SQL計算中間天

Question

假設我有下表，並且我希望提取每個正向和負向運動之間的天數。 這樣，對於每個“ id”，我必須計算每個日期對之間的中間天數以及負數在正數之間的比例，即SQL Teradata。

id   date  money
----------------
1    1-1   10
1    3-1   -5
1    9-1    8
1   10-1   -2
2    3-1   10
2    9-1  -10
2   15-1   20
2   19-1   -15


id  days_in prop
-----------------
1     2     0.5
1     1     0.25
2     6     1
2     4     0.75

Answer 1

您想要這樣的東西：

 select A.id, B.date - A.date as "days_in", (B.money - A.money)  / (b.date - A.date) 
 as "prop"
 from
 (
    select X.id, X.date, min(NextDate.date) as "MinNextDate", X.money
    from [yourTable] X, [yourtable] NextDate
    where
    NextDate.date > X.date
    and NextDate.id = X.id
 ) A,
[YourTable] B
where
   A.id = B.id
   and B.date = A.MinNextDate

我認為，teradata返回的日期差為天數，為整數格式。 如果是dateTime，則可能需要將datetime值的大小寫區分為日期，然后再減去。

Answer 2

如何以略有不同的方式使用自我聯接，但是由於將對每一行進行聯接，因此仍然會產生額外的行。.您可以根據自己的條件進一步限制它

select a.id, (b.date-a.date) as days_in,
abs(b.money)/a.money as prop
from <table> a
inner join <table> b
on a.id=b.id
and a.date<>b.date
where (b.date-a.date)>0 and (abs(b.money)/a.money)>0
and (a.money>0 and b.money<=-1)

Answer 3

要獲取先前的正值，可以使用last_value ：

SELECT id
  ,datecol
  -- ratio between current negative and previous positive money
  ,Abs(Cast(money AS NUMBER)) /
   Last_Value(CASE WHEN money > 0 THEN money end IGNORE NULLS)
   Over (PARTITION BY id
         ORDER BY datecol)
  -- difference between current and previous date
  -- might need a cast to date or interval result if the datecol is a Timestamp
  ,datecol-
   Last_Value(CASE WHEN money > 0 THEN DATE_ end IGNORE NULLS)
   Over (PARTITION BY id
         ORDER BY datecol)
FROM vt AS t
-- return only rows with negative money
QUALIFY money < 0

當然，這假設總是存在具有正負值的交替行。