Question

我目前有一张包含小时和日期的表格。我正在尝试查找“cnt”列不等于0的每个日期的连续小时数

对于这些数据：

 Date       Hour V1 V2 Cnt Cons
 ------------------------------
 2015-12-15  1   a  b   1   3
 2015-12-15  2   a  b   2   3
 2015-12-15  2   b  b   2   1
 2015-12-15  3   a  b   3   3
 2015-12-15  3   b  b   2   1
 2015-12-15  4   a  b   0   0
 2015-12-15  5   a  b   1   2
 2015-12-15  6   a  b   1   2

根据上表，日期2015-12-15有2个实例，其中连续小时cnt不等于0（1-3,5-6）

只有V1和V2不同时才会发生重复小时

我想达到这个输出：

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Answer 1

既然你已经编辑了你的问题，它就表明你不想只在一天内查看和计算，而是在一天之内+ v1 + v2。

如果我一天只选择一个v1 / v2对，除了Cnt = 0，那么我会得到相应的时间进行调查，例如： 1,2,5,6,8,9。因此可能存在间隙（实际间隙或Cnt = 0间隙），但在这样的组内没有重复。如果我从值中减去此结果中的位置，我会连续几小时获得组密钥：

hour   pos  hour-pos=groupkey
1      1    1-1=0
2      2    2-2=0
5      3    5-3=2
6      4    6-4=2
8      5    8-5=3
9      6    9-6=3

这使用我得到：

select 
  date, hour, v1, v2, 
  count(*) over (partition by date, v1, v2, groupkey) as cons
from
(
  select 
    mytable.*, 
    hour - row_number() over (partition by date, v1, v2 order by hour) as groupkey
  from mytable
  where cnt <> 0
) grouped
union all
select date, hour, v1, v2, 0 as cons
from mytable
where cnt = 0
order by date, hour, v1, v2;

根据小时查找连续行并按日期分组

1 个答案: