Python机器学习(十九)决策树之系列二—C4.5原理与代码实现

RememberMePlease 2020-06-17

ID3算法缺点

它一般会优先选择有较多属性值的Feature,因为属性值多的特征会有相对较大的信息增益,信息增益反映的是,在给定一个条件以后,不确定性减少的程度,

这必然是分得越细的数据集确定性更高,也就是条件熵越小,信息增益越大。为了解决这个问题,C4.5就应运而生,它采用信息增益率来作为选择分支的准则。

C4.5算法原理

信息增益率定义为:

              Python机器学习(十九)决策树之系列二—C4.5原理与代码实现

其中,分子为信息增益(信息增益计算可参考上一节ID3的算法原理),分母为属性X的熵。

需要注意的是,增益率准则对可取值数目较少的属性有所偏好。

所以一般这样选取划分属性:选择增益率最高的特征列作为划分属性的依据。

代码实现

与ID3代码实现不同的是:只改变计算香农熵的函数calcShannonEnt,以及选择最优特征索引函数chooseBestFeatureToSplit,具体代码如下:

# -*- coding: utf-8 -*-
"""
Created on Thu Aug  2 17:09:34 2018
决策树ID3,C4.5的实现
@author: weixw
"""
from math import log
import operator
#原始数据
def createDataSet():
    dataSet = [[1, 1, ‘yes‘],
               [1, 1, ‘yes‘],
               [1, 0, ‘no‘],
               [0, 1, ‘no‘],
               [0, 1, ‘no‘]]
    labels = [‘no surfacing‘,‘flippers‘]   
    return dataSet, labels

#多数表决器
#列中相同值数量最多为结果
def majorityCnt(classList):
    classCounts = {}
    for value in classList:
        if(value not in classCounts.keys()):
            classCounts[value] = 0
        classCounts[value] +=1
    sortedClassCount = sorted(classCounts.iteritems(),key = operator.itemgetter(1),reverse =True)
    return sortedClassCount[0][0]
        
    
#划分数据集
#dataSet:原始数据集
#axis:进行分割的指定列索引
#value:指定列中的值
def splitDataSet(dataSet,axis,value):
    retDataSet= []
    for featDataVal in dataSet:
        if featDataVal[axis] == value:
            #下面两行去除某一项指定列的值,很巧妙有没有
            reducedFeatVal = featDataVal[:axis]
            reducedFeatVal.extend(featDataVal[axis+1:])
            retDataSet.append(reducedFeatVal)
    return retDataSet

#计算香农熵
#columnIndex = -1表示获取数据集每一项的最后一列的标签值
#其他表示获取特征列
def calcShannonEnt(columnIndex, dataSet):
    #数据集总项数
    numEntries = len(dataSet)
    #标签计数对象初始化
    labelCounts = {}
    for featDataVal in dataSet:
        #获取数据集每一项的最后一列的标签值
        currentLabel = featDataVal[columnIndex]
        #如果当前标签不在标签存储对象里,则初始化,然后计数
        if currentLabel not in labelCounts.keys():
            labelCounts[currentLabel] = 0
        labelCounts[currentLabel] += 1
    #熵初始化
    shannonEnt = 0.0
    #遍历标签对象,求概率,计算熵
    for key in labelCounts.keys():
        prop = labelCounts[key]/float(numEntries)
        shannonEnt -= prop*log(prop,2)
    return shannonEnt


#通过信息增益,选出最优特征列索引(ID3)
def chooseBestFeatureToSplit(dataSet):
    #计算特征个数,dataSet最后一列是标签属性,不是特征量
    numFeatures = len(dataSet[0])-1
    #计算初始数据香农熵
    baseEntropy = calcShannonEnt(-1, dataSet)
    #初始化信息增益,最优划分特征列索引
    bestInfoGain = 0.0
    bestFeatureIndex = -1
    for i in range(numFeatures):
        #获取每一列数据
        featList = [example[i] for example in dataSet]
        #将每一列数据去重
        uniqueVals = set(featList)
        newEntropy = 0.0
        for value in uniqueVals:
            subDataSet = splitDataSet(dataSet,i,value)
            #计算条件概率
            prob = len(subDataSet)/float(len(dataSet))
            #计算条件熵
            newEntropy +=prob*calcShannonEnt(-1, subDataSet)
        #计算信息增益
        infoGain = baseEntropy - newEntropy
        if(infoGain > bestInfoGain):
            bestInfoGain = infoGain
            bestFeatureIndex = i
    return bestFeatureIndex

#通过信息增益率,选出最优特征列索引(C4.5)
def chooseBestFeatureToSplitOfFurther(dataSet):
    #计算特征个数,dataSet最后一列是标签属性,不是特征量
    numFeatures = len(dataSet[0])-1
    #计算初始数据香农熵H(Y)
    baseEntropy = calcShannonEnt(-1, dataSet)
    #初始化信息增益,最优划分特征列索引
    bestInfoGainRatio = 0.0
    bestFeatureIndex = -1
    for i in range(numFeatures):
        #获取每一特征列香农熵H(X)
        featEntropy = calcShannonEnt(i, dataSet)
        #获取每一列数据
        featList = [example[i] for example in dataSet]
        #将每一列数据去重
        uniqueVals = set(featList)
        newEntropy = 0.0
        for value in uniqueVals:
            subDataSet = splitDataSet(dataSet,i,value)
            #计算条件概率
            prob = len(subDataSet)/float(len(dataSet))
            #计算条件熵
            newEntropy +=prob*calcShannonEnt(-1, subDataSet)
        #计算信息增益
        infoGain = baseEntropy - newEntropy
        #计算信息增益率
        infoGainRatio = infoGain/float(featEntropy)
        if(infoGainRatio > bestInfoGainRatio):
            bestInfoGainRatio = infoGainRatio
            bestFeatureIndex = i
    return bestFeatureIndex
        
#决策树创建
def createTree(dataSet,labels):
    #获取标签属性,dataSet最后一列,区别于labels标签名称
    classList = [example[-1] for example in dataSet]
    #树极端终止条件判断
    #标签属性值全部相同,返回标签属性第一项值
    if classList.count(classList[0]) == len(classList):
        return classList[0]
    #没有特征,只有标签列(1列)
    if len(dataSet[0]) == 1:
        #返回实例数最大的类
        return majorityCnt(classList)
#    #获取最优特征列索引ID3
#    bestFeatureIndex = chooseBestFeatureToSplit(dataSet)
    #获取最优特征列索引C4.5
    bestFeatureIndex = chooseBestFeatureToSplitOfFurther(dataSet)
    #获取最优索引对应的标签名称
    bestFeatureLabel = labels[bestFeatureIndex]
    #创建根节点
    myTree = {bestFeatureLabel:{}}
    #去除最优索引对应的标签名,使labels标签能正确遍历
    del(labels[bestFeatureIndex])
    #获取最优列
    bestFeature = [example[bestFeatureIndex] for example in dataSet]
    uniquesVals = set(bestFeature)
    for value in uniquesVals:
        #子标签名称集合
        subLabels = labels[:]
        #递归
        myTree[bestFeatureLabel][value] = createTree(splitDataSet(dataSet,bestFeatureIndex,value),subLabels)
    return myTree

