get first 2 values in a comma separated string - scala

Question

I am trying to get the first 2 values of a comma separated string in scala. For example
a,b,this is a test
How do i store the values a,b in 2 separate variables?

Answer 1

To keep it easy and clean.
KISS solution:
1.Use split for separation. Then use take which is defined on all ordered sequences to get the elements as needed:
scala> val res = "a,b,this is a test" split ',' take 2
res: Array[String] = Array(a, b)
2.Use Pattern matching to set the variables:
scala> val Array(x,y) = res
x: String = a
y: String = b*

Answer 2

Another solution using Sequence Pattern match in Scalaenter link description here
Welcome to Scala version 2.11.2 (OpenJDK 64-Bit Server VM, Java 1.7.0_65).
Type in expressions to have them evaluated.
Type :help for more information.
scala> val str = "a,b,this is a test"
str: String = a,b,this is a test
scala> val Array(x, y, _*) = str.split(",")
x: String = a
y: String = b
scala> println(s"x = $x, y = $y")
x = a, y = b

Answer 3

Are you looking for the method split ?
"a,b,this is a test".split(',')
res0: Array[String] = Array(a, b, this is a test)
If you want only the first two values you'll need to do something like:
val splitted = "a,b,this is a test".split(',')
val (first, second) = (splitted(0), splitted(1))

Answer 4

There should be some regex options here.
scala> val s = "a,b,this is a test"
s: String = a,b,this is a test
scala> val r = "[^,]+".r
r: scala.util.matching.Regex = [^,]+
scala> r findAllIn s
res0: scala.util.matching.Regex.MatchIterator = non-empty iterator
scala> .toList
res1: List[String] = List(a, b, this is a test)
scala> .take(2)
res2: List[String] = List(a, b)
scala> val a :: b :: _ = res2
a: String = a
b: String = b
but
scala> val a :: b :: _ = (r findAllIn "a" take 2).toList
scala.MatchError: List(a) (of class scala.collection.immutable.$colon$colon)
... 33 elided
or if you're not sure there is a second item, for instance:
scala> val r2 = "([^,]+)(?:,([^,]*))?".r.unanchored
r2: scala.util.matching.UnanchoredRegex = ([^,]+)(?:,([^,]*))?
scala> val (a,b) = "a" match { case r2(x,y) => (x, Option(y)) }
a: String = a
b: Option[String] = None
scala> val (a,b) = s match { case r2(x,y) => (x, Option(y)) }
a: String = a
b: Option[String] = Some(b)
This is a bit nicer if records are long strings.
Footnote: the Option cases look nicer with a regex interpolator.

Answer 5

If your string is short, you may as well just use String.split and take the first two elements.
val myString = "a,b,this is a test"
val splitString = myString.split(',') // Scala adds a split-by-character method in addition to Java's split-by-regex
val a = splitString(0)
val b = splitString(1)
Another solution would be to use a regex to extract the first two elements. I think it's quite elegant.
val myString = "a,b,this is a test"
val regex = """(.*),(.*),.*""".r // all groups (in parenthesis) will be extracted.
val regex(a, b) = myString // a="a", b="b"
Of course, you can tweak the regex to only allow non-empty tokens (or anything else you might need to validate) :
val regex = """(.+),(.+),.+""".r
Note that in my examples I assumed that the string always had at least two tokens. In the first example, you can test the length of the array if needed. The second one will throw a MatchError if the regex doesn't match the string.
I had originally proposed the following solution. I will leave it because it works and doesn't use any class formally marked as deprecated, but the Javadoc for StringTokenizer mentions that it is a legacy class and should no longer be used.
val myString = "a,b,this is a test"
val st = new StringTokenizer(",");
val a = st.nextToken()
val b = st.nextToken()
// You could keep calling st.nextToken(), as long as st.hasMoreTokens is true