使用与周围行数据的间隙距离成比例的值来填充数据中的间隙？

Question

不久之后的某个时候，我将不得不准备几天的物品价格清单。粒度是 1 天，在有商品销售的日子里，我将平均价格以获得当天的平均值。会有几天没有销售，我认为可以通过拉动上一次和下一次的销售来使用足够的近似值，并且在它们之间的每一天都有一个从一个到另一个线性过渡的价格。

假设原始数据是：

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

我可以到达这里：

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？只会思考；我打算做类似的事情：

拉取原始数据
加入 numbers/calendar table 以填充稀疏数据
LAST_VALUE()（或第一个？）OVER ROWS UNBOUNDED PRECEDING/FOLLOWING（带有 nulls-last order 子句）以获取第一个非空 preceding_date、following_date、preceding_price 和 following_price 来自原始数据
DATEDIFF 假日期和 preceding_date 得到天数（这实际上是我们跨越差距的距离，gap_progress）和差距距离（following_date - preceding_date)
计算公式(preceding_price + ((next_price - preceding_price)/gap_distance) * gap_progress)

不过，我想知道是否有更简单的方法，因为我有数百万个项目日，但感觉效率不高..

我发现了很多问题的例子，其中最后一行或下一行的数据被逐字涂抹以填补空白，但我不记得看到过这种尝试某种过渡的情况。也许这种技术可以双重应用，通过向前运行的涂抹，复制最近的值，并在它旁边有一个向后运行的涂抹：

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

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

gap_distance = DATEDIFF(日, DateFwd, DateBak)
gap_progress = DATEDIFF(日, 日期, DateFwd)
next_price = 价格B
preceding_price = PriceF

?

这是我知道我可以获得的数据的 DDL（与日历结合的原始数据 table）

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

我会将您的公式 100 + (150 - 100 / 10) * 9) 等放入标量 UDF 中，并在持久计算列中使用它。

Answer 2

您可以使用 OUTER APPLY 获取价格不为空的上一行和下一行：

select
  d.item,
  d.date,
  case when d.price is null then
    prev.price + ( (next.price - prev.price) /
                   datediff(day, prev.date, next.date) *
                   datediff(day, prev.date, d.date)
                 )
  else
    d.price
  end as price
from data d
outer apply
(
    select top(1) *
    from data d2
    where d2.item = d.item and d2.date < d.date and d2.price is not null
    order by d2.date desc
) prev
outer apply
(
    select top(1) *
    from data d2
    where d2.item = d.item and d2.date > d.date and d2.price is not null
    order by d2.date
) next;

Rextester 演示：http://rextester.com/QBL7472

更新：这可能很慢。将 and d.price is null 添加到子查询中的 where 子句可能有助于向 DBMS 显示当价格不为空时它不必实际查找其他记录。只需查看解释计划，看看是否有帮助。

Answer 3

与价格一起生成缺失缺口更容易

所以我从你的原始原始数据开始

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

Answer 4

这适用于 sql-server-2012+ 测试 table:

DECLARE @t table

(Item char(5), Date date, Price decimal(9,1))

INSERT @t 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)

查询

;WITH CTE as
(
  SELECT
    Item, Date, Price,
    lead(price) over(partition by Item order by Date) nextprice,
    lead(Date) over(partition by Item order by Date) nextDate
  FROM @t
), N(N) as
(
  SELECT 1 FROM(VALUES(1),(1),(1),(1),(1),(1),(1),(1),(1),(1))M(N)
), tally(N) as
(
  SELECT ROW_NUMBER()OVER(ORDER BY N.N)FROM N,N a,N b,N c,N d,N e,N f
)
SELECT 
  dateadd(d, coalesce(r, 0), Date) Date,
  Item, 
  CAST(price + coalesce((nextprice-price) * r 
    / datediff(d, date, nextdate), 0) as decimal(10,1)) Price
FROM CTE
OUTER APPLY
(
  SELECT top(coalesce(datediff(d, date, nextdate), 0)) 
    row_number() over (order by (select 1))-1 r
  FROM N
) z
ORDER BY item, date

结果：

Date    Item    Price
2000-01-01  Bread   10.0
2000-01-02  Bread   9.5
2000-01-03  Bread   9.3
2000-01-04  Bread   9.1
2000-01-01  Sugar   100.0
2000-01-02  Sugar   105.0
2000-01-03  Sugar   110.0
2000-01-04  Sugar   115.0
2000-01-05  Sugar   120.0
2000-01-06  Sugar   125.0
2000-01-07  Sugar   130.0
2000-01-08  Sugar   135.0
2000-01-09  Sugar   140.0
2000-01-10  Sugar   145.0
2000-01-11  Sugar   150.0

使用与周围行数据的间隙距离成比例的值来填充数据中的间隙？

Fill in gaps in data, using a value proportional to the gap distance to data from the surrounding rows?

sql

sql-server

sql-server-2017