I have this List in Scala:
List[String] = List([[aaa|bbb]], [[ccc|ddd]], [[ooo|sss]])
And I want to obtain the same List with the substrings between | and ] removed and | removed too.
So the result would be:
List[String] = List([[aaa]], [[ccc]], [[ooo]])
I tried something making a String with the List and using replaceAll, but I want to conserve the List.
Thanks.
Here is a simple solution that should be quite good in performance:
val list = List("[[aaa|bbb]]", "[[ccc|ddd]]", "[[ooo|sss]]")
list.map(str => str.takeWhile(_ != '|') + "]]" )
It assumes that the format of the strings is:
[ at the beginning,|. 
You can use a simple \|.*?]] regex to match these substrings you need to remove.
Here is a way to perform the replacement in Scala code:
val l = List[String]("[[aaa|bbb]]", "[[ccc|ddd]]", "[[ooo|sss]]")
println(l.map(x => x.replaceAll("""\|.*?(]])""", "$1")))
See the Scala demo
I added a capturing group around ]] and used a $1 backreference in the replacement pattern to insert the ]] back into the result.
Details:
\| - a literal | pi[e symbol (since it is a special char outide of a character class, it must be escaped).*? - any zero or more symbols other than line break symbols(]]) - Group 1 capturing ]] substring (note that ] outside of a character class does not need escaping, it is just the opposite of the case with |). 
replaceAll("""\|[^\]]+""", ""), and it's still accurate if the level of [] nesting changes.]]. It is just a way to show what regex can do."[aa|bb]" or maybe "[[[x|y]]]" then your solution (with or without the capture group) won't balance each [ with a closing ].[[ + some chars other than |, then |, then some chars up to the first ]]. If balanced brackets had been mentioned, I would have never suggested a Java regex solution unless the recursion depth were just 1 or 2 levels.Replace the 3 characters between | and } with ].
regex is "\\|(.{3})\\]" (do not forget to escape | and })
scala> val list = List("[[aaa|bbb]]", "[[ccc|ddd]]", "[[ooo|sss]]")
list: List[String] = List([[aaa|bbb]], [[ccc|ddd]], [[ooo|sss]])
scala> list.map(_.replaceAll("\\|(.{3})\\]", "]"))
res16: List[String] = List([[aaa]], [[ccc]], [[ooo]])
