Doubt 1: entire partition if uses and record spooled as it is in partition
then we get 4 times of source record(If config file is of 4 nodes).
right or wrong??
Doubt 2: Same if uses in first stage then wht would be the default partition.(Because same follows the partition strategy of upcoming stage and if it spplied in first stage the for which partition datastage would go for)??
Doubt 3: Differnce between round robin and random partition(Means random partition known to distribute records in random manner amongest nodes then how random partition manages load balencing)??
Some partition doubts
Moderators: chulett, rschirm, roy
-
- Participant
- Posts: 251
- Joined: Mon Jun 09, 2008 5:52 am
-
- Participant
- Posts: 54607
- Joined: Wed Oct 23, 2002 10:52 pm
- Location: Sydney, Australia
- Contact:
2. Selecting 'same' partition in source/output link will try to use the same partition in the source dataset created by a prior job - avoids repartitioning of data. Though 'same' can be used wtih source database stages, we need to be careful when data is read in parallel from a partitioned DB2 (and i guess in oracle also) table (for eg. using 'current node' clause).
Thanks,
Senthil
Thanks,
Senthil
-
- Participant
- Posts: 1
- Joined: Thu Feb 03, 2011 9:44 am
Re: Some partition doubts
ANSWER: Round Robin and Random are different itself in their distribution.
Lets take one example.
If My data is =(5,8,3,9,4,6,7,5,8,12,45,98,36,14)
Roundrobin for three nodes will be:
First: (5,9,7,12,36)
Second:(8,4,5,45,14)
Third:(3,6,8,98)
But Random could be: (this is one of the possible way)
First: (5,14,5,4,98)
Second:(8,12,7,45,9)
Third:(3,6,8,36)
Hope will be helpful
Lets take one example.
If My data is =(5,8,3,9,4,6,7,5,8,12,45,98,36,14)
Roundrobin for three nodes will be:
First: (5,9,7,12,36)
Second:(8,4,5,45,14)
Third:(3,6,8,98)
But Random could be: (this is one of the possible way)
First: (5,14,5,4,98)
Second:(8,12,7,45,9)
Third:(3,6,8,36)
Hope will be helpful
-
- Premium Member
- Posts: 1044
- Joined: Wed Sep 29, 2004 3:30 am
- Location: Nottingham, UK
- Contact:
Re: Some partition doubts
Sorry to reply to such an old post but this needs to be corrected. That is NOT possible - since 14 is the last input value, it can only ever be the last value out of whichever node it goes to. It can't turn up as the second value in a node, unless that node only received 2 values.datastagesandeep wrote:But Random could be: (this is one of the possible way)
First: (5,14,5,4,98)
Second:(8,12,7,45,9)
Third:(3,6,8,36)
Unless there is also a sort happening on some other unseen value.
Phil Hibbs | Capgemini
Technical Consultant
Technical Consultant