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

Question

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.

Bribery

Answer 1

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

Cobelligerent

I see two problems with this solution. The first is that UUID.randomUUID isn't guaranteed to be unique across a cluster. It uses pseudo-random numbers, which is fine on a single machine, but in a cluster, you could get collisions. Second, the Docs for User Defined Functions say that UDFs must be deterministic. By definitions, UUIDs are not. github.com/apache/spark/blob/… – Withers

For more info about how non-determinism can cause problems, see #42961420 – Beyer

UDFs are bad for your spark code on various levels: for one, Catalyst can't do anything with them because they're not transparent to the optimization engine. – Dedicate

Answer 2

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|
+----+------------------------------------+

Franke

How do you do this in python? I get ('Column is not iterable') when i do expr(uuid()) – Abelabelard

from pyspark.sql.functions import expr df.withColumn("uuid", expr("uuid()")) – Satire

Answer 3

This is how we did in Java, we had a column date and wanted to add another column with month.

Dataset<Row> newData = data.withColumn("month", month((unix_timestamp(col("date"), "MM/dd/yyyy")).cast("timestamp")));

You can use similar technique to add any column.

Dataset<Row> newData1 = newData.withColumn("uuid", lit(UUID.randomUUID().toString()));

Cheers !

Noisy

With the second approach, it add the same ID in all the rows – Volume