在 MySQL 中,怎样盘算一组数据的中位数
发布时间:09/01 来源:未知 浏览:
关键词:
1.将数据排序,并给每一行数据给出其在所有数据中的排行;
2.寻出中位数的排行数字;
3.寻出中心排行对应的值;
下面以某公司员工月收入为例,示例 MySQL 的一些复杂语句的使用。
办法一
创立测试表
第一创立一个收入表,建表语句为:
CREATE TABLE IF NOT EXISTS `employee` ( `id` INT AUTO_INCREMENT PRIMARY KEY, `name` VARCHAR(10) NOT NULL DEFAULT '', `income` INT NOT NULL DEFAULT '0' ) ENGINE = InnoDB DEFAULT CHARSET = utf8; INSERT INTO `employee` (`name`, `income`) VALUES ('麻子', 20000); INSERT INTO `employee` (`name`, `income`) VALUES ('李四', 12000); INSERT INTO `employee` (`name`, `income`) VALUES ('张三', 10000); INSERT INTO `employee` (`name`, `income`) VALUES ('王二', 16000); INSERT INTO `employee` (`name`, `income`) VALUES ('土豪', 40000);
完成任务 1
将数据排序,并给每一行数据给出其在所有数据中的排行:
SELECT t1.name, t1.income, COUNT(*) AS rank FROM employee AS t1, employee AS t2 WHERE t1.income < t2.income OR (t1.income = t2.income AND t1.name <= t2.name) GROUP BY t1.name, t1.income ORDER BY rank;
查询结果为:
完成小任务 2
寻出中位数的排行数字:
SELECT (COUNT(*) + 1) DIV 2 as rank FROM employee;
查询结果为:
完成小任务 3
SELECT income AS median FROM (SELECT t1.name, t1.income, COUNT(*) AS rank FROM employee AS t1, employee AS t2 WHERE t1.income < t2.income OR (t1.income = t2.income AND t1.name <= t2.name) GROUP BY t1.name, t1.income ORDER BY rank) t3 WHERE rank = (SELECT (COUNT(*) + 1) DIV 2 FROM employee)
查询结果为:
至此,我们就寻到了怎样从一组数据中获得中位数的办法。
办法二
下面,来介绍别的一种优化排行语句的办法。
我们都知道怎样给一组数据做排序操纵,在本例中,实现办法如下:
SELECT name, income FROM employee ORDER BY income DESC
查询结果为:
那我们可不成以更进一步,对查询出的结果加一列,这一列的数据为排行呢?
我们可以通过 3 个自定义变量的办法来实现这一目标:
第一个变量用来记载当前行数据的收入
第二个变量用来记载上一行数据的收入
第三个变量用来记载当前行数据的排行
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 employee ORDER BY income DESC
查询结果如下:
然后再寻出中位数的排行数字,进一步寻出收入的中位数:
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 employee ORDER BY income DESC) AS t1 WHERE t1.rank = (SELECT (COUNT(*) + 1) DIV 2 FROM employee)
查询结果为:
至此,我们寻了两种办法来解决中位数的问题。撒花。
引荐:《mysql教程》
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