[英]Why is the error “Unable to find encoder for type stored in a Dataset” when encoding JSON using case classes?
[英]Why is "Unable to find encoder for type stored in a Dataset" when creating a dataset of custom case class?
Spark 2.0(最終版)和 Scala 2.11.8。 以下超級簡單的代碼產生編譯錯誤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.
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)
}
}
Spark Datasets
需要Encoders
用於即將存儲的數據類型。 對於常見類型(原子、產品類型),有許多可用的預定義編碼器,但您必須先從SparkSession.implicits
導入它們才能使其工作:
val sparkSession: SparkSession = ???
import sparkSession.implicits._
val dataset = sparkSession.createDataset(dataList)
或者,您可以直接提供一個明確的
import org.apache.spark.sql.{Encoder, Encoders}
val dataset = sparkSession.createDataset(dataList)(Encoders.product[SimpleTuple])
或隱含的
implicit val enc: Encoder[SimpleTuple] = Encoders.product[SimpleTuple]
val dataset = sparkSession.createDataset(dataList)
存儲類型的Encoder
。
請注意, Encoders
還提供了許多用於原子類型的預定義Encoders
,以及用於復雜類型的Encoders
,可以通過ExpressionEncoder
派生。
進一步閱讀:
Row
對象,您必須在嘗試將數據幀行映射到更新行時顯式提供Encoder
,如編碼器錯誤中所示對於其他用戶(您的用戶是正確的),請注意,在object
范圍之外定義case class
也很重要。 所以:
失敗:
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)
}
}
添加隱式,仍然失敗並出現相同的錯誤:
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)
}
}
作品:
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)
}
}
這是相關的錯誤: https://issues.apache.org/jira/browse/SPARK-13540 ,所以希望它會在 Spark 2 的下一個版本中得到修復。
(編輯:看起來這個錯誤修正實際上是在 Spark 2.0.0 中......所以我不確定為什么這仍然失敗)。
我會用我自己的問題的答案來澄清,如果目標是定義一個簡單的文字 SparkData 框架,而不是使用 Scala 元組和隱式轉換,則更簡單的路線是直接使用 Spark API,如下所示:
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)
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.