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机器学习实战-决策树

机器学习实战-决策树

作者: mov觉得高数好难 | 来源:发表于2017-04-18 20:54 被阅读0次

    第二章介绍的k-近邻算法可以完成很多分类任务,但是它最大的缺点就是无法给出数据内在的含义,决策树的主要优势在于数据形式非常容易理解。

    决策树
    优点:计算复杂度不高,输出结果易于理解,对中间值的缺失不敏感,可以处理不相关特征数据。
    缺点:可能会产生过度匹配的问题
    适用数据类型:数值型和标称型

    先计算给定数据集的香农熵

    在CSDN的Markdown中,第一行居然不居中,简书的是默认居中了

    序号 不浮出水面是否可以生存 是否有脚蹼 属于鱼类
    1 1 1 1
    2 1 1 1
    3 1 0 0
    3 0 1 0
    3 0 1 0
    from math import log
    
    def calcShannonEnt(dataSet):
        numEntries = len(dataSet)
        labelCounts = {}
        #为所有可能的分类创建字典
        for featVec in dataSet:
            currentLabel = featVec[-1]
            if currentLabel not in labelCounts.keys():labelCounts[currentLabel] = 0
            labelCounts[currentLabel] += 1
        shannonEnt = 0.0
        #计算熵
        for key in labelCounts:
            prob = float(labelCounts[key])/numEntries
            #底数为2求对数
            shannonEnt -= prob*log(prob,2)
        return shannonEnt
    
    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
    

    测试数据集的熵

    import trees
    reload(trees)
    myDat,labels = trees.createDataSet()
    
    >>> myDat
    [[1, 1, 'yes'], [1, 1, 'yes'], [1, 0, 'no'], [0, 1, 'no'], [0, 1, 'no']]
    >>> trees.calcShannonEnt(myDat)
    0.9709505944546686
    

    熵越高,则混合的数据也越多
    增加一个测试分类来测试熵的变化:

    >>> myDat[0][-1]='maybe'
    >>> myDat
    [[1, 1, 'maybe'], [1, 1, 'yes'], [1, 0, 'no'], [0, 1, 'no'], [0, 1, 'no']]
    >>> trees.calcShannonEnt(myDat)
    1.3709505944546687
    

    下面,开始划分数据集:

    def splitDataSet(dataSet, axis, value):
        retDataSet = []
        for featVec in dataSet:
            #将符合要求的数据集添加到列表中
            if featVec[axis] == value:
                reducedFeatVec = featVec[:axis]
                reducedFeatVec.extend(featVec[axis+1:])
                retDataSet.append(reducedFeatVec)
        return retDataSet
    

    本段代码使用了三个输入参数:待划分的数据集,划分数据集的特征,需要返回的特征的值。
    然后在Python命令提示符内输入下述命令:

    >>> myDat
    [[1, 1, 'yes'], [1, 1, 'yes'], [1, 0, 'no'], [0, 1, 'no'], [0, 1, 'no']]
    >>> trees.splitDataSet(myDat,0,1)
    [[1, 'yes'], [1, 'yes'], [0, 'no']]
    >>> trees.splitDataSet(myDat,0,0)
    [[1, 'no'], [1, 'no']]
    >>> 
    

    接下来我们开始遍历整个数据集,循环计算香农熵和splitDataSet()函数,找到最好的特征划分方式。

    def chooseBestFeatureToSplit(dataSet):
        numFeatures = len(dataSet[0]) - 1
        baseEntropy = calcShannonEnt(dataSet)#计算原始香农熵
        bastInfoGain = 0.0;baseFeature = -1
        for i in range(numFeatures):
            featList = [example[i] for example in dataSet]
            uniqueVals = set(featList)#创建分类标签集合
            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
            if (infoGain > bestInfogain):
                bestInfoGain = infoGain
                bestFeature = i
            return bestFeature
    

