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Re: Speeding up aggregates

From: Tom Lane <tgl(at)sss(dot)pgh(dot)pa(dot)us>
To: Joe Conway <mail(at)joeconway(dot)com>
Cc: Josh Berkus <josh(at)agliodbs(dot)com>, pgsql-performance(at)postgresql(dot)org
Subject: Re: Speeding up aggregates
Date: 2002-12-09 22:06:50
Message-ID: 20088.1039471610@sss.pgh.pa.us (view raw or flat)
Thread:
Lists: pgsql-performance
Joe Conway <mail(at)joeconway(dot)com> writes:
>> How many rows out if you drop the HAVING clause?

> parts=# set sort_mem to 8000;
> SET
> parts=# explain analyze select i.part_id, sum(w.qty_oh) as total_oh from inv 
> i, iwhs w where i.part_id = w.part_id group by i.part_id;
>                                                           QUERY PLAN
> ----------------------------------------------------------------------------------------------------------------------------
>   HashAggregate  (cost=5617.22..5706.04 rows=35528 width=36) (actual 
> time=1525.93..1627.41 rows=34575 loops=1)
>     ->  Hash Join  (cost=1319.10..5254.45 rows=72553 width=36) (actual 
> time=156.86..1248.73 rows=72548 loops=1)


>> The planner's choice of which to use is dependent on its estimate of the
>> required hashtable size, which is proportional to its guess about how
>> many distinct groups there will be.  The above output doesn't tell us
>> that however, only how many groups passed the HAVING clause.  I'm
>> curious about the quality of this estimate, since the code to try to
>> generate not-completely-bogus group count estimates is all new ...

> If I'm reading it correctly, it looks like the estimate in this case is pretty 
> good.

Better than I had any right to expect ;-).  Thanks.

			regards, tom lane

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