您应该尝试以下方法:

 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()
 

输出将是:

+---+-----+
| 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...|
+---+-----+--------------------+

I want to add a new column to a Dataframe, a UUID generator.

UUID value will look something like 21534cf7-cff9-482a-a3a8-9e7244240da7

My Research:

I've tried with withColumn method in spark.

val DF2 = DF1.withColumn("newcolname", DF1("existingcolname" + 1)

So DF2 will have additional column with newcolname with 1 added to it in all rows.

By my requirement is that I want to have a new column which can generate the UUID.

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...|
+---+-----+--------------------+