第1关：返回分类次数最多的分类名称
import operatordef majorityCnt(classList):classCount = {}for i in classList:if i not in classCount:classCount[i] = 0classCount[i] += 1sortedClassCount = sorted(classCount.items(), key=operator.itemgetter(1), reverse=True)return sortedClassCount[0][0]
第2关：创建树函数
from ex03_lib import majorityCnt,splitDataSet,chooseBestFeatureToSplitdef createTree(dataSet,labels):classList = [example[-1] for example in dataSet] #获取数据集的所有类别#### 请补充完整代码 ####if classList.count(classList[0]) == len(classList):return classList[0]if len(dataSet[0]) == 1: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
第3关：获取叶子节点数目
def getNumLeafs(myTree):numLeafs = 0if type(myTree).__name__ == 'dict':fi = list(myTree.keys())[0]se = myTree[fi]for i in se.keys():if type(se[i]).__name__ == 'dict':numLeafs += getNumLeafs(se[i])else:numLeafs += 1else:numLeafs += 1return numLeafs
第4关：获取树的层数
def getTreeDepth(myTree):maxDepth = 0#### 请补充完整代码 ####fi = list(myTree.keys())[0]se = myTree[fi]for i in se.keys():if type(se[i]).__name__ == 'dict':thisDepth = 1 + getTreeDepth(se[i])else:thisDepth = 1if thisDepth > maxDepth: maxDepth = thisDepth#######################return maxDepth
第5关：注解树节点
import matplotlib
matplotlib.use('Agg')
import matplotlib.pyplot as plt
#定义决策树决策结果的特征，以字典的形式定义
#下面的字典定义也可写作 decisionNode={boxstyle:'sawtooth',fc:'0.8'}
#boxstyle为文本框的类型，sawtooth是锯齿形，fc是边框线粗细
decisionNode = dict(boxstyle="sawtooth", fc="0.8")
leafNode = dict(boxstyle="round4", fc="0.8")
arrow_args = dict(arrowstyle="<-")
def plotNode(nodeTxt, centerPt, parentPt, nodeType):#annotate是关于一个数据点的文本 #nodeTxt为要显示的文本，centerPt为文本的中心点，parentPt为指向文本的点 #### 请补充完整代码 ####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') # 定义一个画布，背景为白色fig.clf() # 把画布清空#createPlot.ax1为全局变量，绘制图像的句柄，subplot为定义了一个绘图，#111表示figure中的图有1行1列，即1个，最后的1代表第一个图 #frameon表示是否绘制坐标轴矩形 #### 请补充完整代码 ####createPlot.ax1 = plt.subplot(111,frameon=False) plotNode('a decision node',(0.2,0.2),(0.6,0.8),decisionNode) plotNode('a leaf node',(0.6,0.1),(0.8,0.8),leafNode) plt.show()#######################
第6关：绘制树形图
import matplotlib
matplotlib.use('Agg')
import matplotlib.pyplot as plt
from ex03_lib import plotNode,getNumLeafs,getTreeDepthdef plotTree(myTree, parentPt, nodeTxt):numLeafs = getNumLeafs(myTree) #当前树的叶子数depth = getTreeDepth(myTree) #没有用到这个变量firstSides = list(myTree.keys())firstStr = firstSides[0]#cntrPt是文本中心点，parentPt指向文本中心点 #### 请补充完整代码 ####cntrPt = (plotTree.xOff + (1.0 + float(numLeafs))/2.0/plotTree.totalW, plotTree.yOff)plotMidText(cntrPt, parentPt, nodeTxt) #画分支上的键plotNode(firstStr, cntrPt, parentPt, decisionNode)secondDict = myTree[firstStr]plotTree.yOff = plotTree.yOff - 1.0/plotTree.totalD #从上往下画for key in secondDict.keys():#如果是字典则是一个判断（内部）结点if type(secondDict[key]).__name__=='dict': plotTree(secondDict[key],cntrPt,str(key)) else: #打印叶子结点plotTree.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#######################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 createPlot(inTree):fig = plt.figure(1, facecolor='white')fig.clf()#### 请补充完整代码 ####axprops = dict(xticks=[], yticks=[]) #定义横纵坐标轴 createPlot.ax1 = plt.subplot(111, frameon=False) plotTree.totalW = float(getNumLeafs(inTree)) #全局变量宽度 = 叶子数plotTree.totalD = float(getTreeDepth(inTree)) #全局变量高度 = 深度#图形的大小是0-1 ，0-1plotTree.xOff = -0.5/plotTree.totalW; #例如绘制3个叶子结点，坐标应为1/3,2/3,3/3#但这样会使整个图形偏右因此初始的，将x值向左移一点。plotTree.yOff = 1.0;plotTree(inTree, (0.5,1.0), '')#######################plt.show()
