有关中位数计算是什么

一.项目背景

中位数是数理统计中一个重要的指标，它可以自动忽略数据极差带来的影响，
能够很好的评估数据，在数理统计中很常用。本文主要介绍在Python中和Mysql
中如何来求中位数,重点让大家掌握SQL计算中位数，也是面试常考题目之一。

二.实现过程

1.Python实现
》》创建DataFrame
》》分组计算中位数

import pandas as pd


#创建DataFrame
data=pd.DataFrame(
    {   'company':['A','A','A','A','B','B','B','B','B'],
        'salary':[1057,1874,2059,2268,6587,6637,6932,7415,7654]
    })
#输出分组统计值
print(data.groupby('company')['salary'].median().reset_index().rename({'salary':'median_salary'},axis=1))

2.SQL实现
》》统计字段长度
》》按照奇偶长度分别计算字段一半
》》按照字段排序统计顺序
》》筛选所需字段并计算

建表语句：
mysql> create table median_val(
    -> company varchar(20),
    -> salary int)
    -> engine=innodb default charset=utf8;
Query OK, 0 rows affected, 1 warning (0.07 sec)

插入数据：
mysql> insert into median_val(company,salary) values("A",1057),("A",1874),("A",2059),("A",2268),("B",7415),("B",7654),("B",6932),("B",6587),("B",6637);
Query OK, 9 rows affected (0.06 sec)
Records: 9  Duplicates: 0  Warnings: 0

计算中位数：
mysql> select
    -> company,
    -> round(avg(salary),1) as median_salary
    -> from
    -> (select
    -> company,
    -> salary,
    -> count(*) over(partition by company) as num_length,
    -> row_number() over(partition by company order by salary) as ranking,
    -> count(*) over(partition by company) /2 as num_company_even,
    -> ceil(count(*) over(partition by company) /2) as num_company_odd
    -> from
    -> median_val)a
    -> where
    -> (mod(num_length,2)=0 and ranking in (num_length/2,num_length/2+1))
    -> or
    -> (mod(num_length,2)=1 and ranking=ceil(num_length/2))
    -> group by
    -> company;
+---------+---------------+
| company | median_salary |
+---------+---------------+
| A       |        1966.5 |
| B       |        6932.0 |
+---------+---------------+
2 rows in set (0.00 sec)