toDF 的值不是 org.apache.spark.rdd.RDD 的成员

Question

例外：

val people = sc.textFile("resources/people.txt").map(_.split(",")).map(p => Person(p(0), p(1).trim.toInt)).toDF()
value toDF is not a member of org.apache.spark.rdd.RDD[Person]

这是

TestApp.scala

文件：

package main.scala

import org.apache.spark.SparkContext
import org.apache.spark.SparkContext._
import org.apache.spark.SparkConf


case class Record1(k: Int, v: String)


object RDDToDataFramesWithCaseClasses {

    def main(args: Array[String]) {
        val conf = new SparkConf().setAppName("Simple Spark SQL Application With RDD To DF")

        // sc is an existing SparkContext.
        val sc = new SparkContext(conf)

        val sqlContext = new SQLContext(sc)

        // this is used to implicitly convert an RDD to a DataFrame.
        import sqlContext.implicits._

        // Define the schema using a case class.
        // Note: Case classes in Scala 2.10 can support only up to 22 fields. To work around this limit,package main.scala

还有

TestApp.scala

import org.apache.spark.SparkContext    
import org.apache.spark.SparkContext._
import org.apache.spark.SparkConf


case class Record1(k: Int, v: String)


object RDDToDataFramesWithCaseClasses {
    def main(args: Array[String]) {
        val conf = new SparkConf().setAppName("RDD To DF")

        // sc is an existing SparkContext.
        // you can use custom classes that implement the Product interface.
        case class Person(name: String, age: Int)

        // Create an RDD of Person objects and register it as a table.
        val people = sc.textFile("resources/people.txt").map(_.split(",")).map(p => Person(p(0), p(1).trim.toInt)).toDF() 
        people.registerTempTable("people")

        // SQL statements can be run by using the sql methods provided by sqlContext.
        val teenagers = sqlContext.sql("SELECT name, age FROM people WHERE age >= 13 AND age <= 19")

        // The results of SQL queries are DataFrames and support all the normal RDD operations.
        // The columns of a row in the result can be accessed by field index:
        teenagers.map(t => "Name: " + t(0)).collect().foreach(println)

        // or by field name:
        teenagers.map(t => "Name: " + t.getAs[String]("name")).collect().foreach(println)

        // row.getValuesMap[T] retrieves multiple columns at once into a Map[String, T]

        teenagers.map(_.getValuesMap[Any](List("name", "age"))).collect().foreach(println)

        // Map("name" -> "Justin", "age" -> 19)

    }
}

和SBT 文件

name := "SparkScalaRDBMS"
version := "1.0"
scalaVersion := "2.11.7"
libraryDependencies += "org.apache.spark" %% "spark-core" % "1.5.1"
libraryDependencies += "org.apache.spark" %% "spark-sql" % "1.5.1"

Answer 1

现在我找到了原因，你应该在对象中定义案例类，并在主函数之外。 看这里

好的，我终于解决了这个问题。需要做 2 件事：

1. 导入隐式：请注意，只有在创建 org.apache.spark.sql.SQLContext 实例后才应执行此操作。应该写成：

   val sqlContext= new org.apache.spark.sql.SQLContext(sc)

   import sqlContext.implicits._

2. 将案例类移到方法之外：案例类（用于定义 DataFrame 的架构）应该在需要它的方法之外定义。您可以在这里阅读更多相关信息：https://issues.scala-lang.org/browse/SI-6649

Answer 2

在 Spark 2 中，您需要从 SparkSession 导入隐式：

val spark = SparkSession.builder().appName(appName).getOrCreate()
import spark.implicits._

有关创建 SparkSession 时的更多选项，请参阅 Spark 文档。

Answer 3

您的代码有两个问题

1. 对于 Spark V 1.0，您需要导入 import sqlContext.implicits._ ；如果您使用 Spark V 2.0 或更高版本，则需要导入 Spark.implicits._

2. 第二种情况 class Record1(k: Int, v: String) 需要位于 main 函数内部但位于 def main(args: Array[String]) { val conf = new SparkConf().setAppName("RDD 转 DF") …

}

Answer 4

scala> case class Employee(id: Int, name: String, age: Int)
defined class Employee
scala> val sqlContext= new org.apache.spark.sql.SQLContext(sc)
warning: there was one deprecation warning; re-run with -deprecation for details
sqlContext: org.apache.spark.sql.SQLContext = org.apache.spark.sql.SQLContext@1f94e3a

scala> import sqlContext.implicits._
import sqlContext.implicits._
scala> var empl1=   empl.map(_.split(",")).map(e=>Employee(e(0).trim.toInt,e(1),e(2).trim.toInt)).toDF
empl1: org.apache.spark.sql.DataFrame = [id: int, name: string ... 1 more field]
scala> val allrecords = sqlContext.sql("SELECT * FROM employee")
allrecords: org.apache.spark.sql.DataFrame = [id: int, name: string ... 1 more field]

scala> allrecords.show();
+----+--------+---+
|  id|    name|age|
+----+--------+---+
|1201|  satish| 25|
|1202| krishna| 28|
|1203|   amith| 39|
|1204|   javed| 23|
|1205|  prudvi| 23|
+----+--------+---+

Answer 5

在 Scala 工作表中运行 Spark 时遇到此问题。基本上，由于工作表的性质，在这些情况下您不能使用 toDF()。相反，请使用

spark.createDataFrame

。