我正在尝试自学 Scala 并使用 IntelliJ IDEA 作为我的 IDE。我已经启动了 IntelliJ 的 sbt shell,运行console
然后输入以下内容:
import org.apache.spark.SparkConf
import org.apache.spark.sql.{DataFrame, SparkSession}
import java.time.LocalDate
object DataFrameExtensions {
implicit class DataFrameExtensions(df: DataFrame){
def featuresGroup1(groupBy: Seq[String], asAt: LocalDate): DataFrame = {df}
def featuresGroup2(groupBy: Seq[String], asAt: LocalDate): DataFrame = {df}
}
}
import DataFrameExtensions._
val spark = SparkSession.builder().config(new SparkConf().setMaster("local[*]")).enableHiveSupport().getOrCreate()
import spark.implicits._
val df = Seq((8, "bat"),(64, "mouse"),(-27, "horse")).toDF("number", "word")
val groupBy = Seq("a","b")
val asAt = LocalDate.now()
val dataFrames = Seq(df.featuresGroup1(groupBy, asAt),df.featuresGroup2(groupBy, asAt))
它在最后一行失败:
scala> val dataFrames = Seq(df.featuresGroup1(groupBy, asAt),df.featuresGroup2(groupBy, asAt))
<console>:25: error: value featuresGroup1 is not a member of
org.apache.spark.sql.DataFrame
val dataFrames = Seq(df.featuresGroup1(groupBy, asAt),df.featuresGroup2(groupBy, asAt))
^
<console>:25: error: value featuresGroup2 is not a member of org.apache.spark.sql.DataFrame
val dataFrames = Seq(df.featuresGroup1(groupBy, asAt),df.featuresGroup2(groupBy, asAt))
^
我已经从其他地方(我知道它可以工作)逐字复制了代码,所以我不知道为什么这不起作用。为什么我的隐式类中定义的函数不能作为 a 上的函数使用DataFrame
?