Cannot convert an RDD to Dataframe - scala

Question

I've converted a dataframe to an RDD:
val rows: RDD[Row] = df.orderBy($"Date").rdd
And now I'm trying to convert it back:
val df2 = spark.createDataFrame(rows)
But I'm getting an error:
Edit:
rows.toDF()
Also produces an error:
Cannot resolve symbol toDF
Even though I included this line earlier:
import spark.implicits._
Full code:
import org.apache.spark._
import org.apache.spark.sql._
import org.apache.spark.sql.expressions._
import org.apache.spark.sql.functions._
import org.apache.spark.sql.types._
import scala.util._
import org.apache.spark.mllib.rdd.RDDFunctions._
import org.apache.spark.rdd._
object Playground {
def main(args: Array[String]): Unit = {
val spark = SparkSession
.builder
.appName("Playground")
.config("spark.master", "local")
.getOrCreate()
import spark.implicits._
val sc = spark.sparkContext
val df = spark.read.csv("D:/playground/mre.csv")
df.show()
val rows: RDD[Row] = df.orderBy($"Date").rdd
val df2 = spark.createDataFrame(rows)
rows.toDF()
}
}

Answer 1

Your IDE is right, SparkSession.createDataFrame needs a second parameter: either a bean class or a schema.
This will fix your problem:
val df2 = spark.createDataFrame(rows, df.schema)