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機器根據數據集創建規則，就是機器學習。

決策樹：從數據集合中提取一系列規則，適用於探索式的知識發現。

決策樹本質：通過一系列規則對數據進行分類的過程。

決策樹算法核心：構建精度高，數據規模小的決策樹。

ID3算法：此算法目的在於減少樹的深度，但是忽略了葉子數目的研究。

C4.5算法：對ID3進行改進，對於預測變量的缺失值處理、剪枝技術、派生規則等方面作了較大改進，既適合分類，又適合回歸。

香農熵：變量的不確定性越大，熵也就越大，把它搞清楚所需要的信息量也就越大。

基尼不純度：一個隨機事件變成它的對立事件的概率。

#coding:utf-8
from math import log
import operator

#創建數據
def createDataSet():
    dataSet = [[1, 1, ‘yes‘],
               [1, 1, ‘yes‘],
               [1, 0, ‘no‘],
               [0, 1, ‘no‘],
               [0, 1, ‘no‘]]
    labels = [‘no surfacing‘,‘flippers‘]
    #change to discrete values
    return dataSet, labels

#計算給定數據集的香農熵
def calcShannonEnt(dataSet):
    numEntries = len(dataSet)
	#輸出列表的的長度 值為：5
    labelCounts = {}
    for featVec in dataSet: 
	#對二維數組進行遍歷
        currentLabel = featVec[-1]
        #把每個子列表的最後一個元素賦值給currentLabel
        if currentLabel not in labelCounts.keys(): labelCounts[currentLabel] = 0
        #如果currentLabel 不在字典中，把字典的值設置為0
        # keys() 函數以列表返回一個字典所有的鍵。
        labelCounts[currentLabel] += 1
        #這一步是統計出現的次數，每次匹配到的話，把對應key的值加1
    #操作完以後labelCounts字典記錄的是yes出現2次  no出現3次
    shannonEnt = 0.0
    for key in labelCounts:
        prob = float(labelCounts[key])/numEntries
        #prob是字典labelCounts每個key的比率
        shannonEnt -= prob * log(prob,2) 
		#log(x, 2) 表示以2為底的對數。
    return shannonEnt
	#熵數字越大，說明混合的數據越多


#按照給定特征劃分數據集    
def splitDataSet(dataSet, axis, value):
	#dataSet:待劃分的數據集；axis：劃分數據集的特征；value：需要返回的特征的值
    retDataSet = []
    for featVec in dataSet:
        if featVec[axis] == value:
            #如果featVec列表的特征值跟valve值一樣
            reducedFeatVec = featVec[:axis] 
            #輸出列表featVec開頭到特征值的列表切片    
            reducedFeatVec.extend(featVec[axis+1:])
            retDataSet.append(reducedFeatVec)
    return retDataSet

#選擇最好的數據集劃分方式    
def chooseBestFeatureToSplit(dataSet):
    numFeatures = len(dataSet[0]) - 1      #the last column is used for the labels
    baseEntropy = calcShannonEnt(dataSet)
    bestInfoGain = 0.0; bestFeature = -1
    for i in range(numFeatures):        #iterate over all the features
        featList = [example[i] for example in dataSet]#create a list of all the examples of this feature
        uniqueVals = set(featList)       #get a set of unique values
        newEntropy = 0.0
        for value in uniqueVals:
            subDataSet = splitDataSet(dataSet, i, value)
            prob = len(subDataSet)/float(len(dataSet))
            newEntropy += prob * calcShannonEnt(subDataSet)     
        infoGain = baseEntropy - newEntropy     #calculate the info gain; ie reduction in entropy
        if (infoGain > bestInfoGain):       #compare this to the best gain so far
            bestInfoGain = infoGain         #if better than current best, set to best
            bestFeature = i
    return bestFeature                      #returns an integer

def majorityCnt(classList):
    classCount={}
    for vote in classList:
        if vote not in classCount.keys(): classCount[vote] = 0
        classCount[vote] += 1
    sortedClassCount = sorted(classCount.iteritems(), key=operator.itemgetter(1), reverse=True)
    return sortedClassCount[0][0]

def createTree(dataSet,labels):
    classList = [example[-1] for example in dataSet]
    if classList.count(classList[0]) == len(classList): 
        return classList[0]#stop splitting when all of the classes are equal
    if len(dataSet[0]) == 1: #stop splitting when there are no more features in dataSet
        return majorityCnt(classList)
    bestFeat = chooseBestFeatureToSplit(dataSet)
    bestFeatLabel = labels[bestFeat]
    myTree = {bestFeatLabel:{}}
    del(labels[bestFeat])
    featValues = [example[bestFeat] for example in dataSet]
    uniqueVals = set(featValues)
    for value in uniqueVals:
        subLabels = labels[:]       #copy all of labels, so trees don‘t mess up existing labels
        myTree[bestFeatLabel][value] = createTree(splitDataSet(dataSet, bestFeat, value),subLabels)
    return myTree    
                            
#使用決策樹的分類函數    
def classify(inputTree,featLabels,testVec):
    firstStr = inputTree.keys()[0]
    secondDict = inputTree[firstStr]
    featIndex = featLabels.index(firstStr)
    key = testVec[featIndex]
    valueOfFeat = secondDict[key]
    if isinstance(valueOfFeat, dict): 
        classLabel = classify(valueOfFeat, featLabels, testVec)
    else: classLabel = valueOfFeat
    return classLabel

#使用pickle模塊存儲決策樹    
def storeTree(inputTree,filename):
    import pickle
    fw = open(filename,‘w‘)
    pickle.dump(inputTree,fw)
    fw.close()
    
def grabTree(filename):
    import pickle
    fr = open(filename)
    return pickle.load(fr)

