如何根据另一列计算后面n行的平均值 - SQL (Oracle)
How to calculate the average value of following n rows based on another column - SQL (Oracle)
我正在尝试按月计算每个 POLICY_ID 的平均每月保费价值，如下所示。当客户将 his/her 年付款频率更新为不同于 12 的值时，我需要手动计算 PREMIUM 的平均每月值。我怎样才能达到 MONTHLY _PREMIUM_DESIRED 中显示的值？
提前致谢。
注：Oracle 版本 12c
我尝试过的：
SELECT
T.*,
SUM(PREMIUM) OVER(PARTITION BY T.POLICY_ID ORDER BY T.POLICY_ID, T.PAYMENT_DATE ROWS BETWEEN CURRENT ROW AND 12/T.YEARLY_PAYMENT_FREQ-1 FOLLOWING ) / (12/T.YEARLY_PAYMENT_FREQ) MONTLY_PREMIUM_CALCULATED
FROM MYTABLE T
;
数据代码：
DROP TABLE MYTABLE;
CREATE TABLE MYTABLE (POLICY_ID NUMBER(11), PAYMENT_DATE DATE, PREMIUM NUMBER(5), YEARLY_PAYMENT_FREQ NUMBER(2),MONTHLY_PREMIUM_DESIRED NUMBER(5));
INSERT INTO MYTABLE VALUES (1, DATE '2014-10-01',120,12,120);
INSERT INTO MYTABLE VALUES (1, DATE '2014-11-01',360,4,120);
INSERT INTO MYTABLE VALUES (1, DATE '2014-12-01',0,4,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-01-01',0,4,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-02-01',360,4,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-03-01',0,4,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-04-01',0,4,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-05-01',720,2,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-06-01',0,2,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-07-01',0,2,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-08-01',0,2,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-09-01',0,2,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-10-01',0,2,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-11-01',120,12,120);
INSERT INTO MYTABLE VALUES (2, DATE '2015-01-01',60,3,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-02-01',0,3,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-03-01',0,3,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-04-01',0,3,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-05-01',180,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-06-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-07-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-08-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-09-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-10-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-11-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-12-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2016-01-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2016-02-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2016-03-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2016-04-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2016-05-01',15,12,15);
INSERT INTO MYTABLE VALUES (2, DATE '2016-06-01',15,12,15);
SELECT * FROM MYTABLE;
编辑：
无论付款频率如何，PREMIUM 金额也可以由客户更改。下面，对于 POLICY_ID = 1，我添加了从“2015/11/01”开始的新记录来演示这种情况。在这种情况下，平均每月保费从 120 增加到 240。
还删除了屏幕截图以使问题更具可读性。
DROP TABLE MYTABLE2;
CREATE TABLE MYTABLE2 (POLICY_ID NUMBER(11), PAYMENT_DATE DATE, PREMIUM NUMBER(5), YEARLY_PAYMENT_FREQ NUMBER(2),MONTHLY_PREMIUM_DESIRED NUMBER(5));
INSERT INTO MYTABLE2 VALUES (1, DATE '2014-10-01',120,12,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2014-11-01',360,4,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2014-12-01',0,4,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-01-01',0,4,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-02-01',360,4,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-03-01',0,4,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-04-01',0,4,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-05-01',720,2,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-06-01',0,2,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-07-01',0,2,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-08-01',0,2,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-09-01',0,2,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-10-01',0,2,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-11-01',240,12,240);
INSERT INTO MYTABLE2 VALUES (1, DATE '2016-12-01',240,12,240); --newly added records
INSERT INTO MYTABLE2 VALUES (1, DATE '2016-01-01',960,4,240); --newly added records
INSERT INTO MYTABLE2 VALUES (1, DATE '2016-02-01',0,4,240); --newly added records
INSERT INTO MYTABLE2 VALUES (1, DATE '2016-03-01',0,4,240); --newly added records
INSERT INTO MYTABLE2 VALUES (1, DATE '2016-04-01',0,4,240); --newly added records
INSERT INTO MYTABLE2 VALUES (1, DATE '2016-05-01',960,4,240); --newly added records
INSERT INTO MYTABLE2 VALUES (1, DATE '2016-06-01',0,4,240); --newly added records
INSERT INTO MYTABLE2 VALUES (1, DATE '2016-07-01',0,4,240); --newly added records
INSERT INTO MYTABLE2 VALUES (1, DATE '2016-08-01',0,4,240); --newly added records
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-01-01',60,3,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-02-01',0,3,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-03-01',0,3,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-04-01',0,3,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-05-01',180,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-06-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-07-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-08-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-09-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-10-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-11-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-12-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2016-01-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2016-02-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2016-03-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2016-04-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2016-05-01',15,12,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2016-06-01',15,12,15);
SELECT * FROM MYTABLE2;
我认为计算是：
select t.*,
premium / (12 / yearly_payment_freq)) as monthly_premium_calculated
from mytable t;
编辑：
我明白了，您还需要中间月份的价差。因此，您可以通过计算非零付款的数量来分配组。那么：
select t.*,
( max(premium) over (partition by policy_id, grp) /
(12 / yearly_payment_freq)
) as monthly_premium_calculated
from (select t.*,
sum(case when premium > 0 then 1 else 0 end) over (partition by policy_id order by payment_date) as grp
