在MySQL中，如何计算一组数据的中位数？

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) a3 WHERE 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。

 

转载于:https://www.cnblogs.com/jxlwqq/p/5868206.html