#获取分类结果
#inputTree:决策树字典
#featLabels:标签列表
#testVec:测试向量  例如:简单实例下某一路径 [1,1]  => yes(树干值组合,从根结点到叶子节点)
def classify(inputTree,featLabels,testVec):
    #获取根结点名称,将dict转化为list
    firstSide = list(inputTree.keys())
    #根结点名称String类型
    firstStr = firstSide[0]
    #获取根结点对应的子节点
    secondDict = inputTree[firstStr]
    #获取根结点名称在标签列表中对应的索引
    featIndex = featLabels.index(firstStr)
    #由索引获取向量表中的对应值
    key = testVec[featIndex]
    #获取树干向量后的对象
    valueOfFeat = secondDict[key]
    #判断是子结点还是叶子节点:子结点就回调分类函数,叶子结点就是分类结果
    #if type(valueOfFeat).__name__==‘dict‘: 等价 if isinstance(valueOfFeat, dict):
    if isinstance(valueOfFeat, dict):
        classLabel = classify(valueOfFeat,featLabels,testVec)
    else:
        classLabel = valueOfFeat
    return classLabel


#将决策树分类器存储在磁盘中,filename一般保存为txt格式
def storeTree(inputTree,filename):
    import pickle
    fw = open(filename,‘wb+‘)
    pickle.dump(inputTree,fw)
    fw.close()
#将瓷盘中的对象加载出来,这里的filename就是上面函数中的txt文件    
def grabTree(filename):
    import pickle
    fr = open(filename,‘rb‘)
    return pickle.load(fr)
决策树算法
‘‘‘
Created on Oct 14, 2010

@author: Peter Harrington
‘‘‘
import matplotlib.pyplot as plt

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

#获取树的叶子节点
def getNumLeafs(myTree):
    numLeafs = 0
    #dict转化为list
    firstSides = list(myTree.keys())
    firstStr = firstSides[0]
    secondDict = myTree[firstStr]
    for key in secondDict.keys():
        #判断是否是叶子节点(通过类型判断,子类不存在,则类型为str;子类存在,则为dict)
        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

#获取树的层数
def getTreeDepth(myTree):
    maxDepth = 0
    #dict转化为list
    firstSides = list(myTree.keys())
    firstStr = firstSides[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)
    firstSides = list(myTree.keys())
    firstStr = firstSides[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
#绘制决策树
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.show()

#绘制树的根节点和叶子节点(根节点形状:长方形,叶子节点:椭圆形)
#def createPlot():
#    fig = plt.figure(1, facecolor=‘white‘)
#    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.show()

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]

#thisTree = retrieveTree(0)
#createPlot(thisTree)
#createPlot() 
#myTree = retrieveTree(0)
#numLeafs =getNumLeafs(myTree)
#treeDepth =getTreeDepth(myTree)
#print(u"叶子节点数目:%d"% numLeafs)
#print(u"树深度:%d"%treeDepth)
绘制决策树
# -*- coding: utf-8 -*-
"""
Created on Fri Aug  3 19:52:10 2018

@author: weixw
"""
import myTrees as mt
import treePlotter as tp
#测试
dataSet, labels = mt.createDataSet()
#copy函数:新开辟一块内存,然后将list的所有值复制到新开辟的内存中
labels1 = labels.copy()
#createTree函数中将labels1的值改变了,所以在分类测试时不能用labels1
myTree = mt.createTree(dataSet,labels1)
#保存树到本地
mt.storeTree(myTree,‘myTree.txt‘)
#在本地磁盘获取树
myTree = mt.grabTree(‘myTree.txt‘)
print(u"采用C4.5算法的决策树结果")
print (u"决策树结构:%s"%myTree)
#绘制决策树
print(u"绘制决策树:")
tp.createPlot(myTree)
numLeafs =tp.getNumLeafs(myTree)
treeDepth =tp.getTreeDepth(myTree)
print(u"叶子节点数目:%d"% numLeafs)
print(u"树深度:%d"%treeDepth)
#测试分类 简单样本数据3列
labelResult =mt.classify(myTree,labels,[1,1])
print(u"[1,1] 测试结果为:%s"%labelResult)
labelResult =mt.classify(myTree,labels,[1,0])
print(u"[1,0] 测试结果为:%s"%labelResult)
测试

 

运行结果

               Python机器学习(十九)决策树之系列二—C4.5原理与代码实现

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