Append a column to Data Frame in Apache Spark 1.3 - scala

Question

Is it possible and what would be the most efficient neat method to add a column to Data Frame?
More specifically, column may serve as Row IDs for the existing Data Frame.
In a simplified case, reading from file and not tokenizing it, I can think of something as below (in Scala), but it completes with errors (at line 3), and anyways doesn't look like the best route possible:
var dataDF = sc.textFile("path/file").toDF()
val rowDF = sc.parallelize(1 to DataDF.count().toInt).toDF("ID")
dataDF = dataDF.withColumn("ID", rowDF("ID"))

Answer 1

It's been a while since I posted the question and it seems that some other people would like to get an answer as well. Below is what I found.
So the original task was to append a column with row identificators (basically, a sequence 1 to numRows) to any given data frame, so the rows order/presence can be tracked (e.g. when you sample). This can be achieved by something along these lines:
sqlContext.textFile(file).
zipWithIndex().
map(case(d, i)=>i.toString + delimiter + d).
map(_.split(delimiter)).
map(s=>Row.fromSeq(s.toSeq))
Regarding the general case of appending any column to any data frame:
The "closest" to this functionality in Spark API are withColumn and withColumnRenamed. According to Scala docs, the former Returns a new DataFrame by adding a column. In my opinion, this is a bit confusing and incomplete definition. Both of these functions can operate on this data frame only, i.e. given two data frames df1 and df2 with column col:
val df = df1.withColumn("newCol", df1("col") + 1) // -- OK
val df = df1.withColumn("newCol", df2("col") + 1) // -- FAIL
So unless you can manage to transform a column in an existing dataframe to the shape you need, you can't use withColumn or withColumnRenamed for appending arbitrary columns (standalone or other data frames).
As it was commented above, the workaround solution may be to use a join - this would be pretty messy, although possible - attaching the unique keys like above with zipWithIndex to both data frames or columns might work. Although efficiency is ...
It's clear that appending a column to the data frame is not an easy functionality for distributed environment and there may not be very efficient, neat method for that at all. But I think that it's still very important to have this core functionality available, even with performance warnings.

Answer 2

not sure if it works in spark 1.3 but in spark 1.5 I use withColumn:
import sqlContext.implicits._
import org.apache.spark.sql.functions._
df.withColumn("newName",lit("newValue"))
I use this when I need to use a value that is not related to existing columns of the dataframe
This is similar to #NehaM's answer but simpler

Answer 3

I took help from above answer. However, I find it incomplete if we want to change a DataFrame and current APIs are little different in Spark 1.6.
zipWithIndex() returns a Tuple of (Row, Long) which contains each row and corresponding index. We can use it to create new Row according to our need.
val rdd = df.rdd.zipWithIndex()
.map(indexedRow => Row.fromSeq(indexedRow._2.toString +: indexedRow._1.toSeq))
val newstructure = StructType(Seq(StructField("Row number", StringType, true)).++(df.schema.fields))
sqlContext.createDataFrame(rdd, newstructure ).show
I hope this will be helpful.

Answer 4

You can use row_number with Window function as below to get the distinct id for each rows in a dataframe.
df.withColumn("ID", row_number() over Window.orderBy("any column name in the dataframe"))
You can also use monotonically_increasing_id for the same as
df.withColumn("ID", monotonically_increasing_id())
And there are some other ways too.