Scala 在列表中搜索以下项以特定方式关联的项
创建功能Scala 在列表中搜索以下项以特定方式关联的项,scala,Scala,创建功能associate（fun，list），其工作方式如下： 例如： def square(x:Int) = x*x val list = List(2,4,16,,5,10,100,105) associate(list,square) Result: List((2,4),(4,16),(10,100)) 我建议： def square(x:Int) = x*x def associate[A, B](fun: A => B, list: List[A])
associate（fun，list） def square(x:Int) = x*x
val list = List(2,4,16,,5,10,100,105)
associate(list,square)
Result: List((2,4),(4,16),(10,100))
def square(x:Int) = x*x
def associate[A, B](fun: A => B, list: List[A]): List[(A, B)] = {
val distinct: List[A] = list.distinct
distinct.zip(distinct.map(fun).foldLeft(List.empty[Option[B]])((maybeOuts, out) => {
maybeOuts :+ distinct.collectFirst { case `out` => out }
}))
.collect { case (in, Some(out)) => in -> out }
}
println(associate(square, List(2, 4, 16, 5, 10, 100, 105)))
println(associate((_: Double).toInt, List[Double](2, 4, 16, 5, 10, 100, 105)))
println(associate(identity[Int], List(2, 4, 16, 5, 10, 100, 105)))
println(associate[Int, Double](Math.pow(_, 2), List(2, 4, 16, 5, 10, 100, 105)))
def square(x:Int) = x*x
def associate[A, B](fun: A => B, list: List[A]): List[(A, B)] = {
val distinct: List[A] = list.distinct
distinct.zip(distinct.map(fun).foldLeft(List.empty[Option[B]])((maybeOuts, out) => {
maybeOuts :+ distinct.collectFirst { case `out` => out }
}))
.collect { case (in, Some(out)) => in -> out }
}
println(associate(square, List(2, 4, 16, 5, 10, 100, 105)))
println(associate((_: Double).toInt, List[Double](2, 4, 16, 5, 10, 100, 105)))
println(associate(identity[Int], List(2, 4, 16, 5, 10, 100, 105)))
println(associate[Int, Double](Math.pow(_, 2), List(2, 4, 16, 5, 10, 100, 105)))
希望有帮助。@LuisMiguelMejíaSuárez不。他们不需要
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