目录

一、数据炸裂

0 问题描述

1 数据准备

2 数据分析

3 小结

二、数据合并

0 问题描述

1 数据准备

2 数据分析

3 小结

一、数据炸裂

0 问题描述

    如何将字符串1-5,16,11-13,9" 扩展成 "1,2,3,4,5,16,11,12,13,9" 且顺序不变。

1 数据准备

with data as (select '1-5,16,11-13,9' as a)

2 数据分析

 步骤一：explode(split(a, ',')) 炸裂 + row_number()排序，一行变多行，且对每行的数据排序，保证有序性。

with data as (select '1-5,16,11-13,9' as a)
selecta,row_number() over () as rn
from (selectexplode(split(a, ',')) as afrom data)tmp1;

输出结果：

步骤二： lateral view explode(split(a, '-'))  、max(b) - min(b) as diff

（1）lateral view +explode 侧写和炸裂，一行变多行，并将源表中每行的输出结果与该行连接；

 （2）group by a, rn .......  select  min(b)   as start_index 得到每个分组的起始值

 （3）max(b) - min(b) 得到每个分组的步长

with data as (select '1-5,16,11-13,9' as a)
selecta,rn,min(b)          as start_data,max(b) - min(b) as diff
from (selecta,rn,bfrom (selecta,row_number() over () as rnfrom (selectexplode(split(a, ',')) as afrom data) tmp1) tmp2lateral view explode(split(a, '-')) table1 as b) tmp3
group by a, rn;

 输出结果是：

步骤三： 根据步长生成索引值，起始值加上索引值获取展开值

（1） lateral view posexplode(split(space(cast (diff as int)), '')) table1 as pos, item;
   侧写和炸裂，根据分组的步长 diff  生成对应的索引值pos

 （2）(start_data + pos) as  str,起始值加上索引值获取展开值

with data as (select '1-5,16,11-13,9' as a)
selecta,rn,cast ((start_data + pos) as int) as str
from (selecta,rn,start_index,diff,posfrom (selecta,rn,min(b) as start_data,max(b) - min(b) as difffrom (selecta,rn,bfrom (selecta,row_number() over () as rnfrom (selectexplode(split(a, ',')) as afrom data) tmp1) tmp2lateral view explode(split(a, '-')) table1 as b) tmp3group by a, rn) tmp4lateral view posexplode(split(space(cast(diff as int)), '')) table1 as pos, val) tmp5order by rn;

输出结果是： 

步骤四： 对a,rn, diff 字段分组，拼接str字符串得到最终结果值

with data as (select '1-5,16,11-13,9' as a)
selectconcat_ws(',', collect_set(cast(str as string))) as result
from (selecta,rn,cast((start_index + pos) as int) as strfrom (selecta,rn,start_index,diff,posfrom (selecta,rn,min(b)  as start_index,max(b) - min(b) as difffrom (selecta,rn,bfrom (selecta,row_number() over () as rnfrom (selectexplode(split(a, ',')) as afrom data) tmp1) tmp2lateral view explode(split(a, '-')) table1 as b) tmp3group by a, rn) tmp4lateral view posexplode(split(space(cast(diff as int)), '')) table1 as pos, val) tmp5) tmp6
group by a,rn,diff;

最终的输出结果：1,2,3,4,5,16,11,12,13,9 

3 小结

   数据炸裂的思路一般是：1.计算区间【a,b】的步长(差值)diff；2.利用split分割函数+ posexplode等 将一行变成 diff+1 行，生成对应的下角标pos(pos的取值为【0,diff】)；3.【a,b】区间的起始值 （a + pos） 将数据平铺开；4.基于平铺开后的数据集进一步加工处理，例如：分组聚合等。

二、数据合并

0 问题描述

   面试题：基于A表的数据生成B表数据

1 数据准备

create table if not exists  tableA
(id        string comment '用户id',name   string comment '用户姓名'
) comment 'A表';insert overwrite table tableA values('1','aa'),('2','aa'),('3','aa'),('4','d'),('5','c'),('6','aa'),('7','aa'),('8','e'),('9','f'),('10','g');create table if not exists  tableC
(id     string comment '用户id',name   string comment '用户姓名'
) comment 'C表';insert overwrite table tableC values('3','aa|aa|aa'),('4','d'),('5','c'),('7','aa|aa'),('8','e'),('9','f'),('10','g');

2 数据分析

 步骤1：寻找满足条件的断点

selectid,name,if(name != lag_name, 1, 0) as flag
from (selectid,name,lag(name, 1, name) over (order by cast(id as int)) as lag_namefrom tableA) tmp1;

输出结果为:

 步骤2：断点处标记为1，非断点处标记为0，并对断点标记值进行累加，构造分组标签

selectid,name,--并对断点标记值进行累加，构造分组标签sum(flag) over (order by cast(id as int)) grp
from (selectid,name,--断点处标记为1，非断点处标记为0if(name != lag_name, 1, 0) flagfrom (selectid,name,lag(name, 1, name) over (order by cast(id as int)) as lag_namefrom tableA) tmp1) tmp2;

输出结果为：

步骤3：按照分组标签进行数据合并，并取得分组中最大值作为id

selectmax_id,
-- collect_list 数据聚合并拼接concat_wsconcat_ws('|', collect_list(name)) as name
from (selectname,grp,max(id) over (partition by grp) max_idfrom (selectid,name,sum(if(name != lag_name, 1, 0)) over (order by cast(id as int)) as grpfrom (selectid,name,lag(name, 1, name) over (order by cast(id as int)) as lag_namefrom tableA) tmp1) tmp2) tmp3
group by max_id, grp;

输出结果为：

通过max_id, grp分组，对name进行 concat_ws('|', collect_list(name)) 聚合拼接，得出最终的结果

3 小结

 断点分组问题的算法总结步骤1：寻找满足条件的断点步骤2：断点处标记值为1，非断点处标记为0步骤3：对断点标记值进行累加 sum(xx)over(order by xx)，构造分组标签步骤4：按照分组标签进行分组求解问题