如何将json转换为RDD [json]

Question

我想在spark中编写json对象但是当我尝试使用sc.parallelize将其转换为RDD时，它再次将其转换回字符串

import scala.util.parsing.json._
import org.apache.spark.sql._
import org.apache.spark.sql.types._
import org.apache.spark.sql.functions.lit
import org.json4s._
import org.json4s.JsonDSL._
import org.json4s.jackson.JsonMethods._

val df = Seq((2012, 8, "Batman", 9.8), 
             (2012, 9, "Batman", 10.0), 
             (2012, 8, "Hero", 8.7),
             (2012, 10, "Hero", 5.7), 
             (2012, 2, "Robot", 5.5), 
             (2011, 7, "Git", 2.0),
             (2010, 1, "Dom", 2.0),
             (2019, 3, "Sri", 2.0)).toDF("year", "month", "title", "rating")

case class Rating(year:Int, month:Int, title:String, rating:Double)


import scala.collection.JavaConversions._
val ratingList = df.as[Rating].collectAsList

import java.io._
val output = for (c <- ratingList) yield
{
      val json = ("record" ->
              ("year" -> c.year) ~
              ("month" -> c.month) ~
              ("title" -> c.title) ~
              ("rating" -> c.rating))
      compact(render(json))
}

output.foreach(println)    

在这个阶段它是一个json对象，一切都很好。但是当我将它转换为RDD时，spark将其视为一个字符串

val outputDF = sc.parallelize(output).toDF("json")
outputDF.show()
outputDF.write.mode("overwrite").json("s3://location/")

输出是：

{"test":{"json":"{\"record\":{\"year\":2012,\"month\":8,\"title\":\"Batman\",\"rating\":9.8}}"}}

Answer 1

当你调用compact时 - 你从渲染的json中创建String。看到：

scala> val json = ("name" -> "joe") ~ ("age" -> 35)
scala> compact(render(json))
res2: String = {"name":"joe","age":35}

这意味着你的output是一系列字符串。当你并行化它 - 你得到RDD [String]。

您可能希望返回render函数的结果以获取JSON对象的集合。但是你需要检查文档。

当然，Spark不知道如何使用toDF()函数将JSON对象从第三方库转换为DataFrame。可能你可以这样做：

val anotherPeopleRDD = sc.parallelize(
  """{"name":"Yin","address":{"city":"Columbus","state":"Ohio"}}""" :: Nil)
val anotherPeople = sqlContext.read.json(anotherPeopleRDD)

所以基本上有RDD [String]然后把它读作JSON。

而BTW

你为什么先这样做：

val ratingList = df.as[Rating].collectAsList
val output = for (c <- ratingList) yield
{
      val json = ("record" ->
              ("year" -> c.year) ~
              ("month" -> c.month) ~
              ("title" -> c.title) ~
              ("rating" -> c.rating))
      compact(render(json))
}

然后：

val outputDF = sc.parallelize(output).toDF("json")

为什么不直接处理集群中的所有数据，如下所示：

df.as[Rating].map{c =>
  val json = ("record" ->
    ("year" -> c.year) ~
      ("month" -> c.month) ~
      ("title" -> c.title) ~
      ("rating" -> c.rating))
  compact(render(json))
}

这样会更有效率。