scala SPARK/SQL：spark 无法解析符号 toDF
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SPARK/SQL:spark can't resolve symbol toDF
提问by yeyimilk
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.
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()，但它会通知我toDF()在编码时无法解析符号。抱歉我toDF()在 Apache Spark 中找不到博士。
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()
回答by zero323
To be able to to use toDFyou have to import sqlContext.implicitsfirst:
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
回答by Vineet Srivastava
This is a very late response to the question but just for the sake of people who are still looking for the answer:
这是对这个问题的一个很晚的回答，但只是为了那些仍在寻找答案的人：
Try the same command on Spark 1.6 it will work.
在 Spark 1.6 上尝试相同的命令，它会起作用。
I was facing the same issue and searched in google and didn't get solution and then I upgraded Spark from 1.5 to 1.6 and it worked.
我遇到了同样的问题并在谷歌搜索并没有得到解决方案，然后我将 Spark 从 1.5 升级到 1.6 并且它起作用了。
If you don't know your Spark version:
如果您不知道您的 Spark 版本：
spark-submit --version (from command prompt)
sc.version (from Scala Shell)
