[英]How to create a Dataset of Maps?
I'm using Spark 2.2 and am running into troubles when attempting to call spark.createDataset
on a Seq
of Map
.我正在使用 Spark 2.2 并且在尝试在Map
的Seq
上调用spark.createDataset
时遇到了麻烦。
Code and output from my Spark Shell session follow:我的 Spark Shell 会话的代码和输出如下:
// createDataSet on Seq[T] where T = Int works
scala> spark.createDataset(Seq(1, 2, 3)).collect
res0: Array[Int] = Array(1, 2, 3)
scala> spark.createDataset(Seq(Map(1 -> 2))).collect
<console>:24: error: 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.
spark.createDataset(Seq(Map(1 -> 2))).collect
^
// createDataSet on a custom case class containing Map works
scala> case class MapHolder(m: Map[Int, Int])
defined class MapHolder
scala> spark.createDataset(Seq(MapHolder(Map(1 -> 2)))).collect
res2: Array[MapHolder] = Array(MapHolder(Map(1 -> 2)))
I've tried import spark.implicits._
, though I'm fairly certain that's implicitly imported by the Spark shell session.我试过import spark.implicits._
,但我相当确定它是由 Spark shell 会话隐式导入的。
Is this is a case not covered by current encoders?这是当前编码器未涵盖的情况吗?
It is not covered in 2.2, but can be easily addressed.它没有在 2.2 中涵盖,但可以很容易地解决。 You can add required Encoder
using ExpressionEncoder
, either explicitly:您可以使用ExpressionEncoder
添加所需的Encoder
,或者明确地:
import org.apache.spark.sql.catalyst.encoders.ExpressionEncoder
import org.apache.spark.sql.Encoder
spark
.createDataset(Seq(Map(1 -> 2)))(ExpressionEncoder(): Encoder[Map[Int, Int]])
or implicitly
:或implicitly
:
implicit def mapIntIntEncoder: Encoder[Map[Int, Int]] = ExpressionEncoder()
spark.createDataset(Seq(Map(1 -> 2)))
Just FYI that the above expression just works in Spark 2.3 (as of this commit if I'm not mistaken).仅供参考,上述表达式仅适用于 Spark 2.3(如果我没记错的话,从这次提交开始)。
scala> spark.version
res0: String = 2.3.0
scala> spark.createDataset(Seq(Map(1 -> 2))).collect
res1: Array[scala.collection.immutable.Map[Int,Int]] = Array(Map(1 -> 2))
I think it's because newMapEncoder
is now part of spark.implicits
.我认为这是因为newMapEncoder
现在是spark.implicits
一部分。
scala> :implicits
...
implicit def newMapEncoder[T <: scala.collection.Map[_, _]](implicit evidence$3: reflect.runtime.universe.TypeTag[T]): org.apache.spark.sql.Encoder[T]
You could "disable" the implicit by using the following trick and give the above expression a try (that will lead to an error).您可以使用以下技巧“禁用”隐式并尝试上述表达式(这将导致错误)。
trait ThatWasABadIdea
implicit def newMapEncoder(ack: ThatWasABadIdea) = ack
scala> spark.createDataset(Seq(Map(1 -> 2))).collect
<console>:26: error: 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.
spark.createDataset(Seq(Map(1 -> 2))).collect
^
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