How to convert ListBuffer to List (immutable collection ) in scala

Question

I have the below code using ListBuffer but I want to use immutable collection like List and achieve the same result

val list1: ListBuffer[String] = ListBuffer("a", "b", "c")
val s= list1
    .foldLeft(mutable.ListBuffer.empty[String]) { (strings, content) =>
      {
        if(StringUtils.isNotBlank(content))
          strings += content
        else
          strings += "-"

      }
    }
    .mkString(";")

Second version

val list1: ListBuffer[String] = ListBuffer("a", "b", "c")
val s= list1
    .foldLeft(mutable.ListBuffer.empty[String]) { (strings, content) =>
      {

       strings += content

      }
    }
    .mkString(";")

.toList? should work

Raman Mishra
– Raman Mishra

2020-05-19 08:53:57 +00:00
Commented May 19, 2020 at 8:53 — Raman Mishra
– Raman Mishra, Commented May 19, 2020 at 8:53

Ivan Stanislavciuc · Accepted Answer · 2020-05-19 08:50:37Z

1

You can use collect

  List("a", "b", "c").collect {
    case c if StringUtils.isNotBlank(c) => c
    case _ => "-"
  }.mkString(";")

answered May 19, 2020 at 8:50

Ivan Stanislavciuc

7,27517 silver badges19 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

coder25 Over a year ago

In case I do not have a blank value check , is there any alternative to collect

Ivan Stanislavciuc Over a year ago

if you don't do any checks, there is no point in foldLeft , this will be sufficient List("a", "b", "c").mkString(";")

Jasper-M Over a year ago

Since you never drop any elements, I think map instead of collect will suffice here.

Collectives™ on Stack Overflow

How to convert ListBuffer to List (immutable collection ) in scala

1 Answer 1

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related