Question

我有一个Hive表，其中的数据如下所示-

每个客户都有对应的帐户，目标是建立内部客户对。配对基于帐户的出生年份相同或名称的前三个字符相同。例如-山姆和塞缪尔。

输出看起来像这样-

理想情况下，不应创建相同的帐户对，例如AA，XX等。一对AC和CA也都相同，因此仅需要一对这样的对。可以在“姓名”和“出生年份”密钥上形成一对，但是这里也只需要一个条目（可以是任何人）。

我应该如何解决这个问题。测试数据进行检查-

create table customer_account(
customer INT NOT NULL,
accounts VARCHAR(100) NOT NULL,
name VARCHAR(40) NOT NULL,
yob DATE,
);

INSERT INTO 
customer_account(customer,accounts,name,yob)
VALUES
(1,"A","John",2001),
(1,"X","Tom",1996),
(1,"C","Harry",2001),
(2,"D","Sam",1994),
(2,"F","Samuel",1995),
(3,"Z","Jake",)1994,
(3,"G","Drake",1998),
(3,"H","Arnold",1993),
(3,"K","Yang",1990)
;

Answer 1

您应该能够将子字符串用于HIVE语言。逻辑应该是合理的，尽管您可能需要根据需要对其进行调整。

您要尝试的是一元（或自我）联接。下面是可以传递的查询类型的示例。本质上，您要加入一个OR条件，并用一个case语句测试该条件以获取“ Pair_Key”。我使用了内部联接，假设您只希望发生匹配的实例。

SELECT 
     t1.customer as Customer1,
     t2.customer as Customer2,
     t1.Accounts as Accounts1,
     t2.Accounts as Accounts2,
     CONCAT(t1.Accounts, t2.Accounts) as Pair_No,
     t1.Name as Name1,
     t2.Name as Name2,
     t1.YOB as YOB1,
     t2.YOB as YOB2,
     CASE
     WHEN t1.YOB = t2.YOB THEN 'YOB'
     WHEN SUBSTR(t1.Name, 3) = SUBSTR(t2.Name, 3) THEN 'Name'
     else 'Issue'
     END as Pair_Key
FROM (SELECT * FROM Table1) as t1
inner join (SELECT * FROM Table1) as t2 --instance 2 of the same table
on (SUBSTR(t1.Name, 3) = SUBSTR(t2.Name, 3) OR t1.YOB = t2.YOB)

没有测试数据，也没有更多关于您走过的距离的详细信息，这是一个开始。

如果客户编号需要相同，则只需调整为：

on (t1.Customer = t2.Customer) and (SUBSTR(t1.Name, 3) = SUBSTR(t2.Name, 3) OR t1.YOB = t2.YOB)

Answer 2

这就是您所描述的：

select t1.*, t2.name, t2.yob
from t t1 join
     t t2
     on t2.customer = t1.customer and
        (t2.yob = t1.yob or
         substr(t2.name, 1, 3) = substr(t1.name, 1, 3)
        ) and
        t2.account > t1.account;

无需两次提取customer。如果需要“相同”对，则将最后一个条件更改为>=。

如何从表中创建对

2 个答案: