Converting A Spark Dataframe To A Scala Map Collection


Answer :

I don't think your question makes sense -- your outermost Map, I only see you are trying to stuff values into it -- you need to have key / value pairs in your outermost Map. That being said:



val peopleArray = df.collect.map(r => Map(df.columns.zip(r.toSeq):_*))


Will give you:



Array(
  Map("age" -> null, "name" -> "Michael"),
  Map("age" -> 30, "name" -> "Andy"),
  Map("age" -> 19, "name" -> "Justin")
)


At that point you could do:



val people = Map(peopleArray.map(p => (p.getOrElse("name", null), p)):_*)


Which would give you:



Map(
  ("Michael" -> Map("age" -> null, "name" -> "Michael")),
  ("Andy" -> Map("age" -> 30, "name" -> "Andy")),
  ("Justin" -> Map("age" -> 19, "name" -> "Justin"))
)


I'm guessing this is really more what you want. If you wanted to key them on an arbitrary Long index, you can do:



val indexedPeople = Map(peopleArray.zipWithIndex.map(r => (r._2, r._1)):_*)


Which gives you:



Map(
  (0 -> Map("age" -> null, "name" -> "Michael")),
  (1 -> Map("age" -> 30, "name" -> "Andy")),
  (2 -> Map("age" -> 19, "name" -> "Justin"))
)


First get the schema from Dataframe



val schemaList = dataframe.schema.map(_.name).zipWithIndex//get schema list from dataframe


Get the rdd from dataframe and mapping with it



dataframe.rdd.map(row =>
  //here rec._1 is column name and rce._2 index
  schemaList.map(rec => (rec._1, row(rec._2))).toMap
 ).collect.foreach(println)


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