Saturday, October 29, 2016

Spark Dataframe encoders

http://stackoverflow.com/questions/36648128/how-to-store-custom-objects-in-a-dataset

http://stackoverflow.com/questions/39433419/encoder-error-while-trying-to-map-dataframe-row-to-updated-row

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)