Add a new column to a Dataframe. New column i want it to be a UUID generator

You can utilize built-in Spark SQL uuid function:

.withColumn("uuid", expr("uuid()"))

A full example in Scala:

import org.apache.spark.sql.SparkSession
import org.apache.spark.sql.functions._
import org.apache.spark.sql.types._

object CreateDf extends App {

  val spark = SparkSession.builder
    .master("local[*]")
    .appName("spark_local")
    .getOrCreate()
  import spark.implicits._

  Seq(1, 2, 3).toDF("col1")
    .withColumn("uuid", expr("uuid()"))
    .show(false)

}

Output:

+----+------------------------------------+
|col1|uuid                                |
+----+------------------------------------+
|1   |24181c68-51b7-42ea-a9fd-f88dcfa10062|
|2   |7cd21b25-017e-4567-bdd3-f33b001ee497|
|3   |1df7cfa8-af8a-4421-834f-5359dc3ae417|
+----+------------------------------------+

You should try something like this:

val sc: SparkContext = ...
val sqlContext = new SQLContext(sc)

import sqlContext.implicits._

val generateUUID = udf(() => UUID.randomUUID().toString)
val df1 = Seq(("id1", 1), ("id2", 4), ("id3", 5)).toDF("id", "value")
val df2 = df1.withColumn("UUID", generateUUID())

df1.show()
df2.show()

Output will be:

+---+-----+
| id|value|
+---+-----+
|id1|    1|
|id2|    4|
|id3|    5|
+---+-----+

+---+-----+--------------------+
| id|value|                UUID|
+---+-----+--------------------+
|id1|    1|f0cfd0e2-fbbe-40f...|
|id2|    4|ec8db8b9-70db-46f...|
|id3|    5|e0e91292-1d90-45a...|
+---+-----+--------------------+