Why is “Unable to find encoder for type stored in a Dataset” when creating a dataset of custom case class?

问题

Spark 2.0 (final) with Scala 2.11.8. The following super simple code yields the compilation error Error:(17, 45) Unable to find encoder for type stored in a Dataset. Primitive types (Int, String, etc) and Product types (case classes) are supported by importing spark.implicits._ Support for serializing other types will be added in future releases.

import org.apache.spark.sql.SparkSession

case class SimpleTuple(id: Int, desc: String)

object DatasetTest {
  val dataList = List(
    SimpleTuple(5, "abc"),
    SimpleTuple(6, "bcd")
  )

  def main(args: Array[String]): Unit = {
    val sparkSession = SparkSession.builder.
      master("local")
      .appName("example")
      .getOrCreate()

    val dataset = sparkSession.createDataset(dataList)
  }
}

回答1:

Spark Datasets require Encoders for data type which is about to be stored. For common types (atomics, product types) there is a number of predefined encoders available but you have to import these first from SparkSession.implicits to make it work:

val sparkSession: SparkSession = ???
import sparkSession.implicits._
val dataset = sparkSession.createDataset(dataList)

Alternatively you can provide directly an explicit

import org.apache.spark.sql.{Encoder, Encoders}

val dataset = sparkSession.createDataset(dataList)(Encoders.product[SimpleTuple])

or implicit

implicit val enc: Encoder[SimpleTuple] = Encoders.product[SimpleTuple]
val dataset = sparkSession.createDataset(dataList)

Encoder for the stored type.

Note that Enocders also provide a number of predefined Encoders for atomic types, and Encoders for complex ones, can derived with ExpressionEncoder.

Further reading:

• For custom objects which are not covered by built-in encoders see How to store custom objects in Dataset?
• For Row objects you have to provide Encoder explicitly as shown in Encoder error while trying to map dataframe row to updated row

回答2:

For other users (yours is correct), note that you it's also important that the case class is defined outside of the object scope. So:

Fails:

object DatasetTest {
  case class SimpleTuple(id: Int, desc: String)

  val dataList = List(
    SimpleTuple(5, "abc"),
    SimpleTuple(6, "bcd")
  )

  def main(args: Array[String]): Unit = {
    val sparkSession = SparkSession.builder
      .master("local")
      .appName("example")
      .getOrCreate()
    val dataset = sparkSession.createDataset(dataList)
  }
}

Add the implicits, still fails with the same error:

object DatasetTest {
  case class SimpleTuple(id: Int, desc: String)

  val dataList = List(
    SimpleTuple(5, "abc"),
    SimpleTuple(6, "bcd")
  )

  def main(args: Array[String]): Unit = {
    val sparkSession = SparkSession.builder
      .master("local")
      .appName("example")
      .getOrCreate()

    import sparkSession.implicits._
    val dataset = sparkSession.createDataset(dataList)
  }
}

Works:

case class SimpleTuple(id: Int, desc: String)

object DatasetTest {   
  val dataList = List(
    SimpleTuple(5, "abc"),
    SimpleTuple(6, "bcd")
  )

  def main(args: Array[String]): Unit = {
    val sparkSession = SparkSession.builder
      .master("local")
      .appName("example")
      .getOrCreate()

    import sparkSession.implicits._
    val dataset = sparkSession.createDataset(dataList)
  }
}

Here's the relevant bug: https://issues.apache.org/jira/browse/SPARK-13540, so hopefully it will be fixed in the next release of Spark 2.

(Edit: Looks like that bugfix is actually in Spark 2.0.0... So I'm not sure why this still fails).

回答3:

I'd clarify with an answer to my own question, that if the goal is to define a simple literal SparkData frame, rather than use Scala tuples and implicit conversion, the simpler route is to use the Spark API directly like this:

  import org.apache.spark.sql._
  import org.apache.spark.sql.types._
  import scala.collection.JavaConverters._

  val simpleSchema = StructType(
    StructField("a", StringType) ::
    StructField("b", IntegerType) ::
    StructField("c", IntegerType) ::
    StructField("d", IntegerType) ::
    StructField("e", IntegerType) :: Nil)

  val data = List(
    Row("001", 1, 0, 3, 4),
    Row("001", 3, 4, 1, 7),
    Row("001", null, 0, 6, 4),
    Row("003", 1, 4, 5, 7),
    Row("003", 5, 4, null, 2),
    Row("003", 4, null, 9, 2),
    Row("003", 2, 3, 0, 1)
  )

  val df = spark.createDataFrame(data.asJava, simpleSchema)

来源：https://stackoverflow.com/questions/38664972/why-is-unable-to-find-encoder-for-type-stored-in-a-dataset-when-creating-a-dat