Re: partitioned table and ORDER BY indexed_field DESC LIMIT 1

On 8/24/07, Anton <anton200(at)gmail(dot)com> wrote:
>
> Hi.
>
> I just created partitioned table, n_traf, sliced by month
> (n_traf_y2007m01, n_traf_y2007m02... and so on, see below). They are
> indexed by 'date_time' column.
> Then I populate it (last value have date 2007-08-...) and do VACUUM
> ANALYZE ON n_traf_y2007... all of it.
>
> Now I try to select latest value (ORDER BY date_time LIMIT 1), but
> Postgres produced the ugly plan:
>
> =# explain SELECT * FROM n_traf ORDER BY date_time DESC LIMIT 1;
> QUERY PLAN
>
> ---------------------------------------------------------------------------------------------------------
> Limit (cost=824637.69..824637.69 rows=1 width=32)
> -> Sort (cost=824637.69..838746.44 rows=5643499 width=32)
> Sort Key: public.n_traf.date_time
> -> Result (cost=0.00..100877.99 rows=5643499 width=32)
> -> Append (cost=0.00..100877.99 rows=5643499 width=32)
> -> Seq Scan on n_traf (cost=0.00..22.30
> rows=1230 width=32)
> -> Seq Scan on n_traf_y2007m01 n_traf
> (cost=0.00..22.30 rows=1230 width=32)
> -> Seq Scan on n_traf_y2007m02 n_traf
> (cost=0.00..22.30 rows=1230 width=32)
> -> Seq Scan on n_traf_y2007m03 n_traf
> (cost=0.00..22.30 rows=1230 width=32)
> -> Seq Scan on n_traf_y2007m04 n_traf
> (cost=0.00..1.01 rows=1 width=32)
> -> Seq Scan on n_traf_y2007m05 n_traf
> (cost=0.00..9110.89 rows=509689 width=32)
> -> Seq Scan on n_traf_y2007m06 n_traf
> (cost=0.00..32003.89 rows=1790489 width=32)
> -> Seq Scan on n_traf_y2007m07 n_traf
> (cost=0.00..33881.10 rows=1895510 width=32)
> -> Seq Scan on n_traf_y2007m08 n_traf
> (cost=0.00..25702.70 rows=1437970 width=32)
> -> Seq Scan on n_traf_y2007m09 n_traf
> (cost=0.00..22.30 rows=1230 width=32)
> -> Seq Scan on n_traf_y2007m10 n_traf
> (cost=0.00..22.30 rows=1230 width=32)
> -> Seq Scan on n_traf_y2007m11 n_traf
> (cost=0.00..22.30 rows=1230 width=32)
> -> Seq Scan on n_traf_y2007m12 n_traf
> (cost=0.00..22.30 rows=1230 width=32)
> (18 rows)
>
>
> Why it no uses indexes at all?
> -------------------------------------------

I'm no expert but I'd guess that the the planner doesn't know which
partition holds the latest time so it has to read them all.

Regards

MP