3

下面的查询将结果分组first到 4 个等间距的日期箱中，并汇总the_value每个箱中的平均值。

WITH first as(
SELECT
    extract(EPOCH FROM foo.t_date) as the_date,
    foo_val as the_value
FROM bar
INNER JOIN foo
ON
    foo.user_id = bar.x_id
    and
    foo.user_name = 'xxxx'
)
SELECT bin, round(sum(bin_sum) OVER w /sum(bin_ct) OVER w, 2) AS running_avg
FROM  (
   SELECT width_bucket(first.the_date
                     , x.min_epoch, x.max_epoch, x.bins) AS bin
        , sum(first.the_value) AS bin_sum
        , count(*)   AS bin_ct
   FROM   first
       , (SELECT MIN(first.the_date) AS min_epoch
               , MAX(first.the_date) AS max_epoch
               , 4 AS bins
          FROM  first
         ) x
   GROUP  BY 1
   ) sub
WINDOW w AS (ORDER BY bin)
ORDER  BY 1;

我希望能够只计算the_value每个箱中最低的 20 的平均值。从 Stackoverflow 上的其他帖子中，我看到这是可能的，而且这也许ORDER BY the_value是rank()最好的方法。但我的挣扎是我不确定我当前的查询应该在哪里修改以实现这一点。

任何见解将不胜感激。

Postgres 9.3 版

4

1 回答 1

2

row_number()在每个垃圾箱上使用。
首先计算行号rn，然后WHERE rn < 21在下一步中应用：

WITH first AS (
   SELECT extract(EPOCH FROM foo.t_date) AS the_date
        , foo_val AS the_value
   FROM bar
   JOIN foo ON foo.user_id = bar.x_id
           AND foo.user_name = 'xxxx'
   )
, x AS (
   SELECT MIN(the_date) AS min_epoch
        , MAX(the_date) AS max_epoch
   FROM  first
   )
, y AS (
   SELECT width_bucket(f.the_date, x.min_epoch, x.max_epoch, 4) AS bin, *
   FROM   first f, x
   )
, z AS (
   SELECT row_number() OVER (PARTITION BY bin ORDER BY the_value) AS rn, *
   FROM   y
   )
SELECT bin, round(sum(bin_sum) OVER w / sum(bin_ct) OVER w, 2) AS running_avg
FROM  (
   SELECT bin
        , sum(the_value) AS bin_sum
        , count(*)       AS bin_ct
   FROM   z
   WHERE  rn < 21   -- max 20 lowest values
   GROUP  BY 1
   ) sub
WINDOW w AS (ORDER BY bin)
ORDER  BY 1;

CTEy和z可能被混为一谈。同样first，x也可以混为一谈。
但这样更清楚。

未经测试，因为我们没有测试数据。

于 2014-07-31T15:20:45.633 回答