我有以下数据，其中包含一组ID的每月目标。目标是2020年每个月的每个id。名为targets的表。 month列表示一年中的月份。
+-------+-------+----+--------+
| month | name | id | target |
+-------+-------+----+--------+
| 1 | Comp1 | 1 | 6000 |
+-------+-------+----+--------+
| 2 | Comp1 | 1 | 6000 |
+-------+-------+----+--------+
| 3 | Comp1 | 1 | 6000 |
+-------+-------+----+--------+
| 1 | Comp2 | 2 | 6000 |
+-------+-------+----+--------+
| 2 | Comp2 | 2 | 6000 |
+-------+-------+----+--------+
| 3 | Comp2 | 2 | 6000 |
+-------+-------+----+--------+
| 1 | Comp3 | 3 | 6000 |
+-------+-------+----+--------+
| 2 | Comp3 | 3 | 6000 |
+-------+-------+----+--------+
| 3 | Comp3 | 3 | 6000 |
+-------+-------+----+--------+
| 1 | Comp4 | 4 | 6000 |
+-------+-------+----+--------+
| 2 | Comp4 | 4 | 6000 |
+-------+-------+----+--------+
| 3 | Comp4 | 4 | 6000 |
+-------+-------+----+--------+
然后我有第二个表，其中包含一组ID的每日数据，并每天进行更新。在我的实际数据集中，我有从2019年1月1日到今天的数据。
+------------+-------+----+--------+--------+
| yyyy_mm_dd | name | id | actual | region |
+------------+-------+----+--------+--------+
| 2019-01-01 | Comp1 | 1 | 1000 | LATAM |
+------------+-------+----+--------+--------+
| 2019-01-01 | Comp1 | 1 | 0 | EU |
+-------------------------------------------+
| 2019-01-02 | Comp1 | 1 | 2000 | EU |
+------------+-------+----+--------+--------+
| 2019-01-03 | Comp1 | 1 | 4000 | EU |
+------------+-------+----+--------+--------+
| 2019-01-01 | Comp2 | 2 | 1000 | EU |
+------------+-------+----+--------+--------+
| 2019-01-02 | Comp2 | 2 | 2000 | EU |
+------------+-------+----+--------+--------+
| 2019-01-03 | Comp2 | 2 | 3000 | EU |
+------------+-------+----+--------+--------+
| 2019-01-01 | Comp3 | 3 | 1000 | EU |
+------------+-------+----+--------+--------+
| 2019-01-02 | Comp3 | 3 | 2000 | EU |
+------------+-------+----+--------+--------+
| 2019-01-03 | Comp3 | 3 | 8000 | EU |
+------------+-------+----+--------+--------+
| 2019-01-01 | Comp4 | 4 | 1000 | EU |
+------------+-------+----+--------+--------+
| 2019-01-02 | Comp4 | 4 | 2000 | EU |
+------------+-------+----+--------+--------+
| 2019-02-03 | Comp4 | 4 | 3000 | EU |
+------------+-------+----+--------+--------+
基于以上两个表，我想创建第三个表并添加一些其他逻辑。最终，我希望引入一个名为payment的新列。除非公司通过了每月目标，否则此列应始终为0。如果达到/通过了每月目标，则付款应为sum actual for that month - monthly target for that month * 1%。
以下是输出数据的外观：
+------------+-------+----+--------+--------+
| yyyy_mm_dd | name | id | actual | payout |
+------------+-------+----+--------+--------+
| 2020-01-01 | Comp1 | 1 | 1000 | 0 |
+------------+-------+----+--------+--------+
