hadoop - How does Hive choose the number of reducers for a job?

Question

Several places say the default # of reducers in a Hadoop job is 1. You can use the mapred.reduce.tasks symbol to manually set the number of reducers.

When I run a Hive job (on Amazon EMR, AMI 2.3.3), it has some number of reducers greater than one. Looking at job settings, something has set mapred.reduce.tasks, I presume Hive. How does it choose that number?

Note: here are some messages while running a Hive job that should be a clue:

...
Number of reduce tasks not specified. Estimated from input data size: 500
In order to change the average load for a reducer (in bytes):
  set hive.exec.reducers.bytes.per.reducer=<number>
In order to limit the maximum number of reducers:
  set hive.exec.reducers.max=<number>
In order to set a constant number of reducers:
  set mapred.reduce.tasks=<number>
...

Answer 1

The default of 1 maybe for a vanilla Hadoop install. Hive overrides it.

In open source hive (and EMR likely)

# reducers = (# bytes of input to mappers)
             / (hive.exec.reducers.bytes.per.reducer)

This post says default hive.exec.reducers.bytes.per.reducer is 1G.

You can limit the number of reducers produced by this heuristic using hive.exec.reducers.max.

If you know exactly the number of reducers you want, you can set mapred.reduce.tasks, and this will override all heuristics. (By default this is set to -1, indicating Hive should use its heuristics.)

In some cases - say 'select count(1) from T' - Hive will set the number of reducers to 1 , irrespective of the size of input data. These are called 'full aggregates' - and if the only thing that the query does is full aggregates - then the compiler knows that the data from the mappers is going to be reduced to trivial amount and there's no point running multiple reducers.