我无法访问包中的SparkConf
。但我已导入import org.apache.spark.SparkConf
。我的代码是:
import org.apache.spark.SparkContext
import org.apache.spark.SparkContext._
import org.apache.spark.SparkConf
import org.apache.spark.rdd.RDD
import org.apache.spark._
import org.apache.spark.streaming._
import org.apache.spark.streaming.StreamingContext._
object SparkStreaming {
def main(arg: Array[String]) = {
val conf = new SparkConf.setMaster("local[2]").setAppName("NetworkWordCount")
val ssc = new StreamingContext( conf, Seconds(1) )
val lines = ssc.socketTextStream("localhost", 9999)
val words = lines.flatMap(_.split(" "))
val pairs_new = words.map( w => (w, 1) )
val wordsCount = pairs_new.reduceByKey(_ + _)
wordsCount.print()
ssc.start() // Start the computation
ssc.awaitTermination() // Wait for the computation to the terminate
}
}
sbt
依赖项是:
name := "Spark Streaming"
version := "1.0"
scalaVersion := "2.10.4"
libraryDependencies ++= Seq(
"org.apache.spark" %% "spark-core" % "1.5.2" % "provided",
"org.apache.spark" %% "spark-mllib" % "1.5.2",
"org.apache.spark" %% "spark-streaming" % "1.5.2"
)
但错误显示无法访问SparkConf
。
[error] /home/cliu/Documents/github/Spark-Streaming/src/main/scala/Spark-Streaming.scala:31: object SparkConf in package spark cannot be accessed in package org.apache.spark
[error] val conf = new SparkConf.setMaster("local[2]").setAppName("NetworkWordCount")
[error] ^
答案 0 :(得分:6)
如果在SparkConf之后添加括号,则编译:
val conf = new SparkConf().setMaster("local[2]").setAppName("NetworkWordCount")
关键是SparkConf是一个类而不是函数,因此您也可以将类名用于范围目的。因此,当您在类名后添加括号时,您确保调用类构造函数而不是作用域功能。以下是Scala shell中的示例,说明了不同之处:
scala> class C1 { var age = 0; def setAge(a:Int) = {age = a}}
defined class C1
scala> new C1
res18: C1 = $iwC$$iwC$C1@2d33c200
scala> new C1()
res19: C1 = $iwC$$iwC$C1@30822879
scala> new C1.setAge(30) // this doesn't work
<console>:23: error: not found: value C1
new C1.setAge(30)
^
scala> new C1().setAge(30) // this works
scala>
答案 1 :(得分:1)
在这种情况下,您不能省略括号,因此它应该是:
val conf = new SparkConf().setMaster("local[2]").setAppName("NetworkWordCount")