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。以下是//
的示例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！
答案 0 :(得分：1)
如果我理解正确，您需要连续一天连续一行，而不仅仅是发生更改的日期。因此，例如ItemID 10007在2011-07-22,2011-07-23,2011-07-24 ......到2011-07-27之间的QTY = 2，当它变为1时。
在这种情况下，请尝试此操作，假设您在R中有一个名为Collection的数据框，并且日期采用字符格式：
@OneToMany(fetch = FetchType.EAGER, cascade = CascadeType.ALL, mappedBy="conversation")
private Set<Question> questions = new HashSet<Question>();
如果您将mydata添加到最后，您应该看到：
library(tidyverse)
mydata %>%
mutate(ADate = as.Date(ADate)) %>%
complete(ADate = seq.Date(min(ADate), max(ADate), by = "day")) %>%
fill(ItemID, QTY)
答案 1 :(得分：0)
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。我假设你有一个@Mapper
public interface MyMapper {
@Mapping(source = "contacts", target = "contactsId")
OrganizationDTO toDto(Organization organization);
default Long fromContact(Contact contact) {
return contact == null ? null : contact.getId();
}
}
表：
numbers答案 2 :(得分：0)
请使用CTE http://rextester.com/live/CDFT20818找到此演示 并且还有许多2012+功能替代。
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;
答案 3 :(得分：0)
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$ADate),]
答案 4 :(得分：0)
您可以使用LEAD（需要sql server 2012+）和CTE生成此日期序列
WITH Orders AS
(
SELECT ItemID,QTY,
LEAD (cADate, 1, NULL) OVER (ORDER BY ItemID,cADate ) AS NExtVal,
cADate AS Number
FROM Table1
-- WHERE QTY > 0 -- remove this if you want
UNION ALL
SELECT ItemID,QTY,NExtVal, DATEADD(DAY, 1, Number)
FROM Orders
WHERE DATEADD(DAY, 1, Number) < NExtVal
)
SELECT ItemID,QTY, Number AS DateSequence FROM Orders
ORDER BY ItemID,Number
以下是示例代码的Link。
答案 5 :(得分：0)
使用R库的tidyverse解决方案，使用@ Brian的数据：
date_seq <- seq(from=min(d$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