How to convert a case-class-based RDD into a DataFrame?

Question

The Spark documentation shows how to create a DataFrame from an RDD, using Scala case classes to infer a schema. I am trying to reproduce this concept using sqlContext

Answer 1

All you need is just

val dogDF = sqlContext.createDataFrame(dogRDD)

Second parameter is part of Java API and expects you class follows java beans convention (getters/setters). Your case class doesn't follow this convention, so no property is detected, that leads to empty DataFrame with no columns.

Answer 2

Case Class Approach won't Work in cluster mode. It'll give ClassNotFoundException to the case class you defined.

Convert it a RDD[Row] and define the schema of your RDD with StructField and then createDataFrame like

val rdd = data.map { attrs => Row(attrs(0),attrs(1)) }  

val rddStruct = new StructType(Array(StructField("id", StringType, nullable = true),StructField("pos", StringType, nullable = true)))

sqlContext.createDataFrame(rdd,rddStruct)

toDF() wont work either

Answer 3

You can create a DataFrame directly from a Seq of case class instances using toDF as follows:

val dogDf = Seq(Dog("Rex"), Dog("Fido")).toDF