sql - Postgres : Get value of a column corresponding to max of other column in a group

Question

我正在尝试编写一个 postgres 查询，它返回组中的最大值、最小值、中值、第一个和最后一个值以及每个聚合值的时间戳列
table

Id Timestamp_utc                  Value
1  2020-11-05 15:36:15.768388     10
1  2020-11-05 15:40:15.768388     20
1  2020-11-05 15:44:15.768388     30
1  2020-11-05 15:45:15.768388.    5
1  2020-11-05 15:59:15.768388     25
1  2020-11-05 16:59:15.768388     25

预期结果

Id Median Median_Timestamp Min Min_Timestamp               Max Max_TimeStamp
1  17.5.  15:44:15.768388  5   2020-11-05 15:45:15.768388  30   2020-11-05 15:44:15.768388

我有这个查询分组数据不包括时间戳

SELECT Id, time_bucket('60', timestamp_utc) AS bucket,
percentile_cont(0.5) within group (order by value) median_value,
min(value) min_value, 
max(value) max_value 
FROM rs.MyTable 
WHERE id IN ( 1111,123)
AND timestamp_utc Between '2020-11-05 10:00:15.748643' and '2020-11-05 16:35:48.750313'
GROUP BY id, bucket 
ORDER BY id, bucket

当值最大时，有没有一种方法可以获取时间戳列以及聚合值(如 timestamp_utc col 数据)？

Answer 1

一种选择是在子查询中使用窗口函数通过递增和递减 value 对时间戳进行排名。 ，然后在外层查询中进行条件聚合，带来相关值

select id, bucket,
    percentile_cont(0.5) within group (order by value) median_value,
    min(value) min_value, 
    max(timestamp_utc) filter(where rn_asc = 1) min_timestamp,
    max(value) max_value,
    max(timestamp_utc) filter(where rn_desc = 1) max_timestamp
from (
    select t.*, 
        row_number() over(partition by id, bucket order by value) rn_asc,
        row_number() over(partition by id, bucket order by value desc) rn_desc
    from (
        select t.*, time_bucket('60', timestamp_utc) as bucket 
        from rs.mytable t
        where 
            id in (1111,123)
            and timestamp_utc between '2020-11-05 10:00:15.748643'::timestamp 
                                  and '2020-11-05 16:35:48.750313'::timestamp
    ) t
) t
group by id, bucket 
order by id, bucket

注意，我们需要先计算bucket，并将其放入窗函数的分区中。