sunhelly commented on pull request #3260: URL: https://github.com/apache/hbase/pull/3260#issuecomment-840352608
> > > I think even with only 1 table, we could still compute the table skew count? > > > For example, we have a table with 100 regions, and there are 3 region servers, but all the regions are placed on a single server? > > > > > > Hi, @Apache9 , when there is only 1 table, just like balance by table (the cluster state only contains distribution info of exactly one table and all the online RSes), the RegionCountSkewCostFunction can replace the TableSkewCostFunction. For the example of [100,0,0], the cost of RegionCountSkewCostFunction is 1, which will generate balance actions in the next steps. > > But when the table regions has already been evenly distributed, for example [1,0,0,0], the cost of TableSkewCostFunction will not be 0, it is 1(the cost of RegionCountSkewCostFunction is 0), but the table does not need to be balanced, it brings many redundant computation steps of actions. > > OK, got your point. So you mean if there is only one table, RegionCountSkewCostFunction is enough, and TableSkewCostFunction will generate unnecesary actions. > > @clarax WDYT? Is this a valid point? Yes, and one table is just a general scenario when we set "hbase.master.loadbalance.bytable" be true. -- This is an automated message from the Apache Git Service. To respond to the message, please log on to GitHub and use the URL above to go to the specific comment. For queries about this service, please contact Infrastructure at: [email protected]
