sql – 为什么以下连接会显着增加查询时间？

我在这里有一个星型模式,我正在查询事实表,并希望加入一个非常小的维度表.我无法解释以下内容：

EXPLAIN ANALYZE SELECT 
  COUNT(impression_id),imp.os_id 
  FROM bi.impressions imp 
  GROUP BY imp.os_id;

                                                                  QUERY PLAN
    --------------------------------------------------------------------------------------------------------------------------------------
     HashAggregate  (cost=868719.08..868719.24 rows=16 width=10) (actual time=12559.462..12559.466 rows=26 loops=1)
       ->  Seq Scan on impressions imp  (cost=0.00..690306.72 rows=35682472 width=10) (actual time=0.009..3030.093 rows=35682474 loops=1)
     Total runtime: 12559.523 ms
    (3 rows)

这需要大约12600毫秒,但当然没有连接数据,所以我无法将imp.os_id“解析”为有意义的东西,所以我添加了一个连接：

EXPLAIN ANALYZE SELECT 
  COUNT(impression_id),imp.os_id,os.os_desc 
  FROM  bi.impressions imp,bi.os_desc os 
  WHERE imp.os_id=os.os_id 
  GROUP BY imp.os_id,os.os_desc;
                                                                     QUERY PLAN
    --------------------------------------------------------------------------------------------------------------------------------------------
     HashAggregate  (cost=1448560.83..1448564.99 rows=416 width=22) (actual time=25565.124..25565.127 rows=26 loops=1)
       ->  Hash Join  (cost=1.58..1180942.29 rows=35682472 width=22) (actual time=0.046..15157.684 rows=35682474 loops=1)
             Hash Cond: (imp.os_id = os.os_id)
             ->  Seq Scan on impressions imp  (cost=0.00..690306.72 rows=35682472 width=10) (actual time=0.007..3705.647 rows=35682474 loops=1)
             ->  Hash  (cost=1.26..1.26 rows=26 width=14) (actual time=0.028..0.028 rows=26 loops=1)
                   Buckets: 1024  Batches: 1  Memory Usage: 2kB
                   ->  Seq Scan on os_desc os  (cost=0.00..1.26 rows=26 width=14) (actual time=0.003..0.010 rows=26 loops=1)
     Total runtime: 25565.199 ms
    (8 rows)

这有效地使我的查询的执行时间加倍.我的问题是,我从画面中遗漏了什么？我认为这么小的查找不会导致查询执行时间的巨大差异.

解决方法

SELECT COUNT(impression_id),os.os_desc 
FROM   bi.impressions imp
JOIN   bi.os_desc os ON os.os_id = imp.os_id
GROUP  BY imp.os_id,os.os_desc;

首先,如果在os_desc中为展示中的每一行找到多于或少于一个匹配项,则第二个查询可能是错误的.
如果你在os_id上有一个外键约束,这可以排除这一点,保证参照完整性,加上bi.impressions.os_id上的NOT NULL约束.
如果是这样,在第一步中,简化为：

SELECT COUNT(*) AS ct,os.os_desc 
FROM   bi.impressions imp
JOIN   bi.os_desc     os USING (os_id)
GROUP  BY imp.os_id,os.os_desc;

count(*)略快于count(列).并为计数添加列别名.
更快,但：

SELECT os_id,os.os_desc,sub.ct
FROM  (
   SELECT os_id,COUNT(*) AS ct
   FROM   bi.impressions
   GROUP  BY 1
   ) sub
JOIN   bi.os_desc os USING (os_id)

先分组,稍后加入.更多细节在这里：
Aggregate a single column in query with many columns
PostgreSQL – order by an array

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