How should I range partition an index with a varchar2 column in Oracle? Is it a bad idea?

Question

I am using Oracle 10g Enterprise edition.

A table in our Oracle database stores the soundex value representation of another text column. We are using a custom soundex implementation in which the soundex values are longer than are generated by traditional soundex algorithms (such as the one Oracle uses). That's really beside the point.

Basically I have a varchar2 column that has values containing a single character followed by a dynamic number of numeric values (e.g. 'A12345', 'S382771', etc). The table is partitioned by another column, but I'd like to add a partitioned index to the soundex column since it is often searched. When trying to add a range partitioned index using the first character of the soundex column it worked great:

create index IDX_NAMES_SOUNDEX on NAMES_SOUNDEX (soundex)
global partition by range (soundex) (
    partition IDX_NAMES_SOUNDEX_PART_A values less than ('B'),  -- 'A%'
    partition IDX_NAMES_SOUNDEX_PART_B values less than ('C'),  -- 'B%'
    ...
);

However, I in order to more evenly distribute the size of the partitions, I want to define some partitions by the first two chars, like so:

create index IDX_NAMES_SOUNDEX on NAMES_SOUNDEX (soundex)
global partition by range (soundex) (
    partition IDX_NAMES_SOUNDEX_PART_A5 values less than ('A5'), -- 'A0% - A4%'
    partition IDX_NAMES_SOUNDEX_PART_A values less than ('B'),   -- 'A4% - A9%'
    partition IDX_NAMES_SOUNDEX_PART_B values less than ('C'),   -- 'B%'
    ...
);

I'm not sure how to properly range partition using varchar2 columns. I'm sure this is a less than ideal choice, so perhaps someone can recommend a better solution. Here's a distribution of the soundex data in my table:

-----------------------------------
|  SUBSTR(SOUNDEX,1,1)  |  COUNT  |
-----------------------------------
|                    A  | 6476349 |
|                    B  |  854880 |
|                    D  |  520676 |
|                    F  | 1200045 |
|                    G  |  280647 |
|                    H  | 3048637 |
|                    J  |  711031 |
|                    K  | 1336522 |
|                    L  |  348743 |
|                    M  | 3259464 |
|                    N  | 1510070 |
|                    Q  |  276769 |
|                    R  | 1263008 |
|                    S  | 3396223 |
|                    V  |  533844 |
|                    W  |  555007 |
|                    Y  |  348504 |
|                    Z  | 1079179 |
-----------------------------------

As you can see, the distribution is not evenly spread, which is why I want to define range partitions using the first two characters instead of just the first character.

Suggestions?

Thanks!

Answer 1

What exactly is your question?

Don't you know how you can split your table in n equal parts to avoid skew?

You can do that with analytic function percentile_disc().

Here an SQL PLUS example with n=100, I admit that it isn't very sophisticated but it will do the job.

set pages 0
set lines 200

drop table random_strings;

create table random_strings 
as 
select upper(dbms_random.string('A', 12)) rndmstr
from dual 
connect by level < 1000;


spool parts

select 'select '||level||'/100,percentile_disc('||level||
       '/100) within group (order by RNDMSTR) from random_strings;' 
       sql_statement
from dual 
connect by level <= 100
/

spool off

This will output in file parts.lst:

select 1/100,percentile_disc(1/100) within group (order by RNDMSTR) from random_strings;                                                                                                                
select 2/100,percentile_disc(2/100) within group (order by RNDMSTR) from random_strings;                                                                                                                
select 3/100,percentile_disc(3/100) within group (order by RNDMSTR) from random_strings;                                                 
...
select 100/100,percentile_disc(100/100) within group (order by RNDMSTR) from random_strings;                                                 

Now you can run script parts.lst to get the partition values. Each partition will contain 1% of the data initially.

Script parts.lst will output:

   ,01 AJUDRRSPGMNP
   ,02 AOMJZQPZASQZ
   ,03 AWDQXVGLLUSJ
   ,04 BIEPUHAEMELR
   .... 
   ,99 ZTMHDWTXUJAR
   1   ZYVJLNATVLOY

Answer 2

Is the table is being searched by the partitioning key in addition to the SOUNDEX value? Or is it being searched just by the SOUNDEX column?

If you are just trying to achieve an even distribution of data among partitions, have you considered using hash partitions rather than range partitions? Assuming you choose a power of 2 for the number of partitions, that should give you a pretty even distribution of data between partitions.

Answer 3

Talk to me! Can you tell me what your reason is for partitioning this table? It sounds like it is an OLTP table and may not need to be partition. We don’t want to partition just to say we are partitioned. Tell me what you are trying to accomplish by partitioning this table and I can help you pick a correct partitioning scheme. Partitioning does not equal faster queries. It actually can cause your queries to be slower in some cases.

I see some of your additional thoughts above and I don’t believe you need to partition your table. If your queries are going to be doing aggregates on entire partitions then you may want to partition. If you are going to have hundreds of millions of rows of data you may want to partition to help with DBA maintenance. If you just want you queries to run fast then the primary key index will suffice. Please let me know

Just create a global index on your desired columns.