CODEWITHSUNDEEP
listscalaprinting
Anton Ashanin
Anton Ashanin

Reputation: 1827

Scala: can not print list elements

I did "mkString" but still can not print list of strings. With input line:

9002194187,2644,54,100,3,4,2,5

I get the following output:

Line: 9002194187,2644,54,100,3,4,2,5
StrArr: 9002194187,2644,54,100,3,4,2,5
Lst: [Ljava.lang.String;@223d2e6c
Lst again: List([Ljava.lang.String;@223d2e6c)
Lst1: [Ljava.lang.String;@223d2e6c
Result: foo

From the code below:

def mkRecord(line: String) : String = {
    val klass = "foo"
    val strArr = line.split(",") // convert string to array of strings
    println("Line: "+line)
    println("StrArr: "+strArr.mkString(","))
    val lst = List(strArr)
    println("Lst: "+lst.mkString(" - "))
    println("Lst again: "+lst)
    val lst1 = lst.tail ++ List(klass) // attribute list except for the first one, plus new klass attribute
    println("Lst1: "+lst.mkString(" , "))
    val result = lst1.mkString(",") // attribute string
    println("Result: "+ result)
    return result
  }

Please, help. I am at complete loss (

Upvotes: 0

Views: 624

Answers (2)

itsbruce
itsbruce

Reputation: 4843

You can turn any array into a list with its toList operator, avoiding this problem (the nature of which Shadowlands has explained). You can do string -> array -> list in one line:

line.split(',').toList

Using a collection's toList method is often going to be faster than extracting all the elements into a sequence and then converting that sequence into a list, not least because you'll be using a method optimised for the source collection. However, that's an optimisation which you can worry about after the priorities of success and clarity.

Upvotes: 0

Shadowlands
Shadowlands

Reputation: 15074

The constructor for List (actually, the apply method on the List companion object) takes parameters in the form of scala's "varargs" equivalent:

def apply[A](xs: A*): List[A] // some irrelevant details have been elided

In java, this would be written something like:

public static List<A> apply(A... args)

In scala this can be called using any Seq (or subclass), but using a special notation. The line you used:

val lst = List(strArr)

creates a List[Array[String]], with a single entry - the array strArr. To tell the compiler to turn the array into a varargs when passing it to the List apply method, add : _* on the end (the space is optional):

val lst = List(strArr: _*)

This change in your code will result in:

scala> mkRecord(chkStr)
Line: 9002194187,2644,54,100,3,4,2,5
StrArr: 9002194187,2644,54,100,3,4,2,5
Lst: 9002194187 - 2644 - 54 - 100 - 3 - 4 - 2 - 5
Lst again: List(9002194187, 2644, 54, 100, 3, 4, 2, 5)
Lst1: 9002194187 , 2644 , 54 , 100 , 3 , 4 , 2 , 5
Result: 2644,54,100,3,4,2,5,foo
res1: String = 2644,54,100,3,4,2,5,foo

Upvotes: 2

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