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首页 / 数据库 / MySQL / 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。本文永久更新链接地址