convert list of elements to tuple5, prevent index out of bounds

Question

I'm trying to create a tuple from a scala list:

.map('path -> ('uri1, 'uri2, 'uri3, 'uri4, 'uri5)) {elems:List[String] =>

  (elems(0), elems(1), elems(2), elems(3), elems(4)) //ouf of bounds!
}

But the elems may have between 1 and 5 elements, so obviously I will hit an index out of bounds exception.

What's the scala/scalding way of doing this? I'm guessing that the proper way is to iterate a range from 1 to 5 and generate the tuple from there.

I'd like to return null (for compatibility reasons) when the elements do not exist.

What would you like to return when `elems` has i.e. 3 elements? — Norbert Radyk, Jun 17 '14 at 10:39
If you're asking what's the Scala way of doing this, then I'd suggest using `Option` for your Tuple values instead of `null`. — Norbert Radyk, Jun 17 '14 at 11:10

score 5 · Answer 1 · answered Jun 17 '14 at 11:08

The approach here really depends on your requirements.

If you're after Tuple5 only when there are 5 elements defined in elem then a simple pattern match is a way to go (and as the function returns Option you can flatten it after map to get an expected result):

  def convert(l: List[String]): Option[(String, String, String, String, String)] = l match {
    case e1 :: e2 :: e3 :: e4 :: e5 :: _ => Some((e1, e2, e3, e4, e5))
    case _ => None
  }

Alternatively if you're after taking all the list sizes and converting them to Tuple5 with some default you can:

Use pattern matching as shown above and handle the remaining 5 cases (0,1,2,3,4 elems in the list) defaulting missing values.
Wrap your elem(n) calls with scala.util.Try and default (i.e. Try(elem(1)).getOrElse(""))
Add the default values to the elem list to fill in the missing values

Example illustrating last option below:

  def convert(l: List[String], default : String): (String, String, String, String, String) = {
    val defaults = (1 to 5 - l.size).map(_ => default).toList
    l ++ defaults match {
      case e1 :: e2 :: e3 :: e4 :: e5 :: _ => (e1, e2, e3, e4, e5)
    }
  }

score 1 · Answer 2 · answered Jun 17 '14 at 15:12

There is a way to do this in 1 line, let stringList be the List[String] and default be some default

new Tuple(stringList.map(_.asInstanceOf[AnyRef]): _*) ++ (stringList.size until 5).map(_ => default)

Problem with this is it's not typesafe, so I gave Norbert an upvote because his answer is - and perhaps easier to read.

score 1 · Accepted Answer · answered Jun 17 '14 at 22:48

1

You can pad the list with nulls as necessary:

.map('path -> ('uri1, 'uri2, 'uri3, 'uri4, 'uri5)) {elems:List[String] =>
  val padded= elems ++ List.fill(5)(null)
  (padded(0), padded(1), padded(2), padded(3), padded(4))  //No ouf of bounds!
}

answered Jun 17 '14 at 22:48

Sasha O

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convert list of elements to tuple5, prevent index out of bounds

3 Answers3