填充缺少的日期值

问题描述 投票：1回答：6

所以我有一个表格，按项目ID显示库存值。我需要找到每个项目缺货的天数和天数。我的表只在每次库存值发生变化时添加一个新行，所以我只有任何给定项目缺货的日子，但没有任何随附的日子它仍然缺货。这是一个它看起来像什么的例子//

ItemID  QTY  ADate 
10007   2   2011-07-22
10007   1   2011-07-27
10007   2   2011-09-01
10007   1   2011-09-23 
10007   2   2011-09-29 
10007   0   2011-10-29
10007   2   2011-11-03
10007   1   2011-11-21
23028   2   2011-07-23 
23028   0   2011-08-03 
23028   2   2011-08-10

我想拥有的是：

ItemID  QTY  ADate
10007   2   2011-07-22 
10007   2   2011-07-23
10007   2   2011-07-24
10007   2   2011-07-25
10007   2   2011-07-26
10007   1   2011-07-27
10007   1   2011-07-28
10007   1   2011-07-29
10007   1   2011-07-30

等...`

然后我可以通过商品ID抓住我缺货的日子。我没有任何代码可以显示，因为坦率地说我甚至不知道从哪里开始。知道怎么解决这个问题吗？非常感谢！

sql sql-server r
6个回答
1
投票

如果我理解正确，您需要连续一天连续一行，而不仅仅是发生变化的日子。因此，例如ItemID 10007在2011-07-22,2011-07-23,2011-07-24 ......到2011-07-27，当它变为1时，QTY = 2。

在这种情况下，试试这个，假设你在R中有一个名为mydata的数据框，并且日期是字符格式：

library(tidyverse)
mydata %>% 
  mutate(ADate = as.Date(ADate)) %>% 
  complete(ADate = seq.Date(min(ADate), max(ADate), by = "day")) %>%
  fill(ItemID, QTY)

如果你添加%>% filter(QTY == 0)到最后，你应该看到：

ADate      ItemID   QTY
   <date>      <int> <int>
 1 2011-08-03  23028     0
 2 2011-08-04  23028     0
 3 2011-08-05  23028     0
 4 2011-08-06  23028     0
 5 2011-08-07  23028     0
 6 2011-08-08  23028     0
 7 2011-08-09  23028     0
 8 2011-10-29  10007     0
 9 2011-10-30  10007     0
10 2011-10-31  10007     0
11 2011-11-01  10007     0
12 2011-11-02  10007     0
0
投票

您似乎希望填写具有最新价值的日子。一种方法是递归CTE。但这实际上是一种痛苦。我假设你有一张numbers表：

select id.itemid, dateadd(day, n.n, t.min_adate) as dte, t.qty
from (select t.itemid, min(adate) as min_adate
      from t
      group by itemid
     ) id join
     numbers n
     on dateadd(day, n.n, t.min_adate) < getdate() outer apply
     (select top 1 t.*
      from t
      where t.itemid = id.itemid and t.date <= dateadd(day, n.n, t.min_adate)
      order by t.date desc
     ) t;
0
投票

请使用CTE http://rextester.com/live/CDFT20818和许多2012+功能替代，找到这个演示。

declare @mytable table (itemid int,qty int,adate date)

insert into @mytable 
values
( 10007,   2,'   2011-07-22'),
( 10007 ,  1,'   2011-07-27'),
( 10008 ,  1,'   2011-07-23'),
( 10008 ,  0,'   2011-07-16'),
( 10008 ,  4,'   2011-07-10')

declare @som date = '2011-07-01';
declare @eom date = eomonth(@som);



;with myCalendar as 
(
   Select  @som startdate
         union all
    Select dateadd(day,1,startdate) startdate
        from myCalendar
        where startdate < @eom
)
,myDistinctList as (
select t.itemid, c.startdate
  from myCalendar c
  cross join 
  (select distinct itemid from @mytable) t
), myFinal as (
select 
l.itemid,
iif(t.qty is null,(select top 1 qty from @mytable x where  x.itemid = l.itemid and x.adate <= l.startdate order by x.adate desc),t.qty) [qty],
l.startdate [adate]
from myDistinctList l
left outer join @mytable t
    on t.itemid = l.itemid
    and t.adate = l.startdate
) 
select * from myFinal 
where qty is not null
order by itemid,adate;
0
投票

这是另一种方式：

tmp <- split(d, d$ItemID)
f <- function(x) {
  b <- seq(min(x$ADate), max(x$ADate), "day")
  a <- do.call(c, mapply(rep, x$QTY[-nrow(x)], diff(x$ADate)))
  a <- c(a, x$QTY[nrow(x)])
  data.frame(x$ItemID[1], a, b)
}
out <- do.call(rbind, lapply(tmp, f))
rownames(out) <- NULL
colnames(out) <- c("ItemID", "QTY", "ADate")

数据

d <- "ItemID  QTY  ADate 
10007   2   2011-07-22
10007   1   2011-07-27
10007   2   2011-09-01
10007   1   2011-09-23 
10007   2   2011-09-29 
10007   0   2011-10-29
10007   2   2011-11-03
10007   1   2011-11-21
23028   2   2011-07-23 
23028   0   2011-08-03 
23028   2   2011-08-10"
d <- read.table(text = d,header = T)
d$ADate <- as.POSIXct(d$ADate,"UTC")
d <- d[order(d$ItemID,d$ADate),]
0
投票

您可以使用LEAD（需要sql server 2012+）和CTE生成此日期序列，如下所示

WITH Orders AS 
( 
  SELECT ItemID,QTY,
      LEAD (cADate, 1, NULL) OVER (ORDER BY ItemID,cADate )  AS NExtVal,
      cADate AS  Number  
  FROM Table1
  -- WHERE QTY > 0 -- remove this  if you want 
  UNION ALL 
  SELECT ItemID,QTY,NExtVal,  DATEADD(DAY, 1, Number)
  FROM Orders  
  WHERE  DATEADD(DAY, 1, Number)  < NExtVal
)     
SELECT ItemID,QTY, Number AS DateSequence FROM Orders
ORDER BY ItemID,Number

这是样本代码的Link。

0
投票

使用R库的tidyverse解决方案，使用@Brian的数据：

date_seq <- seq(from=min(d$ADate),
            to=max(d$ADate),
            by="day")

library(tidyverse)
d %>%
  right_join(expand.grid(ItemID = unique(.$ItemID),ADate=date_seq)) %>%
  arrange(ItemID) %>%
  group_by(ItemID) %>%
  fill(ItemID,QTY) 

# # A tibble: 40 x 3
# # Groups:   ItemID [2]
#    ItemID   QTY      ADate
#     <int> <int>     <dttm>
#  1  10007     2 2011-07-22
#  2  10007     2 2011-07-23
#  3  10007     2 2011-07-24
#  4  10007     2 2011-07-25
#  5  10007     2 2011-07-26
#  6  10007     1 2011-07-27
#  7  10007     1 2011-07-28
#  8  10007     1 2011-07-29
#  9  10007     1 2011-07-30
# 10  10007     1 2011-07-31
# # ... with 30 more rows
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