CODEWITHSUNDEEP
scalaexceptionerror-handlingtry-catch
Soumya Simanta
Soumya Simanta

Reputation: 11741

Converting a List[Try[A]] to List[A] in Scala

I want to filter out bad input from input data. I'm currently using scala.util.Try to wrap any exceptions. Following is a simple example where 3I throws a NumberFormatException. I was wondering if there is a better way of doing this in Scala ? 

val data = List ( ("Joe", "20"), ("James", "30"), ("Pete", "3I") )

scala> val parsedData = data.map{ d => Try{Person( d._1, d._2.toInt ) }}
parsedData: List[scala.util.Try[Person]] = List(Success(Person(Joe,20)), Success(Person(James,30)), Failure(java.lang.NumberFormatException: For input string: "3I"))

scala> val validdata = parsedData.map{ x => x match {
     | case Success(s) => Some(s)
     | case Failure(f) => None }
     | }
validdata: List[Option[Person]] = List(Some(Person(Joe,20)), Some(Person(James,30)), None)

scala> validdata.flatten
res13: List[Person] = List(Person(Joe,20), Person(James,30))

Upvotes: 7

Views: 4436

Answers (2)

elm
elm

Reputation: 20415

One approach that bypasses the use of Try, as follows,

for ( d <- data if d._2.forall(_.isDigit) ) yield Person(d._1, d._2.toInt)

Yet this may not prove as scalable as @m-z approach.

Upvotes: 0

Michael Zajac
Michael Zajac

Reputation: 55569

Use collect to keep only the values that match the pattern you desire:

parsedData collect { case Success(x) => x }

This will also work, though I don't think it's quite as clear:

parsedData.flatMap(_.toOption)

Upvotes: 16

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