1.将数据排序,并给每一行数据给出其在所有数据中的排名;
2.找出中位数的排名数字;
3.找出中心排名对应的值;
下面以某公司员工月收入为例,示例 MySQL 的一些庞杂语句的运用。
要领一
建立测试表
起首建立一个收入表,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);
完成使命 1
将数据排序,并给每一行数据给出其在所有数据中的排名:
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;
查询效果为:
完成小使命 2
找出中位数的排名数字:
SELECT (COUNT(*) + 1) DIV 2 as rank FROM employee;
查询效果为:
完成小使命 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)
查询效果为:
至此,我们就找到了怎样从一组数据中取得中位数的要领。
要领二
下面,来引见别的一种优化排名语句的要领。
我们都晓得怎样给一组数据做排序操纵,在本例中,完成要领以下:
SELECT name, income FROM employee ORDER BY income DESC
查询效果为:
那我们可不能够更进一步,对查询出的效果加一列,这一列的数据为排名呢?
我们能够经由过程 3 个自定义变量的要领来完成这一目的:
第一个变量用来纪录当前行数据的收入
第二个变量用来纪录上一行数据的收入
第三个变量用来纪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 employee ORDER BY income DESC
查询效果以下:
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 employee ORDER BY income DESC) AS t1 WHERE t1.rank = (SELECT (COUNT(*) + 1) DIV 2 FROM employee)
查询效果为:
至此,我们找了两种要领来处理中位数的题目。撒花。
引荐:《mysql教程》
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