Partitioning - default Partition

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kamil
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Partitioning - default Partition

Post by kamil » Tue Nov 24, 2015 10:55 am

Hi,

Im curious if i can manage partitions like in other dbs (postgres,edb etc.)...
I cannot find proper answer in docs and other sources so i will need your help...

My table looks like this:

count | potential_id_key
-----------+------
958800406 | 9
169481205 | 8
13249261 | 7
6070862 | 11
932035 | 22
184952 | 32
120417 | 10
37177 | 3
4090 | 5
395 | 19
227 | 14
192 | 2
99 | 33
48 | -200
18 | 21
6 | 1
1 | 23

So i believe, that the best option will be to partition by potential_id_key...but

Is is possible to create only 3 partitions i.e.
- partition_key=9
- partition_key=8
- partition_key=7
because those 3 keys contains biggest number of elements and other values store in DEFAULT partition?

If something like DEFAULT partition exists in HP VERTICA?

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nnani
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Re: Partitioning - default Partition

Post by nnani » Fri Nov 27, 2015 2:17 pm

No default partition exist in Vertica.
Vertica is a bit different when compared to Postgres in terms of storage.

You will need to partition all your data.

Even if you partition by potential_id_key, All partitions reside on all nodes in cluster.
So every node will have count(distinct(potential_id_key)) as the minimum number of partitions.
Depending on the data, the partitions will be further switched to containers.

Option1
If you don't want to partition rest of the data, dump the required data into another with filter of potential_id_key and partition this new table.
Option2
Partition your fact table by all potential_id_key and then move the partitions for 7,8,9 id_key to a new table.
nnani........
Long way to go

You can check out my blogs at vertica-howto

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