在 impala 中查找连续值

Question

我在下面有一个包含 ID、日期和值的数据集。我想标记连续三天值为0的ID。

id	date	value
1	8/10/2021	1
1	8/11/2021	0
1	8/12/2021	0
1	8/13/2021	0
1	8/14/2021	5
2	8/10/2021	2
2	8/11/2021	3
2	8/12/2021	0
2	8/13/2021	0
2	8/14/2021	6
3	8/10/2021	3
3	8/11/2021	4
3	8/12/2021	0
3	8/13/2021	0
3	8/14/2021	0

输出

id	date	value	Flag
1	8/10/2021	1	Y
1	8/11/2021	0	Y
1	8/12/2021	0	Y
1	8/13/2021	0	Y
1	8/14/2021	5	Y
2	8/10/2021	2	N
2	8/11/2021	3	N
2	8/12/2021	0	N
2	8/13/2021	0	N
2	8/14/2021	6	N
3	8/10/2021	3	Y
3	8/11/2021	4	Y
3	8/12/2021	0	Y
3	8/13/2021	0	Y
3	8/14/2021	0	Y

谢谢。

Answer 1

使用 window count() 函数，您可以计算帧中的 0 [当前行，以下 2 个]（按日期排序） - 为每行计算三个连续行帧：

count(case when value=0 then 1 else null end) over(partition by id order by date_ rows between current row and 2 following ) cnt.

如果 count 恰好等于 3，则意味着找到 3 个连续的 0，case 表达式为 cnt=3 的每一行生成 Y：case when cnt=3 then 'Y' else 'N' end。

要将“Y”标志传播到整个 id 组，请使用 max(...) over (partition by id)

使用您的数据示例进行演示（在 Hive 上测试）：

with mydata as (--Data example, dates converted to sortable format yyyy-MM-dd
select 1 id,'2021-08-10' date_, 1 value union all
select 1,'2021-08-11',0 union all
select 1,'2021-08-12',0 union all
select 1,'2021-08-13',0 union all
select 1,'2021-08-14',5 union all
select 2,'2021-08-10',2 union all
select 2,'2021-08-11',3 union all
select 2,'2021-08-12',0 union all
select 2,'2021-08-13',0 union all
select 2,'2021-08-14',6 union all
select 3,'2021-08-10',3 union all
select 3,'2021-08-11',4 union all
select 3,'2021-08-12',0 union all
select 3,'2021-08-13',0 union all
select 3,'2021-08-14',0
) --End of data example, use your table instead of this CTE

select id, date_, value, 
       max(case when cnt=3 then 'Y' else 'N' end) over (partition by id) flag
from
(
select id, date_, value, 
 count(case when value=0 then 1 else null end) over(partition by id order by date_ rows between current row and 2 following ) cnt
from mydata
)s
  order by id, date_  --remove ordering if not necessary
                      --added it to get result in the same order

结果：

id  date_       value   flag    
1   2021-08-10  1       Y
1   2021-08-11  0       Y
1   2021-08-12  0       Y
1   2021-08-13  0       Y
1   2021-08-14  5       Y
2   2021-08-10  2       N
2   2021-08-11  3       N
2   2021-08-12  0       N
2   2021-08-13  0       N
2   2021-08-14  6       N
3   2021-08-10  3       Y
3   2021-08-11  4       Y
3   2021-08-12  0       Y
3   2021-08-13  0       Y
3   2021-08-14  0       Y

Answer 2

您可以通过比较lag()s 来识别ID。然后将值分布在所有行中。以下获取第三个0上的标志：

select t.*,
       (case when value = 0 and prev_value_date_2 = prev_date_2
             then 'Y' else 'N'
        end) as flag_on_row
from (select t.*,
             lag(date, 2) over (partition by value, id order by date) as prev_value_date_2,
             lag(date, 2) over (partition by id order by date) as prev_date_2
      from t
     ) t;

上述逻辑使用lag()，因此很容易扩展到更长的0s 条纹。 “2”在后面看两行，所以如果滞后值相同，那么一行中有三行具有相同的值。

并传播价值：

select t.*, max(flag_on_row) over (partition by id) as flag
from (select t.*,
             (case when value = 0 and prev_value_date_2 = prev_date_2
                   then 'Y' else 'N'
              end) as flag_on_row
      from (select t.*,
                   lag(date, 2) over (partition by value, id order by date) as prev_value_date_2,
                   lag(date, 2) over (partition by id order by date) as prev_date_2
            from t
           ) t
     ) t;