SQL计算中间天

Question

假设我有下表，并且我希望提取每个正向和负向运动之间的天数。 这样，对于每个“ id”，我必须计算每个日期对之间的中间天数以及负数在正数之间的比例，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

您想要这样的东西：

 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

我认为，teradata返回的日期差为天数，为整数格式。 如果是dateTime，则可能需要将datetime值的大小写区分为日期，然后再减去。

Answer 2

如何以略有不同的方式使用自我联接，但是由于将对每一行进行联接，因此仍然会产生额外的行。.您可以根据自己的条件进一步限制它

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

要获取先前的正值，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

当然，这假设总是存在具有正负值的交替行。