍另外一种优化排名语句的方法。

我们都知道如何给一组数据做排序操作，在本例中，实现方法如下：

SELECT
name,
income
FROM income
ORDER BY income DESC

那我们可不可以更进一步，对查询出的结果加一列，这一列的数据为排名呢？

我们可以通过3个自定义变量的方法来实现这一目标：

第一个变量用来记录当前行数据的收入
第二个变量用来记录上一行数据的收入
第三个变量用来记录当前行数据的排名

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

查询结果如下：

然后再找出中位数的排名数字，进一步找出收入的中位数：

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，我们找了两种方法来解决中位数的问题。撒花。