| 2020-01-02 | Comp1 | 1 | 2000 | 0 |
+------------+-------+----+--------+--------+
| 2020-01-03 | Comp1 | 1 | 4000 | 10 |
+------------+-------+----+--------+--------+
| 2020-01-01 | Comp2 | 2 | 1000 | 0 |
+------------+-------+----+--------+--------+
| 2020-01-02 | Comp2 | 2 | 2000 | 0 |
+------------+-------+----+--------+--------+
| 2020-01-03 | Comp2 | 2 | 3000 | 0 |
+------------+-------+----+--------+--------+
| 2020-01-01 | Comp3 | 3 | 1000 | 0 |
+------------+-------+----+--------+--------+
| 2020-01-02 | Comp3 | 3 | 2000 | 0 |
+------------+-------+----+--------+--------+
| 2020-01-03 | Comp3 | 3 | 8000 | 50 |
+------------+-------+----+--------+--------+
| 2020-01-01 | Comp4 | 4 | 1000 | 0 |
+------------+-------+----+--------+--------+
| 2020-01-02 | Comp4 | 4 | 2000 | 0 |
+------------+-------+----+--------+--------+
| 2020-02-03 | Comp4 | 4 | 3000 | 0 |
+------------+-------+----+--------+--------+
以上数据集中的所有名称/ ID的每月target为6000。因此，当名称/ ID在该月中超过目标时，应仅使用payout。 Comp1和Comp3在1月的第三天都超过了每月目标，因此他们从那一天开始直到月底都将获得付款。然后在2月份重置，因为它是一个新的月，有了新的目标，并且随着该月的进行，我们将获得新的每日数据。
我尝试过的事情：
SELECT
agg.yyyy_mm_dd,
agg.name,
agg.id,
CASE WHEN agg.actual >= targets.target THEN ((agg.actual-targets.target)/100) * 1 ELSE 0 END AS payout
FROM(
SELECT
sum(x.actual) AS actual,
x.yyyy_mm_dd,
x.name,
x.id
FROM(
SELECT
yyyy_mm_dd,
name,
id,
cast(actual as int) as actual
FROM
schema.daily_data
WHERE
yyyy_mm_dd >= '2020-01-01' AND (name = 'Comp1' OR name = 'Comp2')
) x
GROUP BY
2,3,4
) agg
INNER JOIN(
SELECT
id,
month,
target
FROM
schema.targets
) targets ON targets.id = agg.id
GROUP BY
1,2,3,4
但是，以上每个name输出多行。这是因为每日表每天有多次相同公司（预期）。我以为我的小组可以解决这个问题。另外，我认为这并不是整体上最简单的解决方案，我可能对此过于想像/可以提高效率。
答案 0 :(得分：1)
您似乎想将每个公司和月份的actua的累积总和与target进行比较。您可以通过联接和窗口函数来做到这一点：
select
d.yyyy_mm_dd,
case when sum(d.actual) over(partition by d.name, t.month order by d.yyyy_mm_dd) > t.target
then (sum(d.actual) over(partition by d.name, t.month order by d.yyyy_mm_dd) - t.target) / 100.0
else 0
end payout
from schema.targets t
inner join schema.daily_data d
on month(d.yyyy_mm_dd) = t.month
and d.name = t.name
where
d.yyyy_mm_dd >= '2020-01-01'
and d.name in ('Comp1', 'Comp2')
答案 1 :(得分：0)
通过窗口函数可以轻松解决您要求的（部分）总和的要求。不幸的是，我没有使用Hive，所以这是我的Postgres工作解决方案
with t (month, name, id, target) as (values
(1 , 'Comp1', 1 , 6000 ),
(2 , 'Comp1', 1 , 6000 ),
(3 , 'Comp1', 1 , 6000 ),
(1 , 'Comp2', 2 , 6000 ),
(2 , 'Comp2', 2 , 6000 ),
(3 , 'Comp2', 2 , 6000 ),
(1 , 'Comp3', 3 , 6000 ),
(2 , 'Comp3', 3 , 6000 ),
(3 , 'Comp3', 3 , 6000 ),
(1 , 'Comp4', 4 , 6000 ),
