机器学习实战-决策树

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

？？？CSDN的Markdown，第一行居然不居中？还是要加什么？

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 logdef 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 shannonEntdef 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 treesreload(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 operatordef 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.pyimport treesreload(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 pltdecisionNode = 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()

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 numLeafsdef 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 maxDepthdef 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.pyimport treePlottermyTree=treePlotter.retrieveTree(0)treePlotter.createPlot(myTree)

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)

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.pydef 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-1import treesimport treePlottermyDat,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.pydef 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.pyimport treesimport treePlottermyDat,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 treesimport treePlottermyDat,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)