我正在尝试通过地图将 Map(a" -> 2, b" -> 1) 变成 seq(a", a", b")函数,目前我正在尝试运行下面的代码,给我想要的结果.
I am trying to turn a Map("a" -> 2, "b" -> 1) into seq("a","a","b") through the map function, Currently I am trying to run the code below giving me the desired result.
有没有更聪明的方法来做到这一点?可能是通过地图功能的更好方法?
Is there a smarter way to do this? Possibly a better way through the map function?
var multiset : Seq[T] = Seq[T]() var variables : Seq[T] = data.map(x => x._1).toSeq var variableCounts : Seq[Int] = data.map(x => x._2).toSeq for(x <- 0 until variables.length){ for(y <- 0 until variableCounts(x)) multiset = multiset :+ variables(x) } 推荐答案你可以这样做:使用 GenTraversableFactory 的填充方法 def fill[A](n: Int)(elem: => A): CC[A] 从填充的定义我们可以看到它需要一个整数和一个元素.整数告诉我们需要填充给定元素的次数.
you can do something like this: Use fill method of GenTraversableFactory def fill[A](n: Int)(elem: => A): CC[A] from the definition of fill we can see that it takes an integer and an element. Integer tell how many times we need to fill the given element.
object Demo extends App { val x = Map("a" -> 2, "b" -> 1) val p: Seq[String] = x.flatMap { tuple => List.fill(tuple._2)(tuple._1) }.toSeq print(p) //output: List(a, a, b) }希望对你有帮助!!!
如果要避免使用 tuple._1 和 tuple._1 可以使用以下方法.
If you want to avoid tuple._1 and tuple._1 you can use the following approach.
object Demo extends App { val x = Map("a" -> 2, "b" -> 1) val p: Seq[String] = x.flatMap { case (key, value) => List.fill(value)(key) }.toSeq print(p) //output: List(a, a, b) }更多推荐
使用 map 将 Map(“a"
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