SPARK/SQL:spark无法将符号解析为DF [英] SPARK/SQL:spark can't resolve symbol toDF
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问题描述
在我的项目中,我的外部库是spark-assembly-1.3.1-hadoop2.6.0
,如果我按'.',IDE会通知我toDF()
,但它会告诉我在编码时无法解析符号toDF()
.抱歉,我在Apache Spark文档中找不到toDF()
.
In my project, my external library is spark-assembly-1.3.1-hadoop2.6.0
, if I press '.', the IDE inform me toDF()
, but it inform me that can't resolve symbol toDF()
when I code it in. I'm sorry I can't find the toDF()
in Apache Spark doc.
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()
推荐答案
要使用toDF
,您必须先导入sqlContext.implicits
:
To be able to to use toDF
you have to import sqlContext.implicits
first:
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
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