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详细代码及数据已上传到githubhttps://github.com/SONGSONG729/decision-tree
按照书本代码进行实验时的报错:
在使用Matplotlib注解时,出现了错误:Backend Qt5Agg is interactive backend. Turning interactive mode on.
错误原因:matplotlib的默认backend是TkAgg,而FltkAgg, GTK, GTKAgg, GTKCairo, TkAgg , Wx or WxAgg这几个backend都要求有GUI图形界面的,所以在ssh操作的时候会报错。
解决办法:指定不需要GUI的backend(Agg, Cairo, PS, PDF or SVG),在plt.show()
之前,加上plt.switch_backend(‘agg’)
。
在使用pickle模块存储决策树时,出现了错误:write() argument must be str, not bytes
错误原因:fw这个变量的类型应该是bytes的
解决办法:将’w’改为’wb
’。
编译后提示UnicodeDecodeError: 'gbk' codec can't decode byte 0x80 in position 0: illegal multibyte sequence
错误,是由于之前保存的txt文件是用二进制写入的所以在读取时也应该用二进制,即读文件中的fr = open(filename)应该改为fr = open(filename, 'rb')
。
使用决策树预测隐形眼镜类型
# _*_ coding:utf-8 _*_ from math import log import operator import matplotlib.pyplot as plt import treePlotter def calcShannonEnt(dataSet): ''' 计算给定数据集的香农公式 :param dataSet: :return: ''' numEntries = len(dataSet) labelCounts = {} # 为所有可能分类创建字典 for featVec in dataSet: currentLabel = featVec[-1] if currentLabel not in labelCounts.keys(): labelCounts[currentLabel] = 0 labelCounts[currentLabel] += 1 shannonEnt = 0.0 for key in labelCounts: prob = float(labelCounts[key])/numEntries shannonEnt -= prob * log(prob, 2) # 以2为底求对数 return shannonEnt def createDataSet(): ''' 简单鉴定数据集 :return: ''' dataSet = [[1, 1, 'yes'], [1, 1, 'yes'], [1, 0, 'no'], [0, 1, 'no'], [0, 1, 'no']] labels = ['no surfacing', 'flippers'] return dataSet, labels def splitDataSet(dataSet, axis, value): ''' 按照给定特征划分数据集 :param dataSet: 待划分的数据集 :param axis: 划分数据集的特征 :param value: 需要返回的特征值 :return: ''' retDataSet = [] # 创建新的lise对象,不修改原列表 for featVec in dataSet: # 将符合特征的数据抽取出来 if featVec[axis] == value: reduceFeatVec = featVec[:axis] reduceFeatVec.extend(featVec[axis+1:]) retDataSet.append(reduceFeatVec) return retDataSet def chooseBestFeatureToSplit(dataSet): ''' 选择最好的数据集划分方式 :param dataSet: :return: ''' numFeatures = len(dataSet[0]) - 1 baseEntropy = calcShannonEnt(dataSet) bestInfoGain = 0.0 bestFeature = -1 for i in range(numFeatures): # 创建唯一的分类标签列表 featList = [example[i] for example in dataSet] uniqueVals = set(featList) newEntropy = 0.0 # 计算每种划分方式的信息熵 for value in uniqueVals: subDataSet = splitDataSet(dataSet, i, value) prob = len(subDataSet)/float(len(dataSet)) newEntropy += prob * calcShannonEnt(subDataSet) infoGain = baseEntropy - newEntropy # 计算最好的信息增益 if infoGain > bestInfoGain: bestInfoGain = infoGain bestFeature = i return bestFeature def majorityCnt(classList): ''' :param classList: 分类名称的列表 :return: 出现次数最多的分类名称 ''' # key:classList中唯一值的数据字典 # value:classList中每个类标签出现的频率 classCount = {} for vote in classList: if vote not in classCount.keys(): classCount[vote] = 0 classCount[vote] += 1 # 用operator操作键值排序字典 sortedClassCount = sorted(classCount.iteritems(), key=operator.itemgetter(1), reverse=True) return sortedClassCount[0][0] def createTree(dataSet, labels): ''' 创建树 :param dataSet: 数据集 :param labels: 标签列表,包含数据集中所有特征的标签 :return: ''' # classList列表变量包含了数据集的所有类标签 classList = [example[-1] for example in dataSet] # 所有的类标签完全相同则停止划分,返回该类标签 # 第一个元素的值的数量等于整个列表的长度,即说明整个列表都是这个值,所以该数据集类别全部相同了 if classList.count(classList[0]) == len(classList): return classList[0] # 所有特征已经利用完,仍然不能将数据集划分成仅包含唯一类别的分组,返回出现次数最多的类别作为返回值 # 所有特征已经利用完,只剩下标签列,仍然无法区分剩余样本,则采用“少数服从多数”的方案 if len(dataSet[0]) == 1: return majorityCnt() bestFeat = chooseBestFeatureToSplit(dataSet) # 当前数据集选取的最好特征 bestFeatLabel = labels[bestFeat] myTree = {bestFeatLabel: {}} # 存储树的所有信息 # 得到列表包含的所有属性值 del(labels[bestFeat]) featValues = [example[bestFeat] for example in dataSet] uniqueVals = set(featValues) # 遍历当前选择特征包含的所有属性值,在每个数据集划分上递归调用函数createTree(), # 得到的返回值将被插入到字典变量myTree中, # 函数终止时,字典中将会嵌套很多代表叶子节点信息的字典数据 for value in uniqueVals: subLabels = labels[:] # 复制了类标签,并将其存储在新列表变量subLabels中 myTree[bestFeatLabel][value] = createTree(splitDataSet(dataSet, bestFeat, value), subLabels) return myTree def classify(inputTree, featLabels, testVec): ''' 使用决策树的分类函数(递归函数) :param inputTree: :param featLabels: :param testVec: :return: ''' firstStr = list(inputTree.keys())[0] secondDict = inputTree[firstStr] featIndex = featLabels.index(firstStr) # 将标签字符串转换为索引 for key in secondDict.keys(): if testVec[featIndex] == key: if type(secondDict[key]).