西瓜书第四章习题及答案

解答：
每个标记不同的数据特征向量都不相同，即树的每一条枝干（从根节点到叶子结点）就代表一种向量，这样的话决策树与训练集就是一致的了。

解答：
题目4.1中介绍了如果数据不含有冲突数据，那么就会产生绝对的过拟合现象，这也符合最小训练误差的准则，因此使用‘最小训练误差‘作为决策树划分选择准则的会产生在训练集上效果很好，而测试集上效果差的情况，即出现了过拟合现象。

分析：
表4.3属性数据值，有离散的有连续的，而基于信息熵进行属性划分只能是离散属性值，如果使用表4.3的数据集则需要对连续属性进行离散化-对应知识点-机器学习4.4连续与缺失值。
本实验采用去除表4.3中连续值的数据集西瓜数据集2.0。
解答：

第一部分-说在前面：

将数据集进行预处理（手动的呜呜~ 要哭了！)

将离散的属性值进行数字化，将属性用英文代替青绿-0   乌黑-1   浅白-2；蜷缩 -0  稍蜷-1   硬挺-2浊响-0 沉闷-1 清脆-2清晰-0 稍糊-1 模糊-2凹陷-0 稍凹-1 平坦-2硬滑-0  软粘-1是-1 否 -0#texture 纹理 umbilical region---脐部labels = ['color', 'root', 'Knock sound', 'texture','umbilical region','touch']

第二部分-计算机数据集的信息熵：

def calcShannonEnt(dataSet):numEntries = len(dataSet)labelCounts = {}for featVec in dataSet: #the the number of unique elements and their occurancecurrentLabel = featVec[-1]if currentLabel not in labelCounts.keys(): labelCounts[currentLabel] = 0labelCounts[currentLabel] += 1shannonEnt = 0.0for key in labelCounts:prob = float(labelCounts[key])/numEntriesshannonEnt -= prob * log(prob,2) #log base 2return shannonEnt

第三部分-划分数据集

创建数据集函数：

from math import log
import operator
def createDataSet():dataSet = [[0, 0, 0, 0, 0, 0, 'yes'], [1, 0, 1, 0, 0, 0, 'yes'], [1, 0, 0, 0, 0, 0, 'yes'],[0, 0, 1, 0, 0, 0, 'yes'],[2, 0, 0, 0, 0, 0, 'yes'], [0, 1, 0, 0, 1, 1, 'yes'],[1, 1, 0, 1, 1, 1, 'yes'],[1, 1, 0, 0, 1, 0, 'yes'], [1, 1, 1, 1, 1, 0, 'no'], [0, 2, 2, 0, 2, 1, 'no'], [2, 2, 2, 2, 2, 0, 'no'], [2, 0, 0, 2, 2, 1, 'no'], [0, 1, 0, 1, 0, 0, 'no'], [2, 1, 1, 1, 0, 0, 'no'], [1, 1, 0, 0, 1, 1, 'no'], [2, 0, 0, 2, 2, 0, 'no'],[0, 0, 1, 1, 1, 0, 'no']]labels = ['color', 'root', 'Knock sound', 'texture','umbilical region','touch'] #change to discrete valuesreturn dataSet, labels

按照给定特征划分数据集：

def splitDataSet(dataSet, axis, value):retDataSet = []for featVec in dataSet:if featVec[axis] == value:reducedFeatVec = featVec[:axis]     #chop out axis used for splittingreducedFeatVec.extend(featVec[axis+1:])retDataSet.append(reducedFeatVec)return retDataSet

选择最好的数据集划分方式：

def chooseBestFeatureToSplit(dataSet):numFeatures = len(dataSet[0]) - 1      #the last column is used for the labelsbaseEntropy = calcShannonEnt(dataSet)bestInfoGain = 0.0; bestFeature = -1for i in range(numFeatures):        #iterate over all the featuresfeatList = [example[i] for example in dataSet]#create a list of all the examples of this featureuniqueVals = set(featList)       #get a set of unique valuesnewEntropy = 0.0for 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 entropyif (infoGain > bestInfoGain):       #compare this to the best gain so farbestInfoGain = infoGain         #if better than current best, set to bestbestFeature = ireturn bestFeature

