Question

我试图找到地图和数组中的常见字符串以输出相应的值 s（来自地图，这里的值是Map [key - ＆gt; value ]）在Scala中，我试图不使用任何循环。示例：

输入：

Array("Ash","Garcia","Mac") Map("Ash" -> 5, "Mac" -> 4, "Lucas" -> 3)

输出：

Array(5,4)

输出是一个包含5和4的数组，因为Ash和Mac在两种数据结构中都很常见

Answer 1

什么构成循环？

def common(arr: Array[String], m: Map[String,Int]): Array[Int] =
  arr flatMap m.get

用法：

common(Array("Ash","Garcia","Mac")
      ,Map("Ash" -> 5, "Mac" -> 4, "Lucas" -> 3))
// res0: Array[Int] = Array(5, 4)

Answer 2

我认为这是最优雅的解决方案，但如果数组中有重复项，结果可能不符合您的要求。

yourArray.collect(yourMap) // Array(5,4)

Answer 3

scala优雅语法非常简单：

val a = Array("Ash","Garcia","Mac")
val m = Map("Ash" -> 5, "Mac" -> 4, "Lucas" -> 3)
println(m.filter { case (k, v) => a.contains(k)}.map { case (k, v) => v}.toArray)

以下是解决方案！

Answer 4

使用.filter仅查找匹配的条目，然后获取已过滤地图的值。

<强>鉴于

scala> val names = Array("Ash","Garcia","Mac")
names: Array[String] = Array(Ash, Garcia, Mac)

scala> val nameToNumber = Map("Ash" -> 5, "Mac" -> 4, "Lucas" -> 3)
nameToNumber: scala.collection.immutable.Map[String,Int] = Map(Ash -> 5, Mac -> 4, Lucas -> 3)

<强> .filter.map

scala> nameToNumber.filter(x => names.contains(x._1)).map(_._2)
res3: scala.collection.immutable.Iterable[Int] = List(5, 4)

或者，you can use collect，

scala> nameToNumber.collect{case kv if names.contains(kv._1) => kv._2}
res6: scala.collection.immutable.Iterable[Int] = List(5, 4)

此处的复杂性为O(n2)

Answer 5

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从地图类型和数组Scala中查找没有循环的公共字符串

5 个答案: