Randomly shuffle column in Spark RDD or dataframe

问题

Is there anyway I can shuffle a column of an RDD or dataframe such that the entries in that column appear in random order? I'm not sure which APIs I could use to accomplish such a task.

回答1:

What about selecting the column to shuffle, orderBy(rand) the column and zip it by index to the existing dataframe?

import org.apache.spark.sql.functions.rand

def addIndex(df: DataFrame) = spark.createDataFrame(
  // Add index
  df.rdd.zipWithIndex.map{case (r, i) => Row.fromSeq(r.toSeq :+ i)},
  // Create schema
  StructType(df.schema.fields :+ StructField("_index", LongType, false))
)

case class Entry(name: String, salary: Double)

val r1 = Entry("Max", 2001.21)
val r2 = Entry("Zhang", 3111.32)
val r3 = Entry("Bob", 1919.21)
val r4 = Entry("Paul", 3001.5)

val df = addIndex(spark.createDataFrame(Seq(r1, r2, r3, r4)))
val df_shuffled = addIndex(df
  .select(col("salary").as("salary_shuffled"))
  .orderBy(rand))

df.join(df_shuffled, Seq("_index"))
  .drop("_index")
  .show(false) 

+-----+-------+---------------+
|name |salary |salary_shuffled|
+-----+-------+---------------+
|Max  |2001.21|3001.5         |
|Zhang|3111.32|3111.32        |
|Paul |3001.5 |2001.21        |
|Bob  |1919.21|1919.21        |
+-----+-------+---------------+

回答2:

If you don't need a global shuffle across your data, you can shuffle within partitions using the mapPartitions method.

rdd.mapPartitions(Random.shuffle(_));

For a PairRDD (RDDs of type RDD[(K, V)]), if you are interested in shuffling the key-value mappings (mapping an arbitrary key to an arbitrary value):

pairRDD.mapPartitions(iterator => {
  val (keySequence, valueSequence) = iterator.toSeq.unzip
  val shuffledValueSequence = Random.shuffle(valueSequence)
  keySequence.zip(shuffledValueSequence).toIterator
}, true)

The boolean flag at the end denotes that partitioning is preserved (keys are not changed) for this operation so that downstream operations e.g. reduceByKey can be optimized (avoid shuffles).

回答3:

You can add one additional column random generated, and then sort the record based on this random generated column. By this way, you are randomly shuffle your destined column.

In this way, you do not need to have all data in memory, which can easily cause OOM. Spark will take care of sorting and memory limitation issue by spill to disk if necessary.

If you don't want the extra column, you can remove it after sorting.

回答4:

While one can not not just shuffle a single column directly - it is possible to permute the records in an RDD via RandomRDDs. https://spark.apache.org/docs/latest/api/java/org/apache/spark/mllib/random/RandomRDDs.html

A potential approach to having only a single column permuted might be:

use mapPartitions to do some setup/teardown on each Worker task
suck all of the records into memory. i.e. iterator.toList. Make sure you have many (/small) partitions of data to avoid OOME
using the Row object rewrite all back out as original except for the given column
within the mapPartitions create an in-memory sorted list
for the desired column drop its values in a separate collection and randomly sample the collection for replacing each record's entry
return the result as list.toIterator from the mapPartitions

来源：https://stackoverflow.com/questions/37287543/randomly-shuffle-column-in-spark-rdd-or-dataframe

标签

apache-spark

spark-dataframe