(2 , 'Comp4', 4 , 6000 ),
(3 , 'Comp4', 4 , 6000 )
), d (yyyy_mm_dd, name, id, actual, region) as (values
( date '2019-01-01' , 'Comp1' , 1 , 1000 , 'LATAM' ),
( date '2019-01-01' , 'Comp1' , 1 , 0 , 'EU' ),
( date '2019-01-02' , 'Comp1' , 1 , 2000 , 'EU' ),
( date '2019-01-03' , 'Comp1' , 1 , 4000 , 'EU' ),
( date '2019-01-01' , 'Comp2' , 2 , 1000 , 'EU' ),
( date '2019-01-02' , 'Comp2' , 2 , 2000 , 'EU' ),
( date '2019-01-03' , 'Comp2' , 2 , 3000 , 'EU' ),
( date '2019-01-01' , 'Comp3' , 3 , 1000 , 'EU' ),
( date '2019-01-02' , 'Comp3' , 3 , 2000 , 'EU' ),
( date '2019-01-03' , 'Comp3' , 3 , 8000 , 'EU' ),
( date '2019-01-01' , 'Comp4' , 4 , 1000 , 'EU' ),
( date '2019-01-02' , 'Comp4' , 4 , 2000 , 'EU' ),
( date '2019-02-03' , 'Comp4' , 4 , 3000 , 'EU' )
)
select dr.yyyy_mm_dd, dr.name, dr.id, dr.actual,
case when dr.running_sum < t.target then 0 else (dr.running_sum - t.target) / 100 end as payment
from t
join (
select dg.*, sum(actual) over (partition by name order by yyyy_mm_dd) as running_sum
from (
select yyyy_mm_dd, name, id, sum(actual) as actual
from d
group by yyyy_mm_dd, name, id
) dg
) dr on dr.name = t.name
and month(dr.yyyy_mm_dd) = t.month -- edited to hive equivalent of postgres' extract(month from dr.yyyy_mm_dd) = t.month
从日期中提取月份可能会以其他方式进行，但希望您能理解。
答案 2 :(得分：0)
另一种选择是使用开窗SUM函数创建运行总计，然后在CASE语句中使用该总计来获取列值。
SELECT d.yyyy_mm_dd
,d.name
,d.id
,d.actual
,CASE
WHEN
SUM(d.actual)
OVER (PARTITION BY d.id ORDER BY d.yyyy_mm_dd ROWS UNBOUNDED PRECEDING) <= t.target
THEN 0
ELSE
(
SUM(d.actual)
OVER (PARTITION BY d.id ORDER BY d.yyyy_mm_dd ROWS UNBOUNDED PRECEDING) - t.target
) * 0.01
END AS payout
FROM dailies AS d
JOIN targets AS t
ON d.month = MONTH(d.yyyy_mm_dd)
AND d.id = d.id;
我不确定Hive语法是否100％正确，但这非常接近。具体来说，ROWS UNBOUNDED PRECEDING可能不足。您可能需要在其中添加FOLLOWING子句才能正确计算总计。
答案 3 :(得分：0)
我想我现在有一个可行的解决方案。下面给出了预期的输出。因为它不是最快的，所以可能可以对其进行一些优化。
SELECT
x.yyyy_mm_dd,
x.id,
x.name,
x.actual,
x.target,
x.actual_to_date,
CASE WHEN x.actual_to_date > x.target THEN ((x.actual_to_date - x.target) /100) * 1 ELSE 0 END AS payout
FROM(
SELECT
daily.yyyy_mm_dd,
daily.id,
daily.name,
daily.actual,
t.target,
SUM(daily.actual) OVER (PARTITION BY MONTH(daily.yyyy_mm_dd), daily.id ORDER BY daily.yyyy_mm_dd RANGE UNBOUNDED PRECEDING) AS actual_to_date
FROM(
SELECT
yyyy_mm_dd,
id,
name,
sum(cast(actual as int)) as actual
FROM
daily_data_table
WHERE
yyyy_mm_dd >= '2020-01-01'
GROUP BY
1,2,3
) daily
INNER JOIN
monthly_target_table t
ON t.id = daily.id AND t.month = month(daily.yyyy_mm_dd)
WHERE
daily.name = 'Comp1'
) x