[HACKERS] [POC] hash partitioning

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От Yugo Nagata
Тема [HACKERS] [POC] hash partitioning
Дата
Msg-id 20170228233313.fc14d8b6.nagata@sraoss.co.jp
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Ответы Re: [HACKERS] [POC] hash partitioning  (Aleksander Alekseev <a.alekseev@postgrespro.ru>)
Re: [HACKERS] [POC] hash partitioning  (Amit Langote <Langote_Amit_f8@lab.ntt.co.jp>)
Re: [HACKERS] [POC] hash partitioning  (Rushabh Lathia <rushabh.lathia@gmail.com>)
Re: [HACKERS] [POC] hash partitioning  (amul sul <sulamul@gmail.com>)
Re: [HACKERS] [POC] hash partitioning  (Jeff Davis <pgsql@j-davis.com>)
Список pgsql-hackers
Hi all,

Now we have a declarative partitioning, but hash partitioning is not
implemented yet. Attached is a POC patch to add the hash partitioning
feature. I know we will need more discussions about the syntax and other
specifications before going ahead the project, but I think this runnable
code might help to discuss what and how we implement this.

* Description

In this patch, the hash partitioning implementation is basically based
on the list partitioning mechanism. However, partition bounds cannot be
specified explicitly, but this is used internally as hash partition
index, which is calculated when a partition is created or attached.

The tentative syntax to create a partitioned table is as bellow;

 CREATE TABLE h (i int) PARTITION BY HASH(i) PARTITIONS 3 USING hashint4;

The number of partitions is specified by PARTITIONS, which is currently
constant and cannot be changed, but I think this is needed to be changed in
some manner. A hash function is specified by USING. Maybe, specifying hash
function may be ommitted, and in this case, a default hash function
corresponding to key type will be used.

A partition table can be create as bellow;

 CREATE TABLE h1 PARTITION OF h;
 CREATE TABLE h2 PARTITION OF h;
 CREATE TABLE h3 PARTITION OF h;

FOR VALUES clause cannot be used, and the partition bound is
calclulated automatically as partition index of single integer value.

When trying create partitions more than the number specified
by PARTITIONS, it gets an error.

postgres=# create table h4 partition of h;
ERROR:  cannot create hash partition more than 3 for h

An inserted record is stored in a partition whose index equals
abs(hashfunc(key)) % <number_of_partitions>. In the above
example, this is abs(hashint4(i))%3.

postgres=# insert into h (select generate_series(0,20));
INSERT 0 21

postgres=# select *,tableoid::regclass from h;
 i  | tableoid 
----+----------
  0 | h1
  1 | h1
  2 | h1
  4 | h1
  8 | h1
 10 | h1
 11 | h1
 14 | h1
 15 | h1
 17 | h1
 20 | h1
  5 | h2
 12 | h2
 13 | h2
 16 | h2
 19 | h2
  3 | h3
  6 | h3
  7 | h3
  9 | h3
 18 | h3
(21 rows)

* Todo / discussions

In this patch, we cannot change the number of partitions specified
by PARTITIONS. I we can change this, the partitioning rule
(<partition index> = abs(hashfunc(key)) % <number_of_partitions>)
is also changed and then we need reallocatiing records between
partitions.

In this patch, user can specify a hash function USING. However,
we migth need default hash functions which are useful and
proper for hash partitioning. 

Currently, even when we issue SELECT query with a condition,
postgres looks into all partitions regardless of each partition's
constraint, because this is complicated such like "abs(hashint4(i))%3 = 0".

postgres=# explain select * from h where i = 10;
                        QUERY PLAN                        
----------------------------------------------------------
 Append  (cost=0.00..125.62 rows=40 width=4)
   ->  Seq Scan on h  (cost=0.00..0.00 rows=1 width=4)
         Filter: (i = 10)
   ->  Seq Scan on h1  (cost=0.00..41.88 rows=13 width=4)
         Filter: (i = 10)
   ->  Seq Scan on h2  (cost=0.00..41.88 rows=13 width=4)
         Filter: (i = 10)
   ->  Seq Scan on h3  (cost=0.00..41.88 rows=13 width=4)
         Filter: (i = 10)
(9 rows)

However, if we modify a condition into a same expression
as the partitions constraint, postgres can exclude unrelated
table from search targets. So, we might avoid the problem
by converting the qual properly before calling predicate_refuted_by().

postgres=# explain select * from h where abs(hashint4(i))%3 = abs(hashint4(10))%3;
                        QUERY PLAN                        
----------------------------------------------------------
 Append  (cost=0.00..61.00 rows=14 width=4)
   ->  Seq Scan on h  (cost=0.00..0.00 rows=1 width=4)
         Filter: ((abs(hashint4(i)) % 3) = 2)
   ->  Seq Scan on h3  (cost=0.00..61.00 rows=13 width=4)
         Filter: ((abs(hashint4(i)) % 3) = 2)
(5 rows)

Best regards,
Yugo Nagata

-- 
Yugo Nagata <nagata@sraoss.co.jp>

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