Convert RDD to Dataframe in Spark/Scala

Just paste into a spark-shell:

val a = 
  Array(
    Array("4580056797", "0", "2015-07-29 10:38:42", "0", "1", "1"), 
    Array("4580056797", "0", "2015-07-29 10:38:42", "0", "1", "1"))

val rdd = sc.makeRDD(a)

case class X(callId: String, oCallId: String, 
  callTime: String, duration: String, calltype: String, swId: String)

Then map() over the RDD to create instances of the case class, and then create the DataFrame using toDF():

scala> val df = rdd.map { 
  case Array(s0, s1, s2, s3, s4, s5) => X(s0, s1, s2, s3, s4, s5) }.toDF()
df: org.apache.spark.sql.DataFrame = 
  [callId: string, oCallId: string, callTime: string, 
    duration: string, calltype: string, swId: string]

This infers the schema from the case class.

Then you can proceed with:

scala> df.printSchema()
root
 |-- callId: string (nullable = true)
 |-- oCallId: string (nullable = true)
 |-- callTime: string (nullable = true)
 |-- duration: string (nullable = true)
 |-- calltype: string (nullable = true)
 |-- swId: string (nullable = true)

scala> df.show()
+----------+-------+-------------------+--------+--------+----+
|    callId|oCallId|           callTime|duration|calltype|swId|
+----------+-------+-------------------+--------+--------+----+
|4580056797|      0|2015-07-29 10:38:42|       0|       1|   1|
|4580056797|      0|2015-07-29 10:38:42|       0|       1|   1|
+----------+-------+-------------------+--------+--------+----+

If you want to use toDF() in a normal program (not in the spark-shell), make sure (quoted from here):

• To import sqlContext.implicits._ right after creating the SQLContext
• Define the case class outside of the method using toDF()

You need to convert first you Array into Row and then define schema. I made assumption that most of your fields are Long

    val rdd: RDD[Array[String]] = ???
    val rows: RDD[Row] = rdd map {
      case Array(callId, oCallId, callTime, duration, swId) =>
        Row(callId.toLong, oCallId.toLong, callTime, duration.toLong, swId.toLong)
    }

    object schema {
      val callId = StructField("callId", LongType)
      val oCallId = StructField("oCallId", StringType)
      val callTime = StructField("callTime", StringType)
      val duration = StructField("duration", LongType)
      val swId = StructField("swId", LongType)

      val struct = StructType(Array(callId, oCallId, callTime, duration, swId))
    }

    sqlContext.createDataFrame(rows, schema.struct)

Using spark 1.6.1 and scala 2.10

I got the same error error: overloaded method value createDataFrame with alternatives:

For me, gotcha was the signature in createDataFrame, I was trying to use the val rdd : List[Row], but it failed because java.util.List[org.apache.spark.sql.Row] and scala.collection.immutable.List[org.apache.spark.sql.Row] are NOT the same.

The working solution I've found is I would convert val rdd : Array[Array[String]] into RDD[Row] via List[Array[String]]. I find this is the closest to what's in the documentation

import org.apache.spark.sql.Row
import org.apache.spark.sql.types.{StructType,StructField,StringType};
val sqlContext = new org.apache.spark.sql.SQLContext(sc)

val rdd_original : Array[Array[String]] = Array(
    Array("4580056797", "0", "2015-07-29 10:38:42", "0", "1", "1"), 
    Array("4580056797", "0", "2015-07-29 10:38:42", "0", "1", "1"))

val rdd : List[Array[String]] = rdd_original.toList

val schemaString = "callId oCallId callTime duration calltype swId"

// Generate the schema based on the string of schema
val schema =
  StructType(
    schemaString.split(" ").map(fieldName => StructField(fieldName, StringType, true)))

// Convert records of the RDD to Rows.
val rowRDD = rdd.map(p => Row(p: _*)) // using splat is easier
// val rowRDD = rdd.map(p => Row(p(0), p(1), p(2), p(3), p(4), p(5))) // this also works

val df = sqlContext.createDataFrame(sc.parallelize(rowRDD:List[Row]), schema)
df.show