Re: Optimisation deficiency: currval('seq')-->seq scan, constant-->index scan

From: Jules Bean <jules(at)jellybean(dot)co(dot)uk>
To: Tom Lane <tgl(at)sss(dot)pgh(dot)pa(dot)us>
Cc: Tiago Ant?o <tra(at)fct(dot)unl(dot)pt>, pgsql-hackers(at)postgresql(dot)org
Subject: Re: Optimisation deficiency: currval('seq')-->seq scan, constant-->index scan
Date: 2000-08-24 09:11:14
Message-ID: 20000824101113.N17510@grommit.office.vi.net
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On Wed, Aug 23, 2000 at 10:30:30AM -0400, Tom Lane wrote:
> Jules Bean <jules(at)jellybean(dot)co(dot)uk> writes:
> > I have in a table a 'category' column which takes a small number of
> > (basically fixed) values. Here by 'small', I mean ~1000, while the
> > table itself has ~10 000 000 rows. Some categories have many, many
> > more rows than others. In particular, there's one category which hits
> > over half the rows. Because of this (AIUI) postgresql assumes
> > that the query
> > select ... from thistable where category='something'
> > is best served by a seqscan, even though there is an index on
> > category.
>
> Yes, we know about that one. We have stats about the most common value
> in a column, but no information about how the less-common values are
> distributed. We definitely need stats about several top values not just
> one, because this phenomenon of a badly skewed distribution is pretty
> common.

ISTM that that might be enough, in fact.

If you have stats telling you that the most popular value is 'xyz',
and that it constitutes 50% of the rows (i.e. 5 000 000) then you can
conclude that, on average, other entries constitute a mere 5 000
000/999 ~~ 5000 entries, and it would be definitely be enough.
(That's assuming you store the number of distinct values somewhere).

> BTW, if your highly-popular value is actually a dummy value ('UNKNOWN'
> or something like that), a fairly effective workaround is to replace the
> dummy entries with NULL. The system does account for NULLs separately
> from real values, so you'd then get stats based on the most common
> non-dummy value.

I can't really do that. Even if I could, the distribution is very
skewed -- so the next most common makes up a very high proportion of
what's left. I forget the figures exactly.

Jules

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