scala: How to handle Option in a functional way -

i new scala question stupid. if have existing method below. adding these 4 lines method. there better way handle option value?

def processdata(input: string, datamap: map[string, string]): option[string] = {     //4 lines adding.    val data: option[string] = datamap.get(input)    if (data.isempty) {      return none    }     //how avoid line    val datavalue = data.get     //20-25 lines of code in here bunch of pattern matching case statements    cleandata(datavalue)    dosomethingelse("apple", datavalue, "test") } 

essentially want avoid having "data.get" in below code. somehow feels wrong call that. wrote differently using pattern matching below. 20-25 lines of code have bunch of case statements , dont want create layer on top of them.

def processdata(input: string, datamap: map[string, string]): option[string] = {   datamap.get(input) match {     case some(datavalue) => {    //20-25 lines of code in here bunch of pattern matching case statements     cleandata(datavalue)     dosomethingelse("apple", datavalue, "test")   }    case none => none  } 

}

any ideas?

actually second way of functional style, conciseness can use 1 of option higher order functions:

  def processdata(input: string, datamap: map[string, string]): option[string] =      datamap.get(input).map { datavalue =>       cleandata(datavalue)       dosomethingelse("apple", datavalue, "test")     } 

you can avoid dots , parentheses :

  def processdata(input: string, datamap: map[string, string]): option[string] =      datamap input map { datavalue =>       cleandata(datavalue)       dosomethingelse("apple", datavalue, "test")     }