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