I'm working with Apache Spark on a Cluster using HDFS. As far as I understand, HDFS is distributing files on data-nodes. So if a put a "file.txt" on the filesystem, it will be split into partitions. Now I'm calling
rdd = SparkContext().textFile("hdfs://.../file.txt")
from Apache Spark. Has rdd now automatically the same partitions as "file.txt" on the filesystem? What happens when I call
rdd.repartition(x)
where x > then the partitions used by hdfs? Will Spark physically rearrange the data on hdfs to work locally?
Example: I put a 30GB Textfile on the HDFS-System, which is distributing it on 10 nodes. Will Spark a) use the same 10 partitons? and b) shuffle 30GB across the cluster when I call repartition(1000)?
When Spark reads a file from HDFS, it creates a single partition for a single input split. Input split is set by the Hadoop InputFormat
used to read this file. For instance, if you use textFile()
it would be TextInputFormat
in Hadoop, which would return you a single partition for a single block of HDFS (but the split between partitions would be done on line split, not the exact block split), unless you have a compressed text file. In case of compressed file you would get a single partition for a single file (as compressed text files are not splittable).
When you call rdd.repartition(x)
it would perform a shuffle of the data from N
partititons you have in rdd
to x
partitions you want to have, partitioning would be done on round robin basis.
If you have a 30GB uncompressed text file stored on HDFS, then with the default HDFS block size setting (128MB) it would be stored in 235 blocks, which means that the RDD you read from this file would have 235 partitions. When you call repartition(1000)
your RDD would be marked as to be repartitioned, but in fact it would be shuffled to 1000 partitions only when you will execute an action on top of this RDD (lazy execution concept)