scala - Spark /SQL :spark can't resolve symbol toDF

Question

在我的项目中，我的外部库是 spark-assemble-1.3.1-hadoop2.6.0，如果我按“.”，IDE 会通知我 toDF()，但它告诉我，当我编码时无法解析符号 toDF() 。很抱歉，我在 Apache 中找不到 toDF() Spark 文档。

case class Feature(name:String, value:Double, time:String, period:String)
val RESRDD = RDD.map(tuple => {
    var bson=new BasicBSONObject();
    bson.put("name",name);
    bson.put("value",value);
    (null,bson);
})

RESRDD
 .map(_._2)
 .map(f => Feature(f.get("name").toString, f.get("value").toString.toDouble))
 .toDF()

Answer 1

为了能够使用toDF，您必须先导入sqlContext.implicits:

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

case class Foobar(foo: String, bar: Integer)

val foobarRdd = sc.parallelize(("foo", 1) :: ("bar", 2) :: ("baz", -1) :: Nil).
    map { case (foo, bar) => Foobar(foo, bar) } 

val foobarDf = foobarRdd.toDF
foobarDf.limit(1).show