scala return on first Some in list

Question

I have a list l:List[T1] and currently im doing the following:

myfun : T1 -> Option[T2]
val x: Option[T2] = l.map{ myfun(l) }.flatten.find(_=>true)

The myfun function returns None or Some, flatten throws away all the None's and find returns the first element of the list if any.

This seems a bit hacky to me. Im thinking that there might exist some for-comprehension or similar that will do this a bit less wasteful or more clever. For example: I dont need any subsequent answers if myfun returns any Some during the map of the list l.

Answer 1

How about:

l.toStream flatMap (myfun andThen (_.toList)) headOption

Stream is lazy, so it won't map everything in advance, but it won't remap things either. Instead of flattening things, convert Option to List so that flatMap can be used.

Answer 2

In addition to using toStream to make the search lazy, we can use Stream::collectFirst:

List(1, 2, 3, 4, 5, 6, 7, 8).toStream.map(myfun).collectFirst { case Some(d) => d }
// Option[String] = Some(hello)
// given def function(i: Int): Option[String] = if (i == 5) Some("hello") else None

This:

• Transforms the List into a Stream in order to stop the search early.

• Transforms elements using myFun as Option[T]s.

• Collects the first mapped element which is not None and extract it.

Starting Scala 2.13, with the deprecation of Streams in favor of LazyLists, this would become:

List(1, 2, 3, 4, 5, 6, 7, 8).to(LazyList).map(function).collectFirst { case Some(d) => d }