基于一周的范围求和窗口中的列值(黑斑羚)

Question

给定如下表：

client_id   date            connections
---------------------------------------
121438297   2018-01-03      0
121438297   2018-01-08      1
121438297   2018-01-10      3
121438297   2018-01-12      1
121438297   2018-01-19      7
363863811   2018-01-18      0
363863811   2018-01-30      5
363863811   2018-02-01      4
363863811   2018-02-10      0

我正在寻找一种有效的方法来求和当前行(当前行包含在求和中(之后6天内发生的连接数，按client_id进行分区，这将导致：

client_id   date            connections     connections_within_6_days
---------------------------------------------------------------------
121438297   2018-01-03      0               1        
121438297   2018-01-08      1               5     
121438297   2018-01-10      3               4     
121438297   2018-01-12      1               1                       
121438297   2018-01-19      7               7
363863811   2018-01-18      0               0
363863811   2018-01-30      5               9
363863811   2018-02-01      4               4
363863811   2018-02-10      0               0

问题：

1. 我不想添加所有丢失的日期，然后执行滑动窗口计数以下7行，因为我的表已经非常大了。

2. 我使用的是Impala，不支持range between interval '7' days following and current row。

---

编辑：我正在寻找一个通用答案，考虑到我需要将窗口大小更改为更大的数字(例如30天以上(

Answer 1

这回答了问题的原始版本。

Impala并不完全支持range between。不幸的是，这并没有留下太多选择。一种是使用带有大量显式逻辑的lag()：

select t.*,
( (case when lag(date, 6) over (partition by client_id order by date) = date - interval 6 day
then lag(connections, 6) over (partition by client_id order by date)
else 0
end) +
(case when lag(date, 5) over (partition by client_id order by date) = date - interval 6 day
then lag(connections, 5) over (partition by client_id order by date)
else 0
end) +
(case when lag(date, 4) over (partition by client_id order by date) = date - interval 6 day
then lag(connections, 4) over (partition by client_id order by date)
else 0
end) +
(case when lag(date, 3) over (partition by client_id order by date) = date - interval 6 day
then lag(connections, 3) over (partition by client_id order by date)
else 0
end) +
(case when lag(date, 2) over (partition by client_id order by date) = date - interval 6 day
then lag(connections, 2) over (partition by client_id order by date)
else 0
end) +
(case when lag(date, 1) over (partition by client_id order by date) = date - interval 6 day
then lag(connections, 1) over (partition by client_id order by date)
else 0
end) +
connections
) as connections_within_6_days         
from t;

不幸的是，这并不能很好地概括。如果你想多呆几天，你可能想问另一个问题。