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:

MySQL中如何计算一组数据的中位数?

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:

MySQL中如何计算一组数据的中位数?

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。

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