Question

我知道这是一个冗长的问题：）我正在尝试在Scala 2.11中的数据集上实现哈密顿循环，作为此过程的一部分，我试图从值映射表生成邻接矩阵。

说明：

键0到4是不同的城市，因此在“ allRoads”变量下面

0 -> Set(1, 2) Means city0 is connected to city1 and city2
1 -> Set(0, 2, 3, 4) Means City1 is connected to city0,city2,city3,city4
.
.

我需要生成adj矩阵，例如：如果城市连接，我需要生成1，否则我必须生成0，这意味着

for: "0 -> Set(1, 2)", I need to generate: Map(0 -> Array(0,1,1,0,0))

输入-

var allRoads = Map(0 -> Set(1, 2), 1 -> Set(0, 2, 3, 4), 2 -> Set(0, 1, 3, 4), 3 -> Set(2, 4, 1), 4 -> Set(2, 3, 1))

我的代码：

val n: Int = 5
val listOfCities = (0 to n-1).toList
var allRoads = Map(0 -> Set(1, 2), 1 -> Set(0, 2, 3, 4), 2 -> Set(0, 1, 3, 4), 3 -> Set(2, 4, 1), 4 -> Set(2, 3, 1))
var adjmat:Array[Int] = Map()

  for( i <- 0 until allRoads.size;j <- listOfCities) {
    allRoads.get(i) match {
      case Some(elem) => if (elem.contains(j)) adjmat = adjmat:+1 else adjmat = adjmat :+0
      case _ => None
    }
  }

输出：

output: Array[Int] = Array(0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0)

预期的输出-像这样，请建议是否有更好的方法可以生成哈密顿量的输入

Map(0 -> Array(0, 1, 1, 0, 0),1 -> Array(1, 0, 1, 1, 1),2 -> Array(1, 1, 0, 1, 1),3 -> Array(0, 1, 1, 0, 1),4 -> Array(0, 1, 1, 1, 0))

不确定如何将以上输出存储为Map或Plain 2D Array。

Answer 1

尝试

val cities = listOfCities.toSet
allRoads.map { case (city, roads) =>
  city -> listOfCities.map(city => if ((cities diff roads).contains(city)) 0 else 1)
}

输出

Map(0 -> List(0, 1, 1, 0, 0), 1 -> List(1, 0, 1, 1, 1), 2 -> List(1, 1, 0, 1, 1), 3 -> List(0, 1, 1, 0, 1), 4 -> List(0, 1, 1, 1, 0))

从地图生成邻接矩阵

1 个答案: