计算条件适用于两列的连续天数

Question

我有一个类似于下面的表格：

+-------------------------+
¦ ID ¦ Date     ¦ Balance ¦
¦----+----------+---------¦
¦ A  ¦ 20200620 ¦ 150     ¦
¦ A  ¦ 20200621 ¦ -130    ¦
¦ A  ¦ 20200621 ¦ -140    ¦
¦ A  ¦ 20200621 ¦ -200    ¦
¦ A  ¦ 20200622 ¦ 200     ¦
¦ A  ¦ 20200622 ¦ 300     ¦
¦ B  ¦ 20200621 ¦ 350     ¦
¦ B  ¦ 20200621 ¦ 400     ¦
¦ B  ¦ 20200621 ¦ -150    ¦
¦ B  ¦ 20200622 ¦ -200    ¦
¦ B  ¦ 20200622 ¦ -300    ¦
¦ B  ¦ 20200623 ¦ -400    ¦
¦ B  ¦ 20200623 ¦ -500    ¦
+-------------------------+

我需要计算到达 ID 和每个不同日期（包括计算中的日期本身）的“余额 <0”的连续天数。 每个 Id 在给定日期可能有多个余额，无论是正数还是负数。 即使给定日期的一个余额金额为负数，查询也应将这一天考虑在内。 output 结果应该类似于下表：

+--------------------------------------------+
¦ ID ¦ Date     ¦ Number_of_Consecutive_Days ¦
¦----+----------+----------------------------¦
¦ A  ¦ 20200620 ¦ Null                       ¦
¦----+----------+----------------------------¦
¦ A  ¦ 20200621 ¦ 1                          ¦
¦----+----------+----------------------------¦
¦ A  ¦ 20200622 ¦ 1                          ¦
¦----+----------+----------------------------¦
¦ B  ¦ 20200621 ¦ Null                       ¦
¦----+----------+----------------------------¦
¦ B  ¦ 20200622 ¦ 2                          ¦
¦----+----------+----------------------------¦
¦ B  ¦ 20200623 ¦ 3                          ¦
+--------------------------------------------+

你能建议我一种计算方法吗？ 这是高度赞赏。

Answer 1

这是一种与过滤有关的间隙和孤岛问题。 这是一种方法：

select t.*,
       row_number() over (partition by id, dateadd(day, - seqnum, date)
                          order by date
                         ) as Number_of_Consecutive_Days
from (select t.id, date, min(balance) as min_balance,
             row_number() over (partition by id order by date) as seqnum
      from t
      group by t.id, date
      having min(balance) < 0
     ) t;

这仅通过使用负余额的日子来起作用。 然后从日期中减去一个序号。 对于相邻的日子，这是恒定的 - 因此外部row_number()的差异。

编辑：

如果您只想计算到给定日期为止任何负余额的天数，您可以使用：

select t.*,
       sum(case when min_balance < 0 then 1 else 0 end) over (partition by id order by date) as Number_of_Consecutive_Days
from (select t.id, date, min(balance) as min_balance
      from t
      group by t.id, date
     ) t;

Answer 2

请注意，日期之间的任何间隔都将被视为连续天数。

with data as (
    select id, date,
        case when min(balance) >= 0 then 0 else 1 end as tally,
        sum(case when min(balance) >= 0 then 1 else 0 end)
            over (partition by id order by date) as grp
    from t
    group by id, date
)
select id, date,
    sum(tally) over (partition by id, grp, tally order by date) as running_days
from data
order by id, date;

要将缺少的日期视为非连续尝试：

    sum(case when min(balance) >= 0 then 1 else 0 end)
        over (partition by id order by date) +
    datediff(day, min(date) over (partition by id), date) -
    row_number() over (partition by id order by date) + 1 as grp

https://rextester.com/NKBZG48737