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私のクエリは、ファイルから入力を読み取り、scala を使用してファイルのデータ行を List[Map[Int,String]] に変換することです。ここでは、入力としてデータセットを指定します。私のコードは、

  def id3(attrs: Attributes,
      examples: List[Example],
      label: Symbol
       ) : Node = {
level = level+1


  // if all the examples have the same label, return a new node with that label

  if(examples.forall( x => x(label) == examples(0)(label))){
  new Leaf(examples(0)(label))
  } else {
  for(a <- attrs.keySet-label){          //except label, take all attrs
    ("Information gain for %s is %f".format(a,
      informationGain(a,attrs,examples,label)))
  }


  // find the best splitting attribute - this is an argmax on a function over the list

  var bestAttr:Symbol = argmax(attrs.keySet-label, (x:Symbol) =>
    informationGain(x,attrs,examples,label))




  // now we produce a new branch, which splits on that node, and recurse down the nodes.

  var branch = new Branch(bestAttr)

  for(v <- attrs(bestAttr)){


    val subset = examples.filter(x=> x(bestAttr)==v)



    if(subset.size == 0){
      // println(levstr+"Tiny subset!")
      // zero subset, we replace with a leaf labelled with the most common label in
      // the examples
      val m = examples.map(_(label))
      val mostCommonLabel = m.toSet.map((x:Symbol) => (x,m.count(_==x))).maxBy(_._2)._1
      branch.add(v,new Leaf(mostCommonLabel))

    }
    else {
      // println(levstr+"Branch on %s=%s!".format(bestAttr,v))

      branch.add(v,id3(attrs,subset,label))
    }
   }
  level = level-1
  branch
  }
  }
  }
object samplet {
def main(args: Array[String]){

var attrs: sample.Attributes = Map()
attrs += ('0 -> Set('abc,'nbv,'zxc))
attrs += ('1 -> Set('def,'ftr,'tyh))
attrs += ('2 -> Set('ghi,'azxc))
attrs += ('3 -> Set('jkl,'fds))
attrs += ('4 -> Set('mno,'nbh))



val examples: List[sample.Example] = List(
  Map(
    '0 -> 'abc,
    '1 -> 'def,
    '2 -> 'ghi,
    '3 'jkl,
    '4 -> 'mno
  ),
  ........................
  )


// obviously we can't use the label as an attribute, that would be silly!
val label = 'play

println(sample.try(attrs,examples,label).getStr(0))

}
}

しかし、このコードを.csvファイルからの入力を受け入れるように変更するにはどうすればよいですか?

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2 に答える 2