使用与周围行数据的间隙距离成比例的值填充数据中的空白？

Question

在某个时刻,我将不得不准备一个日期的物品价格清单.粒度是1天,当有一个项目的销售时,我将平均价格以获得当天的平均值.有些日子里没有销售,而且我很适合通过拉动销售的前一次和下次销售来使用足够的近似值,并且它们之间的每一天都有一个价格从一个线性转换到另一个的价格.

想象一下原始数据是:

Item   Date       Price
Bread  2000-01-01 10
Bread  2000-01-02 9.5
Bread  2000-01-04 9.1
Sugar  2000-01-01 100
Sugar  2000-01-11 150

我可以到这里:

Item   Date       Price
Bread  2000-01-01 10
Bread  2000-01-02 9.5
Bread  2000-01-03 NULL
Bread  2000-01-04 9.1
Sugar  2000-01-01 100
Sugar  2000-01-02 NULL
Sugar  2000-01-03 NULL
Sugar  2000-01-04 NULL
Sugar  2000-01-05 NULL
Sugar  2000-01-06 NULL
Sugar  2000-01-07 NULL
Sugar  2000-01-08 NULL
Sugar  2000-01-09 NULL
Sugar  2000-01-10 NULL
Sugar  2000-01-11 150

我想去的地方是:

Item   Date       Price
Bread  2000-01-01 10
Bread  2000-01-02 9.5
Bread  2000-01-03 9.3 --being 9.5 + ((9.1 - 9.5 / 2) * 1)
Bread  2000-01-04 9.1
Sugar  2000-01-01 100
Sugar  2000-01-02 105 --being 100 + (150 - 100 / 10) * 1)
Sugar  2000-01-03 110 --being 100 + (150 - 100 / 10) * 2)
Sugar  2000-01-04 115
Sugar  2000-01-05 120
Sugar  2000-01-06 125
Sugar  2000-01-07 130
Sugar  2000-01-08 135
Sugar  2000-01-09 140
Sugar  2000-01-10 145 --being 100 + (150 - 100 / 10) * 9)
Sugar  2000-01-11 150

到目前为止我尝试了什么？只考虑; 我打算做以下事情:

• 拉原始数据
• 加入数字/日历表以填充稀疏数据
• LAST_VALUE()(或第一个？)超过行无限制前进/后续(使用nulls-last order子句)从原始数据中获取第一个非null preceding_date,following_date,preceding_price和following_price
• DATEDIFF假日期和preceding_date获得了几天(这实际上是我们之间的差距,gap_progress)和间隙距离(following_date - preceding_date)
• 获得公式的下一个价格,先前价格和缺口距离(preceding_price +((next_price - preceding_price)/ gap_distance)*gap_progress)

然而,我想知道是否有更简单的方法,因为我有数百万的项目日,这并不觉得它会那么高效..

我发现很多问题的例子,其中来自上一行或下一行的数据被逐字描绘以填补空白,但我不记得看到这种情况,其中尝试了某种转换.也许这种技术可以通过向前运行的涂片,复制最近的值以及向后运行的涂抹来双重应用:

Item   Date       DateFwd    DateBak     PriceF PriceB
Bread  2000-01-01 2000-01-01 2000-01-01  10     10
Bread  2000-01-02 2000-01-02 2000-01-02  9.5    9.5
Bread  2000-01-03 2000-01-02 2000-01-04  9.5    9.1
Bread  2000-01-04 2000-01-04 2000-01-04  9.1    9.1
Sugar  2000-01-01 2000-01-01 2000-01-01  100    100
Sugar  2000-01-02 2000-01-01 2000-01-11  100    150
Sugar  2000-01-03 2000-01-01 2000-01-11  100    150
Sugar  2000-01-04 2000-01-01 2000-01-11  100    150
Sugar  2000-01-05 2000-01-01 2000-01-11  100    150
Sugar  2000-01-06 2000-01-01 2000-01-11  100    150
Sugar  2000-01-07 2000-01-01 2000-01-11  100    150
Sugar  2000-01-08 2000-01-01 2000-01-11  100    150
Sugar  2000-01-09 2000-01-01 2000-01-11  100    150
Sugar  2000-01-10 2000-01-01 2000-01-11  100    150
Sugar  2000-01-11 2000-01-11 2000-01-11  150    150

这些可能为公式提供必要的数据 (preceding_price + ((next_price - preceding_price)/gap_distance) * gap_progress):

• gap_distance = DATEDIFF(day,DateFwd,DateBak)
• gap_progress = DATEDIFF(day,Date,DateFwd)
• next_price = PriceB
• preceding_price = PriceF

？

这是我知道可以访问的数据的DDL(原始数据与日历表结合)

CREATE TABLE Data
([I] varchar(5), [D] date, [P] DECIMAL(10,5))
;

INSERT Data
([I], [D], [P])
VALUES
('Bread', '2000-01-01', 10),
('Bread', '2000-01-02', 9.5),
('Bread', '2000-01-04', 9.1),
('Sugar', '2000-01-01', 100),
('Sugar', '2000-01-11', 150);

CREATE TABLE Cal([D] DATE);
INSERT Cal VALUES
('2000-01-01'),
('2000-01-02'),
('2000-01-03'),
('2000-01-04'),
('2000-01-05'),
('2000-01-06'),
('2000-01-07'),
('2000-01-08'),
('2000-01-09'),
('2000-01-10'),
('2000-01-11');

SELECT d.i as [item], c.d as [date], d.p as [price] FROM
cal c LEFT JOIN data d ON c.d = d.d

Answer 1

更容易一次性生成那些缺失的缺口和价格

所以我从你的原始数据开始

CREATE TABLE t
    ([I] varchar(5), [D] date, [P] DECIMAL(10,2))
;

INSERT INTO t
    ([I], [D], [P])
VALUES
    ('Bread', '2000-01-01 00:00:00', '10'),
    ('Bread', '2000-01-02 00:00:00', '9.5'),
    ('Bread', '2000-01-04 00:00:00', '9.1'),
    ('Sugar', '2000-01-01 00:00:00', '100'),
    ('Sugar', '2000-01-11 00:00:00', '150');

; with
-- number is a tally table. here i use recursive cte to generate 100 numbers
number as
(
    select  n = 0
    union all
    select  n = n + 1
    from    number
    where   n < 99
),
-- a cte to get the Price of next date and also day diff
cte as
(
    select  *, 
            nextP = lead(P) over(partition by I order by D),
            cnt = datediff(day, D, lead(D) over(partition by I order by D)) - 1
    from    t
) 
select  I, 
        D = dateadd(day, n, D), 
        P = coalesce(c.P + (c.nextP - c.P) / ( cnt + 1) * n, c.P)
from    cte c
        cross join number n
where   n.n <= isnull(c.cnt, 0)

drop table t