Question

我有一个选项列表，我想不跳过任何值并返回第一个。

val list = List(None, None, None, Some(3));

list.dropWhile(!_.exists).take(1) // return 3

实际上是这样的：

val list = List(1, 2, 3, 4)
def solve(value: Int): Option[Int] = ???
list.dropWhile(!solve(_).isDefined).take(1) // returns the result of solve method

我需要为每个None结果调用一次solve函数，为最后一个Some结果调用一次。

Answer 1

val list = List(None, None, None, Some(3));
list.flatten.headOption

回答第二个问题：

val list = List(1, 2, 3, 4)
def solve(value: Int): Option[Int] = ???
list.toStream.flatMap(solve).headOption // or call .head if you know what you're doing

Answer 2

像这样：

def solve(hand: Hand): HandValue = all.foldLeft[Option[HandValue]](None) {
  case (None, solver) => solver.solve(hand)
  case (Some(value), _) => Some(value)
} get

Answer 3

def takeFirst[T](list : List[Option[T]]): Option[T] = list.find(_.isDefined).flatten

如何跳过选项值并首先返回一些

3 个答案: