scala - a better way to parse an array of strings into a single string

Question

I have implemented a method that is supposed to convert an array of strings into a single string. But getting an exception when using it with UDF and applying the UDF to a column: val concatUdf = udf(convertArray)

java.lang.ClassCastException: scala.collection.mutable.WrappedArray$ofRef cannot be cast to [Ljava.lang.String;

What should be improved in my current implementation in order to return valid String? I'm new to Scala and probably this is not the most elegant solution.

def convertArray: Array[String] => String =
      (strings: Array[String]) => {
        Option(strings) match {
          case Some(arr) => strings.mkString(", ")
          case Some(Array()) => ""
          case None => ""
        }
      }

Zoltán · Accepted Answer · 2018-08-02 12:19:12Z

1

I believe you could just do

def convertArray(strings: Array[String]): String =
  if (strings.nonEmpty)
    strings.mkString(", ")
  else
    ""

in your code, the second case is unreachable, because your first case will always match, including empty arrays. That said, your code seems to work fine for me on Scala 2.12.6 (apart from the warning about the unreachable code):

scala> def convertArray: Array[String] => String =
     |       (strings: Array[String]) => {
     |         Option(strings) match {
     |               case Some(arr) => strings.mkString(", ")
     |           case Some(Array()) => ""
     |           case None => ""
     |         }
     |       }
<console>:15: warning: unreachable code
                 case Some(Array()) => ""
                                       ^
convertArray: Array[String] => String

scala> convertArray(Array())
res1: String = ""

scala> convertArray(Array("bro"))
res2: String = bro

scala> convertArray(Array("bro", "dude"))
res3: String = bro, dude

edited Aug 2, 2018 at 12:19

answered Aug 2, 2018 at 12:05

Zoltán

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3 Comments

samba Over a year ago

I'm trying to apply convertArray as a udf to a Dataframe column. val concatUdf = udf(convertArray). And getting the ClassCastException as above

Zoltán Over a year ago

@samba please edit your answer then, to add the relevant call. The error is not in your convertArray method, but in the code calling it. (You're likely not passing your method an Array[String], but something else)

samba Over a year ago

Yes, just solved it. I should have used Seq: def convertArray(strings: Seq[String]): String = if (strings.nonEmpty) strings.mkString(", ") else "" Thanks for sharing your knowledge

proximator · Accepted Answer · 2018-08-02 12:50:45Z

1

Just use mkString, no need to re-invent the wheel:

 println(Array().mkString(", "))
 println(Array("hello").mkString(", "))
 println(Array("hello", "world").mkString(", "))

Output:

//empty string
hello
hello, world

answered Aug 2, 2018 at 12:50

proximator

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Comments

teikitel · Accepted Answer · 2018-08-03 07:31:35Z

0

def convertArray(strings: Array[String]): String = Option(strings).getOrElse(Array()).mkString(", ")

Output :

scala> def convertArray(strings: Array[String]): String = Option(strings).getOrElse(Array()).mkString(", ") 
convertArray: (strings: Array[String])String

scala> convertArray(Array("to", "to", "ta", "ta")) 
res16: String = to, to, ta, ta

scala> convertArray(Array()) 
res17: String = ""

scala> convertArray(null) 
res18: String = ""

edited Aug 3, 2018 at 7:31

answered Aug 2, 2018 at 13:25

teikitel

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