hadoop与之前的udf比较

时间:2014-09-13 13:19:27

标签: hive

我的输入文件是

2014-08-23       30000
2014-09-24       20000
2014-10-23       50000
2014-11-24       7000

我想要这样的输出

2014-08-23      30000
2014-09-24     -10000
2014-10-25      30000
2014-11-24      -47000

我想在没有udf的情况下实现这一目标 我试过这段代码

SELECT C.ID ,C.DATE,C.VALUE AS CURRENT_DATE_VALUE,COALESCE(CAST(O.VALUE AS INT),0) AS PREV_DATE_VALUE,(C.VALUE-COALESCE(CAST(O.VALUE as INT),0)) AS DIFF_VALUE 
FROM ITEM O 
LEFT OUTER JOIN 
( SELECT T.ID ,C.DATE,C.VALUE,MAX(UNIX_TIMESTAMP(T.DATE,'dd-MM-yyyy')) AS PREV_DATE 
  FROM ITEM C 
  LEFT OUTER JOIN ITEM T ON(C.ID = T.ID) WHERE   
  UNIX_TIMESTAMP (C.DATE,'dd-MM-yyyy') > UNIX_TIMESTAMP(T.DATE,'dd-MM-yyyy') GROUP BY
  T.ID ,C.DATE,C.VALUE) C 
ON (O.ID = C.ID AND UNIX_TIMESTAMP (O.DATE,'dd-MM-yyyy') = C.PREV_DATE)

1 个答案:

答案 0 :(得分:0)

如果您可以访问HIVE 0.13(具有windowing个功能),则可以使用最少量的代码执行此操作

<强>查询:

select date
    ,(value - prev) as diff
from (
    select date
        ,value
        ,LAG(value, 1, 0) OVER (ORDER BY date) as prev
    from some_database.some_table
    ) x

<强>输出:

2014-08-23    30000
2014-09-24    -10000
2014-10-23    30000
2014-11-24    -43000
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