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Re: [HACKERS] How to change order sort of table in HashJoin

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Thanks for response, sir.

On 11/20/2016 1:18 AM, Tom Lane wrote:
Man Trieu <man.trieu@xxxxxxxxx> writes:
As in the example below, i think the plan which hash table is created on
testtbl2 (the fewer tuples) should be choosen.
The planner usually prefers to hash on the table that has a flatter
MCV histogram, since a hash table with many key collisions will be
inefficient.  You might find it illuminating to read the comments around
estimate_hash_bucketsize().

Thanks, I will read it.

Additional information.
In 9.6 the second table (lesser tuple) was choosen (the same testdata).
There are something (cost estimation?) different  in previous versions.

--- In 9.6.1 ---
postgres=# explain analyze select * from testtbl1 inner join testtbl2 using(c1,c2,c3);
                                                         QUERY PLAN

-----------------------------------------------------------------------------------------------------------------------------
Hash Join (cost=6935.57..60389.58 rows=1 width=60) (actual time=80.214..1165.762 rows=142857 loops=1) Hash Cond: ((testtbl1.c1 = testtbl2.c1) AND (testtbl1.c2 = testtbl2.c2) AND (testtbl1.c3 = testtbl2.c3)) -> Seq Scan on testtbl1 (cost=0.00..21276.00 rows=1000000 width=56) (actual time=0.038..226.324 rows=1000000 loops=1) -> Hash (cost=3039.57..3039.57 rows=142857 width=56) (actual time=79.632..79.632 rows=142857 loops=1)
         Buckets: 65536  Batches: 4  Memory Usage: 3658kB
-> Seq Scan on testtbl2 (cost=0.00..3039.57 rows=142857 width=56) (actual time=0.028..20.646 rows=142857 loops=1)
 Planning time: 0.252 ms
 Execution time: 1174.588 ms
(8 rows)
------

--- In 9.4.10 ---
postgres=# explain analyze select * from testtbl1 inner join testtbl2 using(c1,c2,c3);
                                                           QUERY PLAN

---------------------------------------------------------------------------------------------------------------------------------
Hash Join (cost=48542.00..67353.86 rows=1 width=60) (actual time=880.580..1277.611 rows=142857 loops=1) Hash Cond: ((testtbl2.c1 = testtbl1.c1) AND (testtbl2.c2 = testtbl1.c2) AND (testtbl2.c3 = testtbl1.c3)) -> Seq Scan on testtbl2 (cost=0.00..3039.57 rows=142857 width=56) (actual time=0.016..24.421 rows=142857 loops=1) -> Hash (cost=21276.00..21276.00 rows=1000000 width=56) (actual time=878.296..878.296 rows=1000000 loops=1)
         Buckets: 8192  Batches: 32  Memory Usage: 2839kB
-> Seq Scan on testtbl1 (cost=0.00..21276.00 rows=1000000 width=56) (actual time=0.025..258.193 rows=1000000 loops=1)
 Planning time: 2.683 ms
 Execution time: 1285.868 ms
(8 rows)
------

In general, given a hashtable that fits in memory and light bucket
loading, a hash join is more or less O(M) + O(N); it doesn't matter
so much whether the larger table is on the inside.  It does matter if
the table gets big enough to force batching of the join, but that's
not happening in your example (at least not the first one; it's unclear
to me why it did happen in the second one).  The key thing that will
drive the choice, then, is avoiding a skewed bucket distribution that
causes lots of comparisons for common values.

			regards, tom lane

Thanks and best regards,


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