Spark Dataframe :How to add a index Column : Aka Distributed Data Index

monotonically_increasing_id - The generated ID is guaranteed to be monotonically increasing and unique, but not consecutive.

"I want to add a column from 1 to row's number."

Let say we have the following DF

+--------+-------------+-------+
| userId | productCode | count |
+--------+-------------+-------+
|     25 |        6001 |     2 |
|     11 |        5001 |     8 |
|     23 |         123 |     5 |
+--------+-------------+-------+

To generate the IDs starting from 1

val w = Window.orderBy("count")
val result = df.withColumn("index", row_number().over(w))

This would add an index column ordered by increasing value of count.

+--------+-------------+-------+-------+
| userId | productCode | count | index |
+--------+-------------+-------+-------+
|     25 |        6001 |     2 |     1 |
|     23 |         123 |     5 |     2 |
|     11 |        5001 |     8 |     3 |
+--------+-------------+-------+-------+

With Scala you can use:

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

df.withColumn("id",monotonicallyIncreasingId)

You can refer to this exemple and scala docs.

With Pyspark you can use:

from pyspark.sql.functions import monotonically_increasing_id 

df_index = df.select("*").withColumn("id", monotonically_increasing_id())

NOTE : Above approaches doesn't give a sequence number, but it does give increasing id.

Simple way to do that and ensure the order of indexes is like below.. zipWithIndex.

Sample data.

+-------------------+
|               Name|
+-------------------+
|     Ram Ghadiyaram|
|        Ravichandra|
|              ilker|
|               nick|
|             Naveed|
|      Gobinathan SP|
|Sreenivas Venigalla|
|     Jackela Kowski|
|   Arindam Sengupta|
|            Liangpi|
|             Omar14|
|        anshu kumar|
+-------------------+

---

    package com.example

import org.apache.spark.internal.Logging
import org.apache.spark.sql.SparkSession._
import org.apache.spark.sql.functions._
import org.apache.spark.sql.types.{LongType, StructField, StructType}
import org.apache.spark.sql.{DataFrame, Row}

/**
  * DistributedDataIndex : Program to index an RDD  with
  */
object DistributedDataIndex extends App with Logging {

  val spark = builder
    .master("local[*]")
    .appName(this.getClass.getName)
    .getOrCreate()

  import spark.implicits._

  val df = spark.sparkContext.parallelize(
    Seq("Ram Ghadiyaram", "Ravichandra", "ilker", "nick"
      , "Naveed", "Gobinathan SP", "Sreenivas Venigalla", "Jackela Kowski", "Arindam Sengupta", "Liangpi", "Omar14", "anshu kumar"
    )).toDF("Name")
  df.show
  logInfo("addColumnIndex here")
  // Add index now...
  val df1WithIndex = addColumnIndex(df)
    .withColumn("monotonically_increasing_id", monotonically_increasing_id)
  df1WithIndex.show(false)

  /**
    * Add Column Index to dataframe to each row
    */
  def addColumnIndex(df: DataFrame) = {
    spark.sqlContext.createDataFrame(
      df.rdd.zipWithIndex.map {
        case (row, index) => Row.fromSeq(row.toSeq :+ index)
      },
      // Create schema for index column
      StructType(df.schema.fields :+ StructField("index", LongType, false)))
  }
}

Result :

+-------------------+-----+---------------------------+
|Name               |index|monotonically_increasing_id|
+-------------------+-----+---------------------------+
|Ram Ghadiyaram     |0    |0                          |
|Ravichandra        |1    |8589934592                 |
|ilker              |2    |8589934593                 |
|nick               |3    |17179869184                |
|Naveed             |4    |25769803776                |
|Gobinathan SP      |5    |25769803777                |
|Sreenivas Venigalla|6    |34359738368                |
|Jackela Kowski     |7    |42949672960                |
|Arindam Sengupta   |8    |42949672961                |
|Liangpi            |9    |51539607552                |
|Omar14             |10   |60129542144                |
|anshu kumar        |11   |60129542145                |
+-------------------+-----+---------------------------+

How to get a sequential id column id[1, 2, 3, 4...n]:

from pyspark.sql.functions import desc, row_number, monotonically_increasing_id
from pyspark.sql.window import Window

df_with_seq_id = df.withColumn('index_column_name', row_number().over(Window.orderBy(monotonically_increasing_id())) - 1)

Note that row_number() starts at 1, therefore subtract by 1 if you want 0-indexed column