Scala: Print separators when using output stream - scala

Question

When we need an array of strings to be concatenated, we can use mkString method:
val concatenatedString = listOfString.mkString
However, when we have a very long list of string, getting concatenated string may not be a good choice. In this case, It would be more appropriated to print out to an output stream directly, Writing it to output stream is simple:
listOfString.foreach(outstream.write _)
However, I don't know a neat way to append separators. One thing I tried is looping with an index:
var i = 0
for(str <- listOfString) {
if(i != 0) outstream.write ", "
outstream.write str
i += 1
}
This works, but it is too wordy. Although I can make a function encapsules the code above, I want to know whether Scala API already has a function do the same thing or not.
Thank you.

Answer 1

Here is a function that do what you want in a bit more elegant way:
def commaSeparated(list: List[String]): Unit = list match {
case List() =>
case List(a) => print(a)
case h::t => print(h + ", ")
commaSeparated(t)
}
The recursion avoids mutable variables.
To make it even more functional style, you can pass in the function that you want to use on each item, that is:
def commaSeparated(list: List[String], func: String=>Unit): Unit = list match {
case List() =>
case List(a) => func(a)
case h::t => func(h + ", ")
commaSeparated(t, func)
}
And then call it by:
commaSeparated(mylist, oustream.write _)

Answer 2

I believe what you want is the overloaded definitions of mkString.
Definitions of mkString:
scala> val strList = List("hello", "world", "this", "is", "bob")
strList: List[String] = List(hello, world, this, is, bob)
def mkString: String
scala> strList.mkString
res0: String = helloworldthisisbob
def mkString(sep: String): String
scala> strList.mkString(", ")
res1: String = hello, world, this, is, bob
def mkString(start: String, sep: String, end: String): String
scala> strList.mkString("START", ", ", "END")
res2: String = STARThello, world, this, is, bobEND
EDIT
How about this?
scala> strList.view.map(_ + ", ").foreach(print) // or .iterator.map
hello, world, this, is, bob,

Answer 3

Not good for parallelized code, but otherwise:
val it = listOfString.iterator
it.foreach{x => print(x); if (it.hasNext) print(' ')}

Answer 4

Here's another approach which avoids the var
listOfString.zipWithIndex.foreach{ case (s, i) =>
if (i != 0) outstream write ","
outstream write s }

Answer 5

Self Answer:
I wrote a function encapsulates the code in the original question:
implicit def withSeparator[S >: String](seq: Seq[S]) = new {
def withSeparator(write: S => Any, sep: String = ",") = {
var i = 0
for (str <- seq) {
if (i != 0) write(sep)
write(str)
i += 1
}
seq
}
}
You can use it like this:
listOfString.withSeparator(print _)
The separator can also be assigned:
listOfString.withSeparator(print _, ",\n")
Thank you for everyone answered me. What I wanted to use is a concise and not too slow representation. The implicit function withSeparator looks like the thing I wanted. So I accept my own answer for this question. Thank you again.