From: | Greg Stark <gsstark(at)mit(dot)edu> |
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To: | Markus Schaber <schabios(at)logi-track(dot)com> |
Cc: | PostgreSQL Performance List <pgsql-performance(at)postgresql(dot)org> |
Subject: | Re: Multiple Uniques |
Date: | 2004-09-02 19:33:38 |
Message-ID: | 87isawwjn1.fsf@stark.xeocode.com |
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Lists: | pgsql-performance |
Markus Schaber <schabios(at)logi-track(dot)com> writes:
> logigis=# explain select count(id) from (select ref_in_id as id from streets union select nref_in_id as id from streets) as blubb;
> QUERY PLAN
> ---------------------------------------------------------------------------------------------------------
> Aggregate (cost=16220893.16..16220893.16 rows=1 width=8)
> -> Subquery Scan blubb (cost=15254815.03..16082881.99 rows=55204464 width=8)
> -> Unique (cost=15254815.03..15530837.35 rows=55204464 width=8)
> -> Sort (cost=15254815.03..15392826.19 rows=55204464 width=8)
> Sort Key: id
> -> Append (cost=0.00..6810225.28 rows=55204464 width=8)
> -> Subquery Scan "*SELECT* 1" (cost=0.00..3405112.64 rows=27602232 width=8)
> -> Seq Scan on streets (cost=0.00..3129090.32 rows=27602232 width=8)
> -> Subquery Scan "*SELECT* 2" (cost=0.00..3405112.64 rows=27602232 width=8)
> -> Seq Scan on streets (cost=0.00..3129090.32 rows=27602232 width=8)
You can actually go one step further and do:
select count(distinct id) from (select ... union all select ...) as blubb;
I'm not sure why this is any faster since it still has to do all the same
work, but it's a different code path and it seems to be about 20% faster for
me.
--
greg
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