问题
First, in order to provide full disclosure, I want to point out that this is related to homework in a Machine Learning class. This question is not the homework question and instead is something I need to figure out in order to complete the bigger problem of creating an ID3 Decision Tree Algorithm.
I need to generate tree similar to the following when given a truth table
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 is of type BinaryTree which I've defined as follows:
type BinaryTree =
| Leaf of int
| Node of int * string * BinaryTree * BinaryTree
ID3 algorithms take into account various equations to determine where to split the tree, and I've got all that figured out, I'm just having trouble creating the learned tree from my truth table. For example if I have the following table
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
And I decide to split on attribute A1 I would end up with the following:
(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
Then I would split the left side and split the right side, and continue the recursive pattern until the leaf nodes are pure and I end up with a tree similar to the following based on the splitting.
let learnedTree = Node(0,"A0", Node(2,"A2", Leaf(0), Leaf(1)), Node(1,"A1", Node(2,"A2", Leaf(0), Leaf(1)), Leaf(0)))
Here is what I've kind of "hacked" together thus far, but I think I might be way off:
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"
Should I be using pattern matching instead? Any tips/ideas/help? Thanks a bunch!
回答1:
Edit: I've since posted an implementation of ID3 on my blog at: http://blogs.msdn.com/chrsmith
Hey Jim, I've been wanting to write a blog post implementing ID3 in F# for a while - thanks for giving me an execute. While this code doesn't implement the algorithm full (or correctly), it should be sufficient for getting you started.
In general you have the right approach - representing each branch as a discriminated union case is good. And like Brian said, List.partition is definitely a handy function. The trick to making this work correctly is all in determining the optimal attribute/value pair to split on - and to do that you'll need to calculate information gain via entropy, etc.
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
回答2:
You can use List.partition instead of your two List.choose calls.
http://research.microsoft.com/en-us/um/cambridge/projects/fsharp/manual/FSharp.Core/Microsoft.FSharp.Collections.List.html
(or now http://msdn.microsoft.com/en-us/library/ee353738(VS.100).aspx )
It isn't clear to me that pattern matching will buy you much here; the input type (list of lists) and processing (partitioning and 'pureness' check) doesn't really lend itself to that.
And of course when you finally get the 'end' (a pure list) you need to create a tree, and then presumably this function will create a Leaf when the input only has one 'side' and it's 'pure', but create a Node out of the left-side and right-side results for every other input. Maybe. I didn't quite grok the algorithm completely.
Hopefully that will help steer you a little bit. May be useful to draw up a few smaller sample inputs and outputs to help work out the various cases of the function body.
回答3:
Thanks Brian & Chris! I was actually able to figure this out and I ended up with the following. This calculates the information gain for determining the best place to split. I'm sure there are probably better ways for me to arrive at this solution especially around the chosen data structures, but this is a start. I plan to refine things later.
#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)"
来源：https://stackoverflow.com/questions/1418829/help-needed-creating-a-binary-tree-given-truth-table