CODEWITHSUNDEEP
scaladictionaryrecurrence
GSerum_
GSerum_

Reputation: 75

Scala check value inside mapping

Allright so i don't know if this is possible, but let's say we have the following list:

List(1, 2, 3, 1)

If i want to apply a map over this, is there a way for me to check if i've already had a value before, e.g. on the 4th value (the 2nd 1) it'll say that it already came across the 1 and then throw an error or something.

Upvotes: 3

Views: 211

Answers (3)

user unknown
user unknown

Reputation: 36229

List(1, 2, 3, 1).distinct.map (n => n*n) 
// res163: List[Int] = List(1, 4, 9)

This code removes duplicates, then performs the mappings in a self documenting, brief manner.

Upvotes: 2

jwvh
jwvh

Reputation: 51271

fold is probably the way to go. The problem is that each iteration has to carry both the memory of previous elements as well as the map() results as it is being built.

List(1, 2, 3, 11).foldRight((Set[Int](),List[String]())) {case (i, (st, lst)) =>
  if (st(i)) throw new Error        //duplicate encountered
  else (st + i, i.toString :: lst)  //add to memory and map result
}._2                                //pull the map result from the tuple

Upvotes: 1

Xavier Guihot
Xavier Guihot

Reputation: 61656

This would be the role of a foldLeft stage:

List(1, 2, 3, 1).foldLeft(List[Int]()) {
  // The item has already been encountered:
  case (uniqueItems, b) if uniqueItems.contains(b) => {
    // If as stated, you want to throw an exception, that's where you could do it
    uniqueItems
  }
  // New item not seen yet:
  case (uniqueItems, b) => uniqueItems :+ b
}

foldLeft traverses a sequence while working (at each new element), with a result based on the previous ones.

For each element, the pattern matching (uniqueItems, b) should be understood this way: uniqueItems is the "accumulator" (it's initialized as List[Int]()) and will be updated (or not) for each item of the list. And b if the new item of the list which is currently being processed.

By the way, this example is a (non-efficient) distinct over a list.

Upvotes: 4

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