Hi David, The default partitioner used in map reduce is the hash partitioner. So based on your keys they are send to a particular reducer.
May be in your current data set, the keys that have no values in table are all falling in the same hash bucket and hence being processed by the same reducer. If you are noticing a skew on a particular reducer, sometimes a simple work around like increasing the no of reducers explicitly might help you get pass the hurdle. Also please ensure you have enabled skew join optimization. Regards Bejoy KS Sent from remote device, Please excuse typos -----Original Message----- From: "David Morel" <dmore...@gmail.com> Date: Thu, 24 Jan 2013 18:39:56 To: <user@hive.apache.org>; <bejoy...@yahoo.com> Reply-To: user@hive.apache.org Subject: Re: An explanation of LEFT OUTER JOIN and NULL values On 24 Jan 2013, at 18:16, bejoy...@yahoo.com wrote: > Hi David > > An explain extended would give you the exact pointer. > > From my understanding, this is how it could work. > > You have two tables then two different map reduce job would be > processing those. Based on the join keys, combination of corresponding > columns would be chosen as key from mapper1 and mapper2. So if the > combination of columns having the same value those records from two > set of mappers would go into the same reducer. > > On the reducer if there is a corresponding value for a key from table > 1 to table 2/mapper 2 that value would be populated. If no val for > mapper 2 then those columns from table 2 are made null. > > If there is a key-value just from table 2/mapper 2 and no > corresponding value from mapper 1. That value is just discarded. Hi Bejoy, Thanks! So schematically, something like this, right? mapper1 (bigger table): K1-A, V1A K2-A, V2A K3-A, V3A mapper2 (joined, smaller table): K1-B, V1B reducer1: K1-A, V1A K1-B, V1B returns: K1, V1A, V1B etc reducer2: K2-A, V2A *no* K2-B, V so: K2-B, NULL is created, same for next row. K3-A, V3A returns: K2, V2A, NULL etc K3, V3A, NULL etc I still don't understand why my reducer2 (and only this one, which apparently gets all the keys for which we don't have a row on table B) would become overloaded. Am I completely misunderstanding the whole thing? David > Regards > Bejoy KS > > Sent from remote device, Please excuse typos > > -----Original Message----- > From: "David Morel" <dmore...@gmail.com> > Date: Thu, 24 Jan 2013 18:03:40 > To: user@hive.apache.org<user@hive.apache.org> > Reply-To: user@hive.apache.org > Subject: An explanation of LEFT OUTER JOIN and NULL values > > Hi! > > After hitting the "curse of the last reducer" many times on LEFT OUTER > JOIN queries, and trying to think about it, I came to the conclusion > there's something I am missing regarding how keys are handled in > mapred jobs. > > The problem shows when I have table A containing billions of rows with > distinctive keys, that I need to join to table B that has a much lower > number of rows. > > I need to keep all the A rows, populated with NULL values from the B > side, so that's what a LEFT OUTER is for. > > Now, when transforming that into a mapred job, my -naive- > understanding would be that for every key on the A table, a missing > key on the B table would be generated with a NULL value. If that were > the case, I fail to understand why all NULL valued B keys would end up > on the same reducer, since the key defines which reducer is used, not > the value. > > So, obviously, this is not how it works. > > So my question is: how is this construct handled? > > Thanks a lot! > > D.Morel
