How to split multi-value column into separate rows using typed Dataset?

Question

I am facing an issue of how to split a multi-value column, i.e. List[String], into separate rows.

The initial dataset has following types: Dataset

Answer 1

explode is often suggested, but it's from the untyped DataFrame API and given you use Dataset, I think flatMap operator might be a better fit (see org.apache.spark.sql.Dataset).

flatMap[U](func: (T) ⇒ TraversableOnce[U])(implicit arg0: Encoder[U]): Dataset[U]

(Scala-specific) Returns a new Dataset by first applying a function to all elements of this Dataset, and then flattening the results.

You could use it as follows:

val ds = Seq(
  (0, "Lorem ipsum dolor", 1.0, Array("prp1", "prp2", "prp3")))
  .toDF("id", "text", "value", "properties")
  .as[(Integer, String, Double, scala.List[String])]

scala> ds.flatMap { t => 
  t._4.map { prp => 
    (t._1, t._2, t._3, prp) }}.show
+---+-----------------+---+----+
| _1|               _2| _3|  _4|
+---+-----------------+---+----+
|  0|Lorem ipsum dolor|1.0|prp1|
|  0|Lorem ipsum dolor|1.0|prp2|
|  0|Lorem ipsum dolor|1.0|prp3|
+---+-----------------+---+----+

// or just using for-comprehension
for {
  t <- ds
  prp <- t._4
} yield (t._1, t._2, t._3, prp)

Answer 2

Here's one way to do it:

val myRDD = sc.parallelize(Array(
  (0, "text0", 1.0, List("prp1", "prp2", "prp3")),
  (1, "text1", 2.0, List("prp4", "prp5", "prp6")),
  (2, "text2", 3.0, List("prp7", "prp8", "prp9"))
)).map{
  case (i, t, v, ps) => ((i, t, v), ps)
}.flatMapValues(x => x).map{
  case ((i, t, v), p) => (i, t, v, p)
}

Answer 3

You can use explode:

df.withColumn("property", explode($"property"))

Example:

val df = Seq((1, List("a", "b"))).toDF("A", "B")   
// df: org.apache.spark.sql.DataFrame = [A: int, B: array<string>]

df.withColumn("B", explode($"B")).show
+---+---+
|  A|  B|
+---+---+
|  1|  a|
|  1|  b|
+---+---+