from mytable t
) t;
Here 是一个 db<>fiddle（它使用 Postgres，因为它比 Oracle 更容易设置）。
我正在尝试按月计算每个 POLICY_ID 的平均每月保费价值，如下所示。当客户将 his/her 年付款频率更新为不同于 12 的值时，我需要手动计算 PREMIUM 的平均每月值。我怎样才能达到 MONTHLY _PREMIUM_DESIRED 中显示的值？ 提前致谢。
注：Oracle 版本 12c
我尝试过的：
SELECT
T.*,
SUM(PREMIUM) OVER(PARTITION BY T.POLICY_ID ORDER BY T.POLICY_ID, T.PAYMENT_DATE ROWS BETWEEN CURRENT ROW AND 12/T.YEARLY_PAYMENT_FREQ-1 FOLLOWING ) / (12/T.YEARLY_PAYMENT_FREQ) MONTLY_PREMIUM_CALCULATED
FROM MYTABLE T
;
数据代码：
DROP TABLE MYTABLE;
CREATE TABLE MYTABLE (POLICY_ID NUMBER(11), PAYMENT_DATE DATE, PREMIUM NUMBER(5), YEARLY_PAYMENT_FREQ NUMBER(2),MONTHLY_PREMIUM_DESIRED NUMBER(5));
INSERT INTO MYTABLE VALUES (1, DATE '2014-10-01',120,12,120);
INSERT INTO MYTABLE VALUES (1, DATE '2014-11-01',360,4,120);
INSERT INTO MYTABLE VALUES (1, DATE '2014-12-01',0,4,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-01-01',0,4,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-02-01',360,4,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-03-01',0,4,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-04-01',0,4,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-05-01',720,2,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-06-01',0,2,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-07-01',0,2,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-08-01',0,2,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-09-01',0,2,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-10-01',0,2,120);
INSERT INTO MYTABLE VALUES (1, DATE '2015-11-01',120,12,120);
INSERT INTO MYTABLE VALUES (2, DATE '2015-01-01',60,3,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-02-01',0,3,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-03-01',0,3,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-04-01',0,3,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-05-01',180,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-06-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-07-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-08-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-09-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-10-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-11-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2015-12-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2016-01-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2016-02-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2016-03-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2016-04-01',0,1,15);
INSERT INTO MYTABLE VALUES (2, DATE '2016-05-01',15,12,15);
INSERT INTO MYTABLE VALUES (2, DATE '2016-06-01',15,12,15);
SELECT * FROM MYTABLE;
编辑：
无论付款频率如何，PREMIUM 金额也可以由客户更改。下面，对于 POLICY_ID = 1，我添加了从“2015/11/01”开始的新记录来演示这种情况。在这种情况下，平均每月保费从 120 增加到 240。 还删除了屏幕截图以使问题更具可读性。
DROP TABLE MYTABLE2;
CREATE TABLE MYTABLE2 (POLICY_ID NUMBER(11), PAYMENT_DATE DATE, PREMIUM NUMBER(5), YEARLY_PAYMENT_FREQ NUMBER(2),MONTHLY_PREMIUM_DESIRED NUMBER(5));
INSERT INTO MYTABLE2 VALUES (1, DATE '2014-10-01',120,12,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2014-11-01',360,4,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2014-12-01',0,4,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-01-01',0,4,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-02-01',360,4,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-03-01',0,4,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-04-01',0,4,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-05-01',720,2,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-06-01',0,2,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-07-01',0,2,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-08-01',0,2,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-09-01',0,2,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-10-01',0,2,120);
INSERT INTO MYTABLE2 VALUES (1, DATE '2015-11-01',240,12,240);
INSERT INTO MYTABLE2 VALUES (1, DATE '2016-12-01',240,12,240); --newly added records
INSERT INTO MYTABLE2 VALUES (1, DATE '2016-01-01',960,4,240); --newly added records
INSERT INTO MYTABLE2 VALUES (1, DATE '2016-02-01',0,4,240); --newly added records
INSERT INTO MYTABLE2 VALUES (1, DATE '2016-03-01',0,4,240); --newly added records
INSERT INTO MYTABLE2 VALUES (1, DATE '2016-04-01',0,4,240); --newly added records
INSERT INTO MYTABLE2 VALUES (1, DATE '2016-05-01',960,4,240); --newly added records
INSERT INTO MYTABLE2 VALUES (1, DATE '2016-06-01',0,4,240); --newly added records
INSERT INTO MYTABLE2 VALUES (1, DATE '2016-07-01',0,4,240); --newly added records
INSERT INTO MYTABLE2 VALUES (1, DATE '2016-08-01',0,4,240); --newly added records
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-01-01',60,3,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-02-01',0,3,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-03-01',0,3,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-04-01',0,3,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-05-01',180,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-06-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-07-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-08-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-09-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-10-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-11-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2015-12-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2016-01-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2016-02-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2016-03-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2016-04-01',0,1,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2016-05-01',15,12,15);
INSERT INTO MYTABLE2 VALUES (2, DATE '2016-06-01',15,12,15);
SELECT * FROM MYTABLE2;
我认为计算是：
select t.*,
premium / (12 / yearly_payment_freq)) as monthly_premium_calculated
from mytable t;
编辑：
我明白了，您还需要中间月份的价差。因此，您可以通过计算非零付款的数量来分配组。那么：
select t.*,
( max(premium) over (partition by policy_id, grp) /
(12 / yearly_payment_freq)
) as monthly_premium_calculated
from (select t.*,
sum(case when premium > 0 then 1 else 0 end) over (partition by policy_id order by payment_date) as grp
from mytable t
) t;
Here 是一个 db<>fiddle（它使用 Postgres，因为它比 Oracle 更容易设置）。