第四部分-递归构建决策树

创建树的函数代码：

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 equalif len(dataSet[0]) == 1: #stop splitting when there are no more features in dataSetreturn 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 labelsmyTree[bestFeatLabel][value] = createTree(splitDataSet(dataSet, bestFeat, value),subLabels)return myTree

将结果输出展示：

myDat, labels = createDataSet()
print(myDat)
print(labels)

运行结果如下所示：

Ent1 = calcShannonEnt(myDat)
print(Ent1)bestFeature = chooseBestFeatureToSplit(myDat)
print('最好的划分属性的下标为：{}'.format(bestFeature))

运行结果如下所示：

myTree = createTree(myDat, labels)
myTree

运行结果如下所示：

第五部分-使用matplotlib注解绘制树形图

使用文本注解绘制节点：

import matplotlib.pyplot as plt#定义文本框和箭头的形式
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) #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()

createPlot()#运行结果如下所示

获取叶节点的数目：

def getNumLeafs(myTree):numLeafs = 0firstStr = list(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 nodesnumLeafs += getNumLeafs(secondDict[key])else:   numLeafs +=1return numLeafs

获取树的层数：

#获取树的层数
def getTreeDepth(myTree):maxDepth = 0firstStr = list(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 nodesthisDepth = 1 + getTreeDepth(secondDict[key])else:   thisDepth = 1if thisDepth > maxDepth: maxDepth = thisDepthreturn maxDepth

创建retrieveTree函数输出预先存储的树信息，避免每次测试代码时都要从数据中创建树的麻烦。（就是保结果复制过来保存一下）

def retrieveTree(i):listOfTrees =[{'texture': {0: {'root': {0: 'yes',1: {'color': {0: 'yes', 1: {'touch': {0: 'yes', 1: 'no'}}}},2: 'no'}},1: {'touch': {0: 'no', 1: 'yes'}},2: 'no'}},{'texture': {0: {'root': {0: 'yes',1: {'color': {0: 'yes', 1: {'touch': {0: 'yes', 1: 'no'}}}},2: 'no'}},1: {'touch': {0: 'no', 1: 'yes'}},2: 'no'}}]return listOfTrees[i]

运行上述代码并查看结果：

myTree = retrieveTree(0)
print(myTree)print(myTree.keys())
print(list(myTree.keys())[0])print('树叶节点的数目为：{}'.format(getNumLeafs(myTree)))
print('树的深度为：{}'.format(getTreeDepth(myTree)))

结果如下：
plotTree函数-由三个函数构成，从下往上调用函数。

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 onnumLeafs = getNumLeafs(myTree)  #this determines the x width of this treedepth = getTreeDepth(myTree)firstStr = list(myTree.keys())[0]     #the text label for this node should be thiscntrPt = (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.totalDfor 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))        #recursionelse:   #it's a leaf node print the leaf nodeplotTree.xOff = plotTree.xOff + 1.0/plotTree.totalWplotNode(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 dictdef createPlot(inTree):fig = plt.figure(1, facecolor='white')fig.clf()axprops = dict(xticks=[], yticks=[])createPlot.ax1 = plt.subplot(111, frameon=False, **axprops)    #no ticksplotTree.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), '')###11111111plt.show()

myTree = retrieveTree(0)
createPlot(myTree)

最终运行的结果图：

第六部分-说明

虽然输出的结果丑了点，也确实花费了不少时间（谁让我那么菜呐），所以有写的不好的地方还请各位大佬支出，大家共同进步哈。
今后如果有能力的话，会将最终的决策树美化一下，处理数据的文本样式而不是现在的0 1，再将连续型的属性值也加上去。

第七部分-参考书籍及文章

机器学习-西瓜书
机器学习实战-Peter Harrington