full outer performance problem - Mailing list pgsql-performance

From Kim Bisgaard
Subject full outer performance problem
Date
Msg-id 42A6BC64.4010001@dmi.dk
Whole thread Raw
Responses Re: full outer performance problem
Re: full outer performance problem
List pgsql-performance
Hi,

I'm having problems with the query optimizer and FULL OUTER JOIN on
PostgreSQL 7.4. I cannot get it to use my indexes with full outer joins.
I might be naive, but I think that it should be possible?

I have two BIG tables (virtually identical) with 3 NOT NULL columns
Station_id, TimeObs, Temp_XXXX, with unique indexes on (Station_id,
TimeObs) and valid ANALYSE (set statistics=100). I want to join the two
tables with a FULL OUTER JOIN.

When I specify the query as:

SELECT station_id, timeobs,temp_grass, temp_dry_at_2m
        FROM temp_dry_at_2m a
        FULL OUTER JOIN temp_grass b
        USING (station_id, timeobs)
        WHERE station_id = 52981
          AND timeobs = '2004-1-1 0:0:0'

I get the correct results

 station_id |       timeobs       | temp_grass | temp_dry_at_2m
------------+---------------------+------------+----------------
      52944 | 2004-01-01 00:10:00 |            |           -1.1
(1 row)

BUT LOUSY performance, and the following EXPLAIN:

                                                                                  QUERY PLAN

------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
 Merge Full Join  (cost=1542369.83..1618958.58 rows=6956994 width=32) (actual time=187176.408..201436.264 rows=1
loops=1)
   Merge Cond: (("outer".station_id = "inner".station_id) AND ("outer".timeobs = "inner".timeobs))
   Filter: ((COALESCE("outer".station_id, "inner".station_id) = 52981) AND (COALESCE("outer".timeobs, "inner".timeobs)
='2004-01-01 00:00:00'::timestamp without time zone)) 
   ->  Sort  (cost=1207913.44..1225305.93 rows=6956994 width=16) (actual time=145748.253..153851.607 rows=6956994
loops=1)
         Sort Key: a.station_id, a.timeobs
         ->  Seq Scan on temp_dry_at_2m a  (cost=0.00..117549.94 rows=6956994 width=16) (actual time=0.049..54226.770
rows=6956994loops=1) 
   ->  Sort  (cost=334456.38..340472.11 rows=2406292 width=16) (actual time=31668.876..34491.123 rows=2406292 loops=1)
         Sort Key: b.station_id, b.timeobs
         ->  Seq Scan on temp_grass b  (cost=0.00..40658.92 rows=2406292 width=16) (actual time=0.052..5484.489
rows=2406292loops=1) 
 Total runtime: 201795.989 ms
(10 rows)

If I change the query (note the "b."s)

explain analyse SELECT b.station_id, b.timeobs,temp_grass, temp_dry_at_2m
        FROM temp_dry_at_2m a
        FULL OUTER JOIN temp_grass b
        USING (station_id, timeobs)
        WHERE b.station_id = 52981
          AND b.timeobs = '2004-1-1 0:0:0'

I seem to destroy the FULL OUTER JOIN and get wrong results (nothing)
If I had happend to use "a.", and not "b.", I would have gotten correct
results (by accident).
The "a." variant gives this EXPLAIN:

                                                                 QUERY PLAN

--------------------------------------------------------------------------------------------------------------------------------------------
 Nested Loop Left Join  (cost=0.00..11.97 rows=1 width=20) (actual time=0.060..0.067 rows=1 loops=1)
   ->  Index Scan using temp_dry_at_2m_idx on temp_dry_at_2m a  (cost=0.00..5.99 rows=1 width=16) (actual
time=0.033..0.036rows=1 loops=1) 
         Index Cond: ((station_id = 52981) AND (timeobs = '2004-01-01 00:00:00'::timestamp without time zone))
   ->  Index Scan using temp_grass_idx on temp_grass b  (cost=0.00..5.96 rows=1 width=16) (actual time=0.018..0.021
rows=1loops=1) 
         Index Cond: (("outer".station_id = b.station_id) AND ("outer".timeobs = b.timeobs))
 Total runtime: 0.140 ms
(6 rows)

Why will PostgreSQL not use the same plan for both these queries - they
are virtually identical??

I have tried to formulate the problem with left joins, but this demands
from me that I know which table has all the values (and thus has to go
first), and in practice no such table excists.

TIA,
Kim Bisgaard.


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