__name__ == 'dict': classLabel = classify(secondDict[key], featLabels, testVec) else: classLabel = secondDict[key] return classLabel def storeTree(inputTree, filename): import pickle fw = open(filename, 'wb') pickle.dump(inputTree, fw) fw.close() def grabTree(filename): import pickle fr = open(filename, 'rb') return pickle.load(fr) def main(): ''' myDat, labels = createDataSet() print(myDat) print(calcShannonEnt(myDat)) ''' ''' myDat, labels = createDataSet() print(myDat) print(splitDataSet(myDat, 0, 1)) print(splitDataSet(myDat, 0, 0)) ''' ''' myDat, labels = createDataSet() print(chooseBestFeatureToSplit(myDat)) print(myDat) ''' ''' myDat, labels = createDataSet() myTree = createTree(myDat, labels) print(myTree) ''' """ myDat, labels = createDataSet() print(labels) myTree = treePlotter.retrieveTree(0) print(myTree) print(classify(myTree, labels, [1, 0])) print(classify(myTree, labels, [1, 1])) """ ''' myTree = treePlotter.retrieveTree(0) storeTree(myTree, 'classifierStorage.txt') grabTree('classifierStorage.txt') ''' # with open('lenses.txt') as fr: fr = open('lenses.txt') lenses = [inst.strip().split('\t') for inst in fr.readlines()] lensesLables = ['age', 'prescript', 'astigmatic', 'tearRate'] lensesTree = createTree(lenses, lensesLables) print(lensesTree) treePlotter.createPlot(lensesTree) if __name__ == '__main__': main()
import matplotlib.pyplot as plt from pylab import mpl # 用以显示中文 mpl.rcParams['font.sans-serif'] = ['FangSong'] ''' 使用文本注解绘制树节点 ''' 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.axl.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.axl = plt.subplot(111, frameon=True) plotNode('决策节点', (0.5, 0.1), (0.1, 0.5), decisionNode) plotNode('叶节点', (0.8, 0.1), (0.3, 0.8), leafNode) plt.switch_backend('agg') plt.show() def getNumLeafs(myTree): ''' 遍历整棵树,累计叶子节点的个数,并返回该值 :param myTree: :return: ''' numLeafs = 0 # in the py3, type(myTree.keys()) is dict_keys. # It can be directed used for iteration, but index is not suitable. # in the py2, type(myTree.keys()) is list. firstStr = list(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): ''' 计算遍历过程中遇到判断节点的个数 :param myTree: :return: ''' maxDepth = 0 firstStr = list(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] def plotMidText(cntrPt, parentPt, txtString): ''' 在父子节点间填充文本信息 :param cntrPt: :param parentPt: :param txtString: :return: ''' xMid = (parentPt[0] - cntrPt[0])/2.0 + cntrPt[0] yMid = (parentPt[1] - cntrPt[1])/2.0 + cntrPt[1] createPlot.axl.text(xMid, yMid, txtString) def plotTree(myTree, parentPt, nodeTxt): ''' 绘制树形图 plotTree.xOff、plotTree.yOff:全局变量,追踪已绘制的节点位置 :param myTree: :param parentPt: :param nodeTxt: :return: ''' numLeafs = getNumLeafs(myTree) depth = getTreeDepth(myTree) # 计算宽与高 firstStr = list(myTree.keys())[0] cntrPt = (plotTree.xOff + (1.0 + float(numLeafs))/2.0/plotTree.totalW, plotTree.yOff) plotMidText(cntrPt, parentPt, nodeTxt) # 标记子节点属性值 plotNode(firstStr, cntrPt, parentPt, decisionNode) secondDict = myTree[firstStr] plotTree.yOff = plotTree.yOff - 1.0/plotTree.totalD # 减少y偏移,自顶向下绘制图形,一次递减y的坐标 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 def createPlot(inTree): ''' 创建绘图区,计算树图形的尺寸,并调用递归函数plotTree() :param inTree: :return: ''' fig = plt.figure(1, facecolor='white') fig.clf() axprops = dict(xticks=[], yticks=[]) createPlot.axl = plt.subplot(111, frameon=False, **axprops) 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()
运行tree.py后的结果
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