How to build Encoder for Regex class in Scala w/ Spark

Question

I'd like to build a structure that links a regex pattern to a description of a feature within some text.

Example: "^.* horses .$" maps to 'horses'; "^. pigs .*$" maps to 'pigs' and so on

There are thousands of possible descriptions for this text, so grouping a compiled regex pattern w/ its description would allow me to search efficiently. Below is the key part of my code:

import org.apache.spark.sql.functions._
import org.apache.spark.sql.types._
import org.apache.spark.sql.SparkSession
import org.apache.spark.sql.{Encoder, Encoders}
import scala.util.matching.Regex

object GlueApp {
    case class RegexMetadata(regexName: String, pattern: scala.util.matching.Regex)
    def main(sysArgs: Array[String]) {
      val args = GlueArgParser.getResolvedOptions(sysArgs, Seq("JOB_NAME").toArray)
      val sc: SparkContext = new SparkContext()
      val glueContext: GlueContext = new GlueContext(sc)
      val spark = glueContext.getSparkSession
      import spark.implicits._
      Job.init(args("JOB_NAME"), glueContext, args.asJava)

      implicit val regexEncoder = Encoders.kryo[scala.util.matching.Regex]
      implicit val regexMetadataEncoder = Encoders.product[RegexMetadata]
      Job.commit()
}

}

When I run this, I get the following error: java.lang.UnsupportedOperationException No Encoder found for scala.util.matching.Regex

It compiles and runs fine when I don't have the "implicit val regexMetadataEncoder" line. This seems to work on Databricks, but not on AWS Glue.

Some searching found these similar questions, but I can't solve my problem w/ them:

scala generic encoder for spark case class

Thank you for your help!

Spark 2.x scala 2.1.1 custom encoder class type mismatch

score 1 · Answer 1 · answered Nov 27 '19 at 18:23

Got it working. I had an issue with declaring my encoders correctly. Below is the key section of my working code:

import org.apache.spark.sql.functions._
import org.apache.spark.sql.types._
import org.apache.spark.sql.SparkSession
import org.apache.spark.sql.{Encoder, Encoders}
import org.apache.spark.sql.catalyst.encoders.ExpressionEncoder
import scala.collection.mutable.WrappedArray
import java.util.regex.Pattern
object GlueApp {
    /* RegexConfig -- maps a regex pattern string to a value */
    case class RegexConfig(value: String, patternRegex: String)
    /* RegexMetadata -- maps a compiled regex pattern to a regex config */
    case class RegexMetadata(config: RegexConfig, pattern: java.util.regex.Pattern)
    def main(sysArgs: Array[String]) {
        val args = GlueArgParser.getResolvedOptions(sysArgs, Seq("JOB_NAME").toArray)
        val sc: SparkContext = new SparkContext()
        val glueContext: GlueContext = new GlueContext(sc)
        val spark = glueContext.getSparkSession
        import spark.implicits._
        Job.init(args("JOB_NAME"), glueContext, args.asJava)
        implicit val regexMetadataEncoder = Encoders.kryo[RegexMetadata]
        val regexEncoder = Encoders.product[RegexConfig]
        << read file w/ regex patterns and put into regexConfigArray >>
        val regexLocal = for (config <- regexConfigArray) yield 
            RegexMetadata(config, Pattern.compile(config.patternRegex, 
                               Pattern.CASE_INSENSITIVE))
        Job.commit()
    }
}

How to build Encoder for Regex class in Scala w/ Spark

1 Answers1