Hivesql/Python求相同老师学生连续上多节课的数据

表结构：

表结构

方式一：

HIVESQL:

1.首先按照学生和上课时间排序（升序），并新增一列row值当作参考列

SELECT uid,`datetime`,teacher,id,ROW_NUMBER( ) OVER(order by `uid`,`datetime`) as rank from class_table结果如下：

2.利用lag(a,b,c)开窗函数，共三个参数,a为选中的滞后列，b为滞后的行数，c为空时默认值（可以不指定）

暂定连续上4节课

SELECT uid, `datetime`, teacher, `rank`

, lag(`rank`, 3) OVER (PARTITION BY uid, teacher ORDER BY `datetime`) AS `rank2`

FROM (

SELECT uid, `datetime`, teacher, ROW_NUMBER() OVER (ORDER BY `uid`, `datetime`) AS rank

FROM class_table

) tmp

结果如下：

3.通过判断列rank和rank2的关系即可找出相同老师学生连续上多节课的数据

SELECT *

FROM (

SELECT uid, `datetime`, teacher, `rank`

, lag(`rank`, 3) OVER (PARTITION BY uid, teacher ORDER BY `datetime`) AS `rank2`

FROM (

SELECT uid, `datetime`, teacher, ROW_NUMBER() OVER (ORDER BY `uid`, `datetime`) AS rank

FROM class_table

) tmp

) tmp2

WHERE `rank` = `rank2` + 3

调节滞后量即可查询其他连续数量

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方式二：

原理都是一样的，pandas需要额外加一列自增列

data.sort_values(by=['uid','datetime'],inplace=True)

data['row'] = range(len(data))

data['rank'] = data.sort_values('datetime').groupby(['uid','teacher'])['row'].shift(3)

data[data['row']==data['rank']+3]

希望能帮助到大家

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