Question

我有一个三级数据结构（缩进和换行符以便于阅读）：

scala> import scala.collection.mutable.Map
import scala.collection.mutable.Map

scala> val m = Map("normal" -> Map("home" -> Map("wins" -> 0, "scores" -> 0),
                                   "away" -> Map("wins" -> 0, "scores" -> 0)))
m: scala.collection.mutable.Map[java.lang.String,
   scala.collection.mutable.Map[java.lang.String,
   scala.collection.mutable.Map[java.lang.String,Int]]] = 
Map((normal,Map(away -> Map(wins -> 0, scores -> 0),
     home -> Map(wins -> 0, scores -> 0))))

访问最里面的数据（分数）需要大量输入：

import org.scalatest.{Assertions, FunSuite}

class MapExamplesSO extends FunSuite with Assertions {
  test("Update values in a mutable map of map of maps") {
    import scala.collection.mutable.Map
    // The m map is essentially an accumulator
    val m = Map("normal" -> 
                Map("home" -> Map("wins" -> 0, "scores" -> 0),
                    "away" -> Map("wins" -> 0, "scores" -> 0)
                  )
          )
    //
    // Is there a less verbose way to increment the scores ?
    //
    assert(m("normal").apply("home").apply("scores") === 0)

    val s1 = m("normal").apply("home").apply("scores") + 1
    m("normal").apply("home").update("scores", s1)

    assert(m("normal").apply("home").apply("scores") === 1)

    val s2 = m("normal").apply("home").apply("scores") + 2
    m("normal").apply("home").update("scores", s2)

    assert(m("normal").apply("home").apply("scores") === 3)
  }
}

是否有一种不那么冗长的方法来修改分数值？

我是Scala新手，所以对上述代码的所有其他观察结果也是受欢迎的。

Answer 1

您不必使用“apply”，只需使用“（）”

即可

m("normal")("home")("scores") = 1

Answer 2

你可以写

m("normal").apply("home").apply("scores")

作为

m("normal")("home")("scores")

但是我不确定这样的结构是不是一个好主意。也许您应该考虑将此功能封装在专门的类中。

Answer 3

添加本地帮助函数始终是减少代码重复的好方法：

class MapExamplesSO {
  def test {
    import scala.collection.mutable.Map
    // The m map is essentially an accumulator
    var m = Map("normal" -> 
                Map("home" -> Map("wins" -> 0, "scores" -> 0),
                    "away" -> Map("wins" -> 0, "scores" -> 0)))


    //Local Helper returns (Old, New)
    def updateScore(k1 : String,k2 : String,k3 : String)
                   (f : Int => Int) : (Int, Int) = {
      val old = m(k1)(k2)(k3)
      m(k1)(k2)(k3) = f(old)
      (old, m(k1)(k2)(k3))
    }

    assert(m("normal")(home")("scores") === 0)
    assert(updateScore("normal","home","scores")(_+1)._2 === 1)
    assert(updateScore("normal","home","scores")(_+2)._2 === 3)
  }
}

[编辑代码更紧]

Answer 4

少详细：

assert(m("normal")("home")("scores") === 0)

val s1 = m("normal")("home")("scores") + 1
m("normal")("home")("scores") = s1

assert(m("normal")("home")("scores") === 1)

val s2 = m("normal")("home")("scores") + 2
m("normal")("home")("scores") = s2

assert(m("normal")("home")("scores") === 3)

如上所述，这利用了apply和update都具有语法糖的事实。更短的时间：

// On REPL, put both these definitions inside an object instead
// of entering them on different lines
def scores = m("normal")("home")("scores")
def scores_=(n: Int) = m("normal")("home")("scores") = n

assert(scores === 0)

val s1 = scores + 1
scores = s1

assert(scores === 1)

val s2 = scores + 2
scores = s2

// Just so you see these updates are being made to the map:
assert(m("normal")("home")("scores") === 3)

利用getter和setter的语法糖（getter定义必须才能使setter定义生效）。

如何访问和更新地图地图的可变地图中的值

4 个答案: