需要帮助创建一个给定真值的二叉树

Question

首先，为了提供完整的信息，我想指出这与机器学习课程中的作业有关。这个问题不是作业问题，而是我需要弄清楚的，以便完成创建ID3决策树算法的更大问题。

我需要生成类似于树给定的真值表

let learnedTree = Node(0,"A0", Node(2,"A2", Leaf(0), Leaf(1)), Node(1,"A1", Node(2,"A2", Leaf(0), Leaf(1)), Leaf(0))) 

learnedTree是我定义如下类型二叉树的时候以下几点：

type BinaryTree = 
    | Leaf of int 
    | Node of int * string * BinaryTree * BinaryTree 

ID3算法考虑到各种方程来确定在哪里拆分树，我已经搞清楚了，我只是无法从我的真值表中创建学习树。例如，如果我有以下表

A1 | A2 | A3 | Class 
1  0 0  1 
0  1 0  1 
0  0 0  0 
1  0 1  0 
0  0 0  0 
1  1 0  1 
0  1 1  0 

我决定拆就属性A1我将结束与以下：

   (A1 = 1) A1 (A1 = 0) 
    A2 | A3 | Class    A2 | A3 | Class 
    0  0  1    1  0  1 
    0  1  0    0  0  0 
    1  0  1    0  0  0 
           0  1  1 

然后我会分裂左侧和分裂右侧，并继续递归模式，直到叶节点是纯粹的，我最终得到一棵类似于以下的基于分裂的树。

let learnedTree = Node(0,"A0", Node(2,"A2", Leaf(0), Leaf(1)), Node(1,"A1", Node(2,"A2", Leaf(0), Leaf(1)), Leaf(0))) 

这里就是我这样的“黑客”一起迄今为止，但我想我可能是遥远：

let rec createTree (listToSplit : list<list<float>>) index = 
    let leftSideSplit = 
     listToSplit |> List.choose (fun x -> if x.Item(index) = 1. then Some(x) else None) 
    let rightSideSplit = 
     listToSplit |> List.choose (fun x -> if x.Item(index) = 0. then Some(x) else None) 
    if leftSideSplit.Length > 0 then 
     let pureCheck = isListPure leftSideSplit 
     if pureCheck = 0 then 
      printfn "%s" "Pure left node class 0" 
      createTree leftSideSplit (index + 1) 
     else if pureCheck = 1 then 
      printfn "%s" "Pure left node class 1" 
      createTree leftSideSplit (index + 1) 
     else 
      printfn "%s - %A" "Recursing Left" leftSideSplit 
      createTree leftSideSplit (index + 1) 
    else printfn "%s" "Pure left node class 0" 

我应该使用模式，而不是匹配？任何提示/想法/帮助？谢谢一堆！

Answer 1

编辑：因为我已经在贴ID3的实现，在我的博客： http://blogs.msdn.com/chrsmith

嗨，吉姆，我一直想要写在博客上的F＃实现ID3一会儿 - 感谢给我一个执行。尽管这段代码没有实现完整（或正确）的算法，但它应该足以让你开始。

一般来说，你有正确的方法 - 代表每个分支作为歧视的工会案例是好的。就像Brian所说的，List.partition绝对是一个方便的功能。诀窍使这项工作正确的是所有在确定最佳的属性/值对分裂的 - 要做到这一点，你需要通过熵计算信息增益等

type Attribute = string 
type Value = string 

type Record = 
    { 
     Weather : string 
     Temperature : string 
     PlayTennis : bool 
    } 
    override this.ToString() = 
     sprintf 
      "{Weather = %s, Temp = %s, PlayTennis = %b}" 
      this.Weather 
      this.Temperature 
      this.PlayTennis 

type Decision = Attribute * Value 

type DecisionTreeNode = 
    | Branch of Decision * DecisionTreeNode * DecisionTreeNode 
    | Leaf of Record list 

// ------------------------------------ 

// Splits a record list into an optimal split and the left/right branches. 
// (This is where you use the entropy function to maxamize information gain.) 
// Record list -> Decision * Record list * Record list 
let bestSplit data = 
    // Just group by weather, then by temperature 
    let uniqueWeathers = 
     List.fold 
      (fun acc item -> Set.add item.Weather acc) 
      Set.empty 
      data 

    let uniqueTemperatures = 
     List.fold 
      (fun acc item -> Set.add item.Temperature acc) 
      Set.empty 
      data 

    if uniqueWeathers.Count = 1 then 
     let bestSplit = ("Temperature", uniqueTemperatures.MinimumElement) 
     let left, right = 
      List.partition 
       (fun item -> item.Temperature = uniqueTemperatures.MinimumElement) 
       data 
     (bestSplit, left, right) 
    else 
     let bestSplit = ("Weather", uniqueWeathers.MinimumElement) 
     let left, right = 
      List.partition 
       (fun item -> item.Weather = uniqueWeathers.MinimumElement) 
       data 
     (bestSplit, left, right) 

let rec determineBranch data = 
    if List.length data < 4 then 
     Leaf(data) 
    else 
     // Use the entropy function to break the dataset on 
     // the category/value that best splits the data 
     let bestDecision, leftBranch, rightBranch = bestSplit data 
     Branch(
      bestDecision, 
      determineBranch leftBranch, 
      determineBranch rightBranch) 

// ------------------------------------  

let rec printID3Result indent branch = 
    let padding = new System.String(' ', indent) 
    match branch with 
    | Leaf(data) -> 
     data |> List.iter (fun item -> printfn "%s%s" padding <| item.ToString()) 
    | Branch(decision, lhs, rhs) -> 
     printfn "%sBranch predicate [%A]" padding decision 
     printfn "%sWhere predicate is true:" padding 
     printID3Result (indent + 4) lhs 
     printfn "%sWhere predicate is false:" padding 
     printID3Result (indent + 4) rhs 


// ------------------------------------  

let dataset = 
    [ 
     { Weather = "windy"; Temperature = "hot"; PlayTennis = false } 
     { Weather = "windy"; Temperature = "cool"; PlayTennis = false } 
     { Weather = "nice"; Temperature = "cool"; PlayTennis = true } 
     { Weather = "nice"; Temperature = "cold"; PlayTennis = true } 
     { Weather = "humid"; Temperature = "hot"; PlayTennis = false } 
    ] 

printfn "Given input list:" 
dataset |> List.iter (printfn "%A") 

printfn "ID3 split resulted in:" 
let id3Result = determineBranch dataset 
printID3Result 0 id3Result 

Answer 2

您可以使用List.partition而不是两个List.choose调用。

http://research.microsoft.com/en-us/um/cambridge/projects/fsharp/manual/FSharp.Core/Microsoft.FSharp.Collections.List.html

（或现在http://msdn.microsoft.com/en-us/library/ee353738(VS.100).aspx）

目前尚不清楚对我来说，模式匹配会买你在这里多;输入类型（列表清单）和处理（分区和“纯粹”检查）并不适用于此。当你最终得到'end'（一个纯粹的列表）时，你需要创建一棵树，然后假设这个函数会在输入只有一个'side'并且它是'pure'时创建一个Leaf， ，但为其他输入的左侧和右侧结果创建一个节点。也许。我没有完全理解算法。

希望这会有助于引导你一点点。可能有助于制定一些较小的样本输入和输出来帮助解决功能体的各种情况。

Answer 3

感谢布赖恩&克里斯！我实际上能够弄清楚这一点，并最终得出以下结论。这将计算用于确定最佳拆分位置的信息增益。我相信对于我来说，可能有更好的方式来实现这个解决方案，特别是在选定的数据结构周围，但这是一个开始。我打算稍后再改进。

#light 
open System 

let trainList = 
    [ 
    [1.;0.;0.;1.;]; 
    [0.;1.;0.;1.;]; 
    [0.;0.;0.;0.;]; 
    [1.;0.;1.;0.;]; 
    [0.;0.;0.;0.;]; 
    [1.;1.;0.;1.;]; 
    [0.;1.;1.;0.;]; 
    [1.;0.;0.;1.;]; 
    [0.;0.;0.;0.;]; 
    [1.;0.;0.;1.;]; 
    ] 

type BinaryTree = 
    | Leaf of int 
    | Node of int * string * BinaryTree * BinaryTree 

let entropyList nums = 
    let sumOfnums = 
     nums 
     |> Seq.sum 
    nums 
    |> Seq.map (fun x -> if x=0.00 then x else (-((x/sumOfnums) * Math.Log(x/sumOfnums, 2.)))) 
    |> Seq.sum 

let entropyBinaryList (dataListOfLists:list<list<float>>) = 
    let classList = 
     dataListOfLists 
     |> List.map (fun x -> x.Item(x.Length - 1)) 
    let ListOfNo = 
     classList 
     |> List.choose (fun x -> if x = 0. then Some(x) else None) 
    let ListOfYes = 
     classList 
     |> List.choose (fun x -> if x = 1. then Some(x) else None) 
    let numberOfYes : float = float ListOfYes.Length 
    let numberOfNo : float = float ListOfNo.Length 
    let ListOfNumYesAndSumNo = [numberOfYes; numberOfNo] 
    entropyList ListOfNumYesAndSumNo 

let conditionalEntropy (dataListOfLists:list<list<float>>) attributeNumber = 
    let NoAttributeList = 
     dataListOfLists 
     |> List.choose (fun x -> if x.Item(attributeNumber) = 0. then Some(x) else None) 
    let YesAttributeList = 
     dataListOfLists 
     |> List.choose (fun x -> if x.Item(attributeNumber) = 1. then Some(x) else None) 
    let numberOfYes : float = float YesAttributeList.Length 
    let numberOfNo : float = float NoAttributeList.Length 
    let noConditionalEntropy = (entropyBinaryList NoAttributeList) * (numberOfNo/(numberOfNo + numberOfYes)) 
    let yesConditionalEntropy = (entropyBinaryList YesAttributeList) * (numberOfYes/(numberOfNo + numberOfYes)) 
    [noConditionalEntropy; yesConditionalEntropy] 

let findBestSplitIndex(listOfInstances : list<list<float>>) = 
    let IGList = 
     [0..(listOfInstances.Item(0).Length - 2)] 
     |> List.mapi (fun i x -> (i, (entropyBinaryList listOfInstances) - (List.sum (conditionalEntropy listOfInstances x)))) 
    IGList 
    |> List.maxBy snd 
    |> fst 

let isListPure (listToCheck : list<list<float>>) = 
    let splitList = listToCheck |> List.choose (fun x -> if x.Item(x.Length - 1) = 1. then Some(x) else None) 
    if splitList.Length = listToCheck.Length then 1 
    else if splitList.Length = 0 then 0 
    else -1 

let rec createTree (listToSplit : list<list<float>>) = 
     let pureCheck = isListPure listToSplit 
     if pureCheck = 0 then 
      printfn "%s" "Pure - Leaf(0)" 
     else if pureCheck = 1 then 
      printfn "%s" "Pure - Leaf(1)" 
     else 
      printfn "%A - is not pure" listToSplit 
      if listToSplit.Length > 1 then // There are attributes we can split on 
       // Chose best place to split list 
       let splitIndex = findBestSplitIndex(listToSplit) 
       printfn "spliting at index %A" splitIndex 
       let leftSideSplit = 
        listToSplit |> List.choose (fun x -> if x.Item(splitIndex) = 1. then Some(x) else None) 
       let rightSideSplit = 
        listToSplit |> List.choose (fun x -> if x.Item(splitIndex) = 0. then Some(x) else None) 
       createTree leftSideSplit 
       createTree rightSideSplit 
      else 
       printfn "%s" "Not Pure, but can't split choose based on heuristics - Leaf(0 or 1)"