    从列表中新建集合是Python语言得到列表中唯一元素的最快方法

    下面开始测试上面代码的实际输出结果

    >>> trees.chooseBestFeatureToSplit(myDat)
    0
    >>> myDat
    [[1, 1, 'yes'], [1, 1, 'yes'], [1, 0, 'no'], [0, 1, 'no'], [0, 1, 'no']]
    

    代码告诉我们,第0个特征是最好的用于划分数据集的特征

    下面我们开始采用递归的方式处理数据集

    如果数据集已经处理了所有属性,但是类标签依然不是唯一的,此时我们需要决定如何定义该叶子节点,在这种情况下,我们通常会采用多数表决的方法决定该叶子节点的分类

    import operator
    def majorityCnt(classList):
        classCount = {}
        for vote in classList:#统计数据字典classList每一个类标签出现的频率
            if vote not in classCount.keys(): classCount[vote] = 0
            classCount[vote] += 1
        sortedClassCount = sorted(classCount.iteritems(), key=operator.itemgette(1), reverse=True)#排序
        return sortedClassCount[0][0]#返回次数最多的
    

    下面开始加入创建树的函数代码

    def createTree(dataSet, labels):#数据集,标签列表
        classList = [example[-1] for example in dataSet]#最后一个属性加入列表变量classList
        #类别完全相同则停止划分
        if classList.count(classList[0]) == len(classList):
            return classList[0]
        if len(dataSet[0]) == 1:#遍历完所有特征,返回次数最多的类别
            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[:]
            myTree[bestFeatLabel][value] = createTree(splitDataSet(dataSet, bestFeat, value),subLabels)
        return myTree
    

    subLabels = labels[:],这行代码复制了类标签,并将其存储在新列表变量subLabels中。之所以这样做,是因为在Python语言中,函数参数是列表类型时,参数是按照引用方式传递的。为了保证每次调用函数createTree()时不改变原始列表的内容,使用新变量subLabels代替原始列表。

    下面是测试实际输出结果

    #trees-1.py
    import trees
    reload(trees)
    myDat,labels = trees.createDataSet()
    myTree = trees.createTree(myDat,labels)
    
    >>> myTree
    {'no surfacing': {0: 'no', 1: {'flippers': {0: 'no', 1: 'yes'}}}}
    

    下面开始学习使用Matplotlib画图

    #-*- coding=utf-8 -*-
    
    import matplotlib.pyplot as plt
    
    
    decisionNode = dict(boxstyle="sawtooth",fc="0.8")
    leafNode = dict(boxstyle="round4",fc="0.8")
    arrow_args = dict(arrowstyle="<-")
    
    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 createPlot():
        fig = plt.figure(1,facecolor='white')
        fig.clf()
        createPlot.ax1 = plt.subplot(111,frameon=False)
    
        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()
    
    >>> import treePlotter;treePlotter.createPlot()
    
    函数plotNode例图

    下面开始获取叶节点的数目和树的层数

    def getNumLeafs(myTree):
        numLeafs = 0
        firstStr = myTree.keys()[0]
        secondDict = myTree[firstStr]
        for key in secondDict.keys():
            if type(secondDict[key]).__name__=='dict':
                 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':
                thisDepth = 1 + getTreeDepth(secondDict[key])
            else:   thisDepth = 1
            if thisDepth > maxDepth: maxDepth = thisDepth
        return maxDepth
    
    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]
    
    >>> import treePlotter
    >>> treePlotter.retrieveTree(1)
    {'no surfacing': {0: 'no', 1: {'flippers': {0: {'head': {0: 'no', 1: 'yes'}}, 1: 'no'}}}}
    >>> myTree = treePlotter.retrieveTree(0)
    >>> treePlotter.getNumLeafs(myTree)
    3
    >>> treePlotter.getTreeDepth(myTree)
    2
    

    更新部分代码,开始尝试画图

    def plotMidText(cntrPt, parentPt, txtString):#计算父节点和子节点的中间位置
        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):
        numLeafs = getNumLeafs(myTree)
        depth = getTreeDepth(myTree)
        firstStr = myTree.keys()[0]
        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':
               plotTree(secondDict[key],cntrPt,str(key))
            else:
                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
    

