Question

我必须对具有400+百万行的表进行一些数据分析。我将其用于一个小样本，但我确定它将在生产中耗尽内存。

表结构如下（对于数百万个序列号）：

+------------+---------------+------------+----------+
|    date    | serial_number |  status_1  | status_2 |
+------------+---------------+------------+----------+
| 10/1/2018  |           123 | warehouse  | v        |
| 10/10/2018 |           123 | warehouse  | w        |
| 10/20/2018 |           123 | warehouse  | x        |
| 11/2/2018  |           123 | in transit | y        |
+------------+---------------+------------+----------+

我需要获取日期，其中前一个日期为status_1 =“运输中”当前，而status_2 =“ x”。看起来应该像这样：

+-----------+---------------+------------+----------+------------+
|  date_1   | serial_number |  status_1  | status_2 |   date_2   |
+-----------+---------------+------------+----------+------------+
| 11/2/2018 |           123 | in transit | x        | 10/20/2018 |
+-----------+---------------+------------+----------+------------+

我使用两个等级函数得到了它，但这可能会在一个大桌子上使它窒息。

with transit as (
select 
*
from (
    select *,
    rank() over(partition by serial_number order by date desc) rnk
    from sample_t 
    order by serial_number, date asc
    ) 
where rnk=1 and status_1 = 'in transit'
),
x_type as (
select 
*
from (
    select *,
    rank() over(partition by serial_number order by date desc) rnk
    from sample_t 
    order by serial_number, date asc
    ) 
where rnk>1 and status_2 = 'x'
)
select tr.date date_1,
tr.serial_number,
tr.status_1,
x.status_2,
x.date date_2
from transit tr left join x_type x on tr.serial_number = x.serial_number

我看不到如何使用一个等级函数执行此操作。有没有更好，更有效的方法？

Answer 1

您可以使用lag来完成此操作。

select *
from (select t.*
      ,lag(status_2) over(partition by serial_no order by date) as prev_status_2
      ,lag(date) over(partition by serial_no order by date) as prev_date
      from tbl t  
     ) t 
where status_1 = 'in_transit' and prev_status_2 = 'x'

将窗口函数应用于大数据集（如何优化？）

1 个答案: