spark.default.parallelism for Parallelize RDD defaults to 2 for spark submit

7
votes

Spark standalone cluster with a master and 2 worker nodes 4 cpu core on each worker. Total 8 cores for all workers.

When running the following via spark-submit (spark.default.parallelism is not set)

val myRDD = sc.parallelize(1 to 100000)
println("Partititon size - " + myRDD.partitions.size)
val totl = myRDD.reduce((x, y) => x + y)
println("Sum - " + totl)

It returns value 2 for partition size.

When using spark-shell by connecting to spark standalone cluster the same code returns correct partition size 8.

What can be the reason ?

Thanks.

1
scalaapache-spark

1 Answers

6
votes

spark.default.parallelism defaults to the number of all cores on all machines. The parallelize api has no parent RDD to determine the number of partitions, so it uses the spark.default.parallelism.

When running spark-submit, you're probably running it locally. Try submitting your spark-submit with the same start up configs as you do the spark-shell.

Pulled this from the documentation:

spark.default.parallelism

For distributed shuffle operations like reduceByKey and join, the largest number of partitions in a parent RDD. For operations like parallelize with no parent RDDs, it depends on the cluster manager:

Local mode: number of cores on the local machine

Mesos fine grained mode: 8

Others: total number of cores on all executor nodes or 2, whichever is larger

Default number of partitions in RDDs returned by transformations like join, reduceByKey, and parallelize when not set by user.