    并更新creaePlot()部分的代码

    def createPlot(inTree):
        fig = plt.figure(1,facecolor='white')
        fig.clf()
        axprops = dict(xticks=[], yticks=[])
        createPlot.ax1 = plt.subplot(111,frameon=False, **axprops)
        plotTree.totalW = float(getNumLeafs(inTree))#宽度
        plotTree.totalD = float(getTreeDepth(inTree))#高度
        plotTree.xOff = -0.5/plotTree.totalW;plotTree.yOff = 1.0;
        plotTree(inTree, (0.5,1.0),'')
        plt.show()
    

    开始画图

    #treePlotter-1.py
    import treePlotter
    myTree=treePlotter.retrieveTree(0)
    treePlotter.createPlot(myTree)
    
    图3-6

    接着添加字典的内容,重新绘制图片

    >>> myTree['no surfacing'][3] = 'maybe'
    >>> myTree
    {'no surfacing': {0: 'no', 1: {'flippers': {0: 'no', 1: 'yes'}}, 3: 'maybe'}}
    >>> treePlotter.createPlot(myTree)
    
    图3-7

    下面开始重点讲如何利用决策树执行数据分类
    在执行数据分类时,需要使用决策树以及用于构造决策树的标签向量。然后,程序比较测试数据与决策树上的数值,递归执行该过程直到进入叶子节点;最后将测试数据定义为叶子节点所属的类型。

    #使用决策树的分类函数
    #添加到trees.py
    def classify(inputTree, featLabels, testVec):
        firstStr = inputTree.keys()[0]
        secondDict = inputTree[firstStr]
        featIndex = featLabels.index(firstStr)#将标签字符串转化为索引
        for key in secondDict.keys():
            if testVec[featIndex] == key:
                if type(secondDict[key]).__name__=='dict':#递归遍历
                    classLabel = classify(secondDict[key],featLabels,testVec)
                else: classLabel = secondDict[key]
        #key = testVec[featIndex]
        #valueOfFeat = secondDict[key]
        #if isinstance(valueOfFeat, dict):#是否是实例
        #    classLabel = classify(valueOfFeat, featLabels, testVec)
        #else: classLabel = valueOfFeat
        return classLabel
    
    #trees-1
    import trees
    import treePlotter
    myDat,labels = trees.createDataSet()
    myTree=treePlotter.retrieveTree(0)
    
    >>> labels
    ['no surfacing', 'flippers']
    >>> myTree
    {'no surfacing': {0: 'no', 1: {'flippers': {0: 'no', 1: 'yes'}}}}
    >>> trees.classify(myTree,labels,[1,0])
    'no'
    >>> trees.classify(myTree,labels,[1,1])
    'yes'
    

    下面开始在硬盘上存储决策树分类器

    #trees.py
    def storeTree(inputTree,filename):
        import pickle#重点
        fw = open(filename,'w')
        pickle.dump(inputTree,fw)
        fw.close()
        
    def grabTree(filename):
        import pickle
        fr = open(filename)
        return pickle.load(fr)
    
    #tree-1.py
    import trees
    import treePlotter
    myDat,labels = trees.createDataSet()
    myTree = trees.createTree(myDat,labels)
    trees.storeTree(myTree,'classifierStorage.txt')
    
    >>> trees.grabTree('classifierStorage.txt')
    {'no surfacing': {0: 'no', 1: {'flippers': {0: 'no', 1: 'yes'}}}}
    
    import trees
    import treePlotter
    myDat,labels = trees.createDataSet()
    myTree = trees.createTree(myDat,labels)
    
    fr=open('lenses.txt')
    lenses=[inst.strip().split('\t') for inst in fr.readlines()]
    lensesLabels=['age','prescript','astigmatic','tearRate']
    lensesTree = trees.createTree(lenses,lensesLabels)
    
    >>> treePlotter.createPlot(lensesTree)
    
    ID3算法产生的决策树

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