2

面对查询设计问题，不确定我解决问题的方法是否过于复杂：

我有一个事实表：

       Column   |            Type             |                       Modifiers                       
------------+-----------------------------+-------------------------------------------------------
 id         | integer                     | not null default nextval('messages_id_seq'::regclass)
 type       | character varying(255)      | 
 ts         | numeric                     | 
 text       | text                        | 
 score      | double precision            | 
 user_id    | integer                     | 
 channel_id | integer                     | 
 time_id    | integer                     | 
 created_at | timestamp without time zone | 
 updated_at | timestamp without time zone |

我目前正在针对它运行一些分析查询，其中之一（例如）是：

  with intervals as (
  select 
    (select '09/27/2014'::date) + (n      || ' minutes')::interval start_time,
    (select '09/27/2014'::date) + ((n+60) || ' minutes')::interval end_time
      from generate_series(0, (24*60*7), 60 * 4) n
  )
  select 
    extract(epoch from i.start_time)::numeric * 1000 as ts, 
    extract(epoch from i.end_time)::numeric * 1000 as end_ts,
    sum(avg(messages.score)) over (order by i.start_time) as score

  from messages
  right join intervals i
    on messages.timestamp >= i.start_time and messages.timestamp < i.end_time

  where messages.timestamp between '09/27/2014' and '10/04/2014'

  group by i.start_time, i.end_time 
  order by i.start_time

正如你们可能知道的那样 - 此查询计算给定时间段分布的消息的“分数”属性的平均值，然后计算跨段的累积值（使用窗口）。

我接下来要做的是找到messages.text最接近每个桶的平均值的前 5 个（例如）。

现在，我唯一的计划是：

1) Join messages with the time-buckets
2) Compute a score - avg(score) over (partition by start_time) as deviation and save it against each record of the joined relation
3) Compute a rank() over (order by deviation) as rank
4) Select where rank between 1 and 5

我之所以把这个命令式地按步骤写下来，是因为我第一次尝试设计一个涉及在窗口函数中使用窗口函数的设计(rank() over (partition by start_time, order by score - avg(score) over (partition by start_time))，我什至不打算尝试看看它是否可行。

我可以就我是否朝着正确的方向寻求一些建议吗？

4

2 回答 2

0

您应该前进的方向（这只是我的建议）：

  1. 获得平均分（所有记录）
  2. 操作MINUS上(row score, avg(score))

-- This will leave you with values also positive and negative

  1. abs()在相同的计算中使用步骤 2 中的每个操作
  2. 适当地使用rank()和订购它们
  3. WHERE rank BETWEEN 1 AND 5
于 2014-10-06T09:48:08.157 回答
0

Whelp - 这是我所拥有的并且似乎可以工作：

现在可以批评的是我的查询的结构、性能优化和冗余！^_^ （减去直接生成时间序列而不是我最终会修复的所有扭曲间隔数学！）

with intervals as (
    select 
        (select '09/29/2014'::date) + (n      || ' minutes')::interval start_time,
        (select '09/29/2014'::date) + ((n+60) || ' minutes')::interval end_time
        from generate_series(0, (24*60*7), 60 * 4) n
), intervaled_messages as (
    select
        extract(epoch from i.start_time)::numeric * 1000 as ts, 
        extract(epoch from i.end_time)::numeric * 1000 as end_ts,
        abs(score - avg(score) over (partition by i.start_time)) as deviation
    from messages
    right join intervals i
        on messages.timestamp >= i.start_time and messages.timestamp < i.end_time
    where messages.timestamp between '09/29/2014' and '10/06/2014'
), ranked_messages as (
    select ts, end_ts, deviation, 
    rank() over (partition by ts order by deviation) as rank,
    row_number() over (partition by ts order by deviation) as row_number
    from intervaled_messages
)
select ts, end_ts, deviation, rank 
from ranked_messages 
where rank between 1 and 5
  and row_number between 1 and 5
order by ts;
于 2014-10-06T23:52:29.887 回答