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在oracle中计算滚动加权平均值

oracle12c oracle sql

tblznbits · 技术社区 · 7 年前

create table tbl_1 as (
    select * from (
        select trunc(sysdate - (rownum - 1)) as call_dt,
               rownum as calls,
               to_char(trunc(sysdate - (rownum - 1)), 'DAY') as dow
        from dual connect by rownum <= 22
    )
    where dow like '%MONDAY%'
    order by call_dt
)
;

 call_dt  | calls | dow
-------------------------
17-SEP-18    22    MONDAY   
24-SEP-18    15    MONDAY   
01-OCT-18    8     MONDAY   
08-OCT-18    1     MONDAY

我还有另一张表,上面有未来的日期,如下所示:

create table tbl_2 as (
    select * from (
        select  trunc(sysdate + (rownum - 1)) as call_dt, 
                0 as calls,
                to_char(trunc(sysdate + (rownum - 1)), 'DAY') as dow
        from dual
        connect by rownum <= 15
    )
    where dow like '%MONDAY%'
)
;

 call_dt  | calls | dow
-------------------------
15-OCT-18     0    MONDAY   
22-OCT-18     0    MONDAY

我试图把我未来的日期,附加到我的历史数据中,然后计算一个滚动加权平均值。我目前正在使用以下查询执行此操作。

select  call_dt,
        case when calls = 0 then (
            (1 * lag1) + (0.8 * lag2) + (0.5 * lag3) + (0.3 * lag4))
             else calls 
             end as calls,
        dow
from (
    select  call_dt, calls, dow,
            lag(calls, 4) OVER (partition by dow order by call_dt) as lag4,
            lag(calls, 3) OVER (partition by dow order by call_dt) as lag3,
            lag(calls, 2) OVER (partition by dow order by call_dt) as lag2,
            lag(calls, 1) OVER (partition by dow order by call_dt) as lag1
    from (
        select * from tbl_1
        union
        select * from tbl_2
    )
    order by dow, call_dt
)
;

 call_dt  | calls | dow
-------------------------
17-SEP-18    22    MONDAY   
24-SEP-18    15    MONDAY   
01-OCT-18    8     MONDAY   
08-OCT-18    1     MONDAY 
15-OCT-18    46    MONDAY   
22-OCT-18    24    MONDAY

lag* 变量等于0,因此该值处于禁用状态。以下是我希望实现的目标:

 call_dt  | calls | dow
-------------------------
17-SEP-18    22    MONDAY   
24-SEP-18    15    MONDAY   
01-OCT-18    8     MONDAY   
08-OCT-18    1     MONDAY 
15-OCT-18    46    MONDAY   
22-OCT-18    70    MONDAY

我看着 this question ,似乎能给我想要的?但使用的窗口函数关键字对我来说是陌生的。我还看了 this tutorial ,但这些滚动平均函数似乎假定了非零项。有可能达到这些结果吗?

1 回复 | 直到 7 年前

Ponder Stibbons 7 年前

使用递归查询,它取最后一个 calls 作为 lag1 移动所有其他 lag 回顾过去:

with 
  s as (
    select  rn, call_dt, calls, 
            lag(calls, 4) OVER (partition by dow order by call_dt) as lag4,
            lag(calls, 3) OVER (partition by dow order by call_dt) as lag3,
            lag(calls, 2) OVER (partition by dow order by call_dt) as lag2,
            lag(calls, 1) OVER (partition by dow order by call_dt) as lag1
    from (
        select 0 rn, tbl_1.* from tbl_1 union all
        select row_number() over (order by call_dt), tbl_2.* from tbl_2)),
  c(rn, call_dt, calls, lag1, lag2, lag3, lag4) as (
    select rn, call_dt, (1 * lag1) + (0.8 * lag2) + (0.5 * lag3) + (0.3 * lag4), 
           lag1, lag2, lag3, lag4 
      from s where rn = 1
    union all
    select s.rn, s.call_dt, (1 * c.calls) + (0.8 * c.lag1) + (0.5 * c.lag2) + (0.3 * c.lag3), 
           c.calls, c.lag1, c.lag2, c.lag3
      from s join c on c.rn+1 = s.rn)
select * from c

s -基本上是您的查询,我添加了行编号。 c CTE在哪里 rn = 1 是我们的锚,第一步。然后我们逐行添加下一步, 向右

结果:

    RN CALL_DT          CALLS       LAG1       LAG2       LAG3       LAG4
------ ----------- ---------- ---------- ---------- ---------- ----------
     1 2018-10-15        21,5          1          8         15         22
     2 2018-10-22        30,8       21,5          1          8         15