How to split strings into characters in Scala
StringScalaCharacter EncodingSplitCharacterString Problem Overview
For example, there is a string val s = "Test"
. How do you separate it into t, e, s, t
?
String Solutions
Solution 1 - String
Do you need characters?
"Test".toList // Makes a list of characters
"Test".toArray // Makes an array of characters
Do you need bytes?
"Test".getBytes // Java provides this
Do you need strings?
"Test".map(_.toString) // Vector of strings
"Test".sliding(1).toList // List of strings
"Test".sliding(1).toArray // Array of strings
Do you need UTF-32 code points? Okay, that's a tougher one.
def UTF32point(s: String, idx: Int = 0, found: List[Int] = Nil): List[Int] = {
if (idx >= s.length) found.reverse
else {
val point = s.codePointAt(idx)
UTF32point(s, idx + java.lang.Character.charCount(point), point :: found)
}
}
UTF32point("Test")
Solution 2 - String
You can use toList
as follows:
scala> s.toList
res1: List[Char] = List(T, e, s, t)
If you want an array, you can use toArray
scala> s.toArray
res2: Array[Char] = Array(T, e, s, t)
Solution 3 - String
Actually you don't need to do anything special. There is already implicit conversion in Predef
to WrappedString
and WrappedString
extends IndexedSeq[Char]
so you have all goodies that available in it, like:
"Test" foreach println
"Test" map (_ + "!")
###Edit###
Predef
has augmentString
conversion that has higher priority than wrapString
in LowPriorityImplicits
. So String end up being StringLike[String]
, that is also Seq
of chars.
Solution 4 - String
Additionally, it should be noted that if what you actually want isn't an actual list object, but simply to do something which each character, then Strings can be used as iterable collections of characters in Scala
for(ch<-"Test") println("_" + ch + "_") //prints each letter on a different line, surrounded by underscores