toDF不是Seq的成员，toDS不是Seq的成员

Question

我正在尝试使用以下代码将序列转换为数据框或数据集。这似乎很简单，但是这样做时却出现了异常。不确定我犯了什么错误。我试图找出解决方案。类似的问题，并且按照我将类定义移出main之外的方法，但是仍然遇到问题。下面是代码

package sparkWCExample.spWCExample
      import org.apache.log4j.Level
      import org.apache.spark.sql.{Dataset, SparkSession , DataFrame , Row , Encoders }
      import org.apache.spark.sql.functions._
      import org.apache.spark.SparkContext
      import org.apache.spark.SparkConf
      import org.apache.spark.sql.Row
      import org.apache.spark.sql.Dataset



      // Create the case classes for our domain
case class Department(id: String, name: String)
case class Person(name: String, age: Long)


object DatasetExample  {
             def  main(args: Array[String]){
          println("Start now")
          val conf = new SparkConf().setAppName("Spark Scala WordCount Example").setMaster("local[1]")
        val spark = SparkSession.builder().config(conf).appName("CsvExample").master("local").getOrCreate()
        val sqlContext = new org.apache.spark.sql.SQLContext(spark.sparkContext)
        import sqlContext.implicits._
        import spark.implicits._

//val df = spark.read.options(Map("inferSchema"->"true","delimiter"->",","header"->"true")).csv("C:\\Sankha\\Study\\data\\salary.csv")

// Create the Departments
val department1 = new Department("123456", "Computer Science")
val department2 = new Department("789012", "Mechanical Engineering")
val department3 = new Department("345678", "Theater and Drama")
val department4 = new Department("901234", "Indoor Recreation")

val caseClassDS = Seq(Person("Andy", 32)).toDS()
val df = Seq(department1,department2,department3,department4).toDF



        }
}

我正在使用spark 2.4.5和scala 2.12，上面的代码是用scala ide编写的，下面是例外

toDF不是Seq [sparkWCExample.spWCExample.Department的成员toDS不是Seq [sparkWCExample.spWCExample.Person]的成员

[我正在尝试使用以下代码将序列转换为数据框或数据集。这似乎很简单，但是这样做时却出现异常。不确定我犯了什么错误。我试图...

Answer 1

删除：

Answer 2

这就是它的意思。类Seq没有方法toDF或toDS。我认为您应该在RDD类中寻找那些方法。