Encoder error while trying to map dataframe row to updated row

Question

When I m trying to do the same thing in my code as mentioned below

dataframe.map(row => {
  val row1 = row.getAs[String](1)
  val make = if (row1.toLowerCase == "tesla") "S" else row1
  Row(row(0),make,row(2))
})

I have taken the above reference from here: Scala: How can I replace value in Dataframs using scala But I am getting encoder error as

Unable to find encoder for type stored in a Dataset. Primitive types (Int, S tring, etc) and Product types (case classes) are supported by importing spark.im plicits._ Support for serializing other types will be added in future releases.

Note: I am using spark 2.0!

Answer 1

There is nothing unexpected here. You're trying to use code which has been written with Spark 1.x and is no longer supported in Spark 2.0:

• in 1.x DataFrame.map is ((Row) ⇒ T)(ClassTag[T]) ⇒ RDD[T]
• in 2.x Dataset[Row].map is ((Row) ⇒ T)(Encoder[T]) ⇒ Dataset[T]

To be honest it didn't make much sense in 1.x either. Independent of version you can simply use DataFrame API:

import org.apache.spark.sql.functions.{when, lower}

val df = Seq(
  (2012, "Tesla", "S"), (1997, "Ford", "E350"),
  (2015, "Chevy", "Volt")
).toDF("year", "make", "model")

df.withColumn("make", when(lower($"make") === "tesla", "S").otherwise($"make"))

If you really want to use map you should use statically typed Dataset:

import spark.implicits._

case class Record(year: Int, make: String, model: String)

df.as[Record].map {
  case tesla if tesla.make.toLowerCase == "tesla" => tesla.copy(make = "S")
  case rec => rec
}

or at least return an object which will have implicit encoder:

df.map {
  case Row(year: Int, make: String, model: String) => 
    (year, if(make.toLowerCase == "tesla") "S" else make, model)
}

Finally if for some completely crazy reason you really want to map over Dataset[Row] you have to provide required encoder:

import org.apache.spark.sql.catalyst.encoders.RowEncoder
import org.apache.spark.sql.types._
import org.apache.spark.sql.Row

// Yup, it would be possible to reuse df.schema here
val schema = StructType(Seq(
  StructField("year", IntegerType),
  StructField("make", StringType),
  StructField("model", StringType)
))

val encoder = RowEncoder(schema)

df.map {
  case Row(year, make: String, model) if make.toLowerCase == "tesla" => 
    Row(year, "S", model)
  case row => row
} (encoder)

Answer 2

For scenario where dataframe schema is known in advance answer given by @zero323 is the solution

but for scenario with dynamic schema / or passing multiple dataframe to a generic function: Following code has worked for us while migrating from 1.6.1 from 2.2.0

import org.apache.spark.sql.Row

val df = Seq(
   (2012, "Tesla", "S"), (1997, "Ford", "E350"),
   (2015, "Chevy", "Volt")
 ).toDF("year", "make", "model")

val data = df.rdd.map(row => {
  val row1 = row.getAs[String](1)
  val make = if (row1.toLowerCase == "tesla") "S" else row1
  Row(row(0),make,row(2))
})

this code executes on both the versions of spark.

disadvantage : optimization provided by spark on dataframe/datasets api wont be applied.

Answer 3

Just to add a few other important-to-know points in order to well understand the other answers (especially the final point of @zero323's answer about map over Dataset[Row]):

• First of all, Dataframe.map gives you a Dataset (more specifically, Dataset[T], rather than Dataset[Row])!
• And Dataset[T] always requires an encoder, that's what this sentence "Dataset[Row].map is ((Row) ⇒ T)(Encoder[T]) ⇒ Dataset[T]" means.
• There are indeed lots of encoders predefined already by Spark (which can be imported by doing import spark.implicits._), but still the list would not be able to cover many domain specific types that developers may create, in which case you need to create encoders yourself.
• In the specific example on this page, df.map returns a Row type for Dataset, and hang on a minute, Row type is not within the list of types that have encoders predefined by Spark, hence you are going to create one on your own.
• And I admit that creating an encoder for Row type is a bit different than the approach described in the above link, and you have to use RowEncoder which takes StructType as param describing type of a row, like what @zero323 provides above:

// this describes the internal type of a row
val schema = StructType(Seq(StructField("year", IntegerType), StructField("make", StringType), StructField("model", StringType)))

// and this completes the creation of encoder
// for the type `Row` with internal schema described above
val encoder = RowEncoder(schema)

Answer 4

In my case of spark 2.4.4 version, I had to import implicits. This is a general answer

val spark2 = spark
import spark2.implicits._

val data = df.rdd.map(row => my_func(row))

where my_func did some operation.