#coding:utf-8
import matplotlib as mpl
mpl.use(‘Agg‘)
import matplotlib.pyplot as plt

decisionNode = dict(boxstyle="sawtooth", fc="0.8")
leafNode = dict(boxstyle="round4", fc="0.8")
arrow_args = dict(arrowstyle="<-")

def getNumLeafs(myTree):
    numLeafs = 0
    firstStr = myTree.keys()[0]
    secondDict = myTree[firstStr]
    for key in secondDict.keys():
        if type(secondDict[key]).__name__==‘dict‘:#test to see if the nodes are dictonaires, if not they are leaf nodes
            numLeafs += getNumLeafs(secondDict[key])
        else:   numLeafs +=1
    return numLeafs

def getTreeDepth(myTree):
    maxDepth = 0
    firstStr = myTree.keys()[0]
    secondDict = myTree[firstStr]
    for key in secondDict.keys():
        if type(secondDict[key]).__name__==‘dict‘:#test to see if the nodes are dictonaires, if not they are leaf nodes
            thisDepth = 1 + getTreeDepth(secondDict[key])
        else:   thisDepth = 1
        if thisDepth > maxDepth: maxDepth = thisDepth
    return maxDepth

def plotNode(nodeTxt, centerPt, parentPt, nodeType):
    createPlot.ax1.annotate(nodeTxt, xy=parentPt,  xycoords=‘axes fraction‘,
             xytext=centerPt, textcoords=‘axes fraction‘,
             va="center", ha="center", bbox=nodeType, arrowprops=arrow_args )
    
def plotMidText(cntrPt, parentPt, txtString):
    xMid = (parentPt[0]-cntrPt[0])/2.0 + cntrPt[0]
    yMid = (parentPt[1]-cntrPt[1])/2.0 + cntrPt[1]
    createPlot.ax1.text(xMid, yMid, txtString, va="center", ha="center", rotation=30)

def plotTree(myTree, parentPt, nodeTxt):#if the first key tells you what feat was split on
    numLeafs = getNumLeafs(myTree)  #this determines the x width of this tree
    depth = getTreeDepth(myTree)
    firstStr = myTree.keys()[0]     #the text label for this node should be this
    cntrPt = (plotTree.xOff + (1.0 + float(numLeafs))/2.0/plotTree.totalW, plotTree.yOff)
    plotMidText(cntrPt, parentPt, nodeTxt)
    plotNode(firstStr, cntrPt, parentPt, decisionNode)
    secondDict = myTree[firstStr]
    plotTree.yOff = plotTree.yOff - 1.0/plotTree.totalD
    for key in secondDict.keys():
        if type(secondDict[key]).__name__==‘dict‘:#test to see if the nodes are dictonaires, if not they are leaf nodes   
            plotTree(secondDict[key],cntrPt,str(key))        #recursion
        else:   #it‘s a leaf node print the leaf node
            plotTree.xOff = plotTree.xOff + 1.0/plotTree.totalW
            plotNode(secondDict[key], (plotTree.xOff, plotTree.yOff), cntrPt, leafNode)
            plotMidText((plotTree.xOff, plotTree.yOff), cntrPt, str(key))
    plotTree.yOff = plotTree.yOff + 1.0/plotTree.totalD
#if you do get a dictonary you know it‘s a tree, and the first element will be another dict


def createPlot(inTree):
    fig = plt.figure(1, facecolor=‘white‘)
    fig.clf()
    axprops = dict(xticks=[], yticks=[])
    createPlot.ax1 = plt.subplot(111, frameon=False, **axprops)    #no ticks
    #createPlot.ax1 = plt.subplot(111, frameon=False) #ticks for demo puropses 
    plotTree.totalW = float(getNumLeafs(inTree))
    plotTree.totalD = float(getTreeDepth(inTree))
    plotTree.xOff = -0.5/plotTree.totalW; plotTree.yOff = 1.0;
    plotTree(inTree, (0.5,1.0), ‘‘)
    plt.savefig(‘trees.png‘)
‘‘‘
def createPlot():
    fig = plt.figure(1, facecolor=‘white‘)
	#創建第一張圖，設置背景色是白色
    fig.clf()
    createPlot.ax1 = plt.subplot(111, frameon=False) #ticks for demo puropses 
    plotNode(‘a decision node‘, (0.5, 0.1), (0.1, 0.5), decisionNode)
    plotNode(‘a leaf node‘, (0.8, 0.1), (0.3, 0.8), leafNode)
    plt.savefig(‘trees.png‘)
‘‘‘
def retrieveTree(i):
    listOfTrees =[{‘no surfacing‘: {0: ‘no‘, 1: {‘flippers‘: {0: ‘no‘, 1: ‘yes‘}}}},
                  {‘no surfacing‘: {0: ‘no‘, 1: {‘flippers‘: {0: {‘head‘: {0: ‘no‘, 1: ‘yes‘}}, 1: ‘no‘}}}}
                  ]
    return listOfTrees[i]

小結：

一、一棵最優決策樹主要應解決以下三個問題：

① 生成最少數目的葉子節點

② 生成每個葉子節點的深度最小

③ 生成的決策樹葉子節點最少且每個葉子節點深度最小

二、如果葉子節點只能增加少許信息，則可以刪除該節點，將它並入到其它節點中去。

三、ID3算法無法直接處理數值型數據。

四、采用文本方式很難展示決策樹，所以要用Matplotlib註解繪制樹型圖。

本文出自 “付煒超” 博客，謝絕轉載！

機器學習實戰精讀--------決策樹