忽略postgresql rank()窗口函数中的空值

时间:2013-08-12 18:26:44

标签: sql postgresql

我正在使用PostgreSQL编写SQL查询,需要对在某个位置“到达”的人进行排名。然而,不是每个人都到了。我正在使用rank()窗口函数来生成到达等级,但是在到达时间为空的地方,而不是返回空等级,rank()聚合函数只是将它们视为它们之后到达其他所有人。我想要发生的是,这些没有出现的排名为NULL,而不是这个估算的排名。

这是一个例子。假设我有一个表dinner_show_up,如下所示:

 | Person | arrival_time | Restaurant |
 +--------+--------------+------------+
 | Dave   |     7        | in_and_out |
 | Mike   |     2        | in_and_out | 
 | Bob    |   NULL       | in_and_out | 
鲍勃从不露面。我正在写的查询是:

select Person, 
       rank() over (partition by Restaurant order by arrival_time asc) 
               as arrival_rank
from dinner_show_up; 

结果将是

 | Person | arrival_rank | 
 +--------+--------------+
 | Dave   |     2        | 
 | Mike   |     1        | 
 | Bob    |     3        |  

我想要发生的是:

 | Person | arrival_rank | 
 +--------+--------------+
 | Dave   |     2        | 
 | Mike   |     1        | 
 | Bob    |     NULL     |  

3 个答案:

答案 0 :(得分:14)

只需在case附近使用rank()语句:

select Person, 
       (case when arrival_time is not null
             then rank() over (partition by Restaurant order by arrival_time asc) 
        end) as arrival_rank
from dinner_show_up; 

答案 1 :(得分:7)

对于所有聚合函数而言,更通用的解决方案,不仅是rank(),还要按“到达时间”进行分区,而不是空的'在over()子句中。这将导致所有null arrival_time行被放入同一个组并给定相同的排名,使非空行仅相对于彼此排名。

为了一个有意义的例子,我模拟了一个CTE,其行数比初始问题集多。请原谅宽行,但我认为他们更好地对比不同的技术。

with dinner_show_up("person", "arrival_time", "restaurant") as (values
   ('Dave' ,    7, 'in_and_out')
  ,('Mike' ,    2, 'in_and_out')
  ,('Bob'  , null, 'in_and_out')
  ,('Peter',    3, 'in_and_out')
  ,('Jane' , null, 'in_and_out')
  ,('Merry',    5, 'in_and_out')
  ,('Sam'  ,    5, 'in_and_out')
  ,('Pip'  ,    9, 'in_and_out')
)

select 
   person
  ,case when arrival_time is not null then         rank() over (                                      order by arrival_time) end as arrival_rank_without_partition
  ,case when arrival_time is not null then         rank() over (partition by arrival_time is not null order by arrival_time) end as arrival_rank_with_partition
  ,case when arrival_time is not null then percent_rank() over (                                      order by arrival_time) end as arrival_pctrank_without_partition
  ,case when arrival_time is not null then percent_rank() over (partition by arrival_time is not null order by arrival_time) end as arrival_pctrank_with_partition
from dinner_show_up

此查询为arrival_rank_with / without_partition提供相同的结果。但是,percent_rank()的结果确实不同:without_partition是错误的,范围从0%到71.4%,而with_partition正确地给出pctrank()范围从0%到100%。

同样的模式也适用于ntile()聚合函数。

它的工作原理是将所有空值与非空值分开,以便进行排名。这确保Jane和Bob被排除在0%到100%的百分位排名之外。

 |person|arrival_rank_without_partition|arrival_rank_with_partition|arrival_pctrank_without_partition|arrival_pctrank_with_partition|
 +------+------------------------------+---------------------------+---------------------------------+------------------------------+
 |Jane  |null                          |null                       |null                             |null                          |
 |Bob   |null                          |null                       |null                             |null                          |
 |Mike  |1                             |1                          |0                                |0                             |
 |Peter |2                             |2                          |0.14                             |0.2                           |
 |Sam   |3                             |3                          |0.28                             |0.4                           |
 |Merry |4                             |4                          |0.28                             |0.4                           |
 |Dave  |5                             |5                          |0.57                             |0.8                           |
 |Pip   |6                             |6                          |0.71                             |1.0                           |

答案 2 :(得分:0)

select Person, 
   rank() over (partition by Restaurant order by arrival_time asc) 
           as arrival_rank
from dinner_show_up
where arrival_time is not null
union 
select Person,NULL as arrival_rank
from dinner_show_up
where arrival_time is null;