建议使用蜂巢或猪的最优化方式

Suggest most optimized way using hive or pig

问题陈述 假设有一个日志文本文件。以下是文件中的字段。

日志文件

userID
productID
action

Action 是其中之一 –

Browse, Click, AddToCart, Purchase, LogOut

Select 位执行了 AddToCart 操作但未执行 Purchase 操作的用户。

('1001','101','201','Browse'),
('1002','102','202','Click'),
('1001','101','201','AddToCart'),
('1001','101','201','Purchase'),
('1002','102','202','AddToCart')

任何人都可以建议使用具有优化性能的配置单元或猪来获取此信息

Hive: 使用not in

select * from table 
where action='AddtoCart' and 
      userID not in (select distinct userID from table where action='Purchase')

Pig: 使用 action 过滤 id 并进行左连接并检查 id 是否为 null

A = LOAD '\path\file.txt' USING PigStorage(',') AS (userID:int,b:int,c:int,action:chararray) -- Note I am assuming the first 3 columns are int.You will have to figure out the loading without the quotes.
B = FILTER A BY (action='AddToCart');
C = FILTER A BY (action='Purchase');
D = JOIN B BY userID LEFT OUTER,C BY userID;
E = FILTER D BY C.userID is null;
DUMP E;

这可以使用 sum() 或分析 sum() 来完成,具体取决于单个 table 扫描中的确切要求。如果用户将两种产品添加到购物车,但只购买了一种怎么办?

对于用户+产品:

select userID, productID 
  from
(
select 
       userID,
       productID,
       sum(case when action='AddToCart' then 1 else 0 end) addToCart_cnt,
       sum(case when action='Purchase' then 1 else 0 end)  Purchase_cnt
  from table
  group by userID, productID
)s
where addToCart_cnt>0 and Purchase_cnt=0