python实现C4.5决策树算法
C4.5算法使用信息增益率来代替ID3的信息增益进行特征的选择,克服了信息增益选择特征时偏向于特征值个数较多的不足。信息增益率的定义如下:
# -*- coding: utf-8 -*- from numpy import * import math import copy import cPickle as pickle class C45DTree(object): def __init__(self): # 构造方法 self.tree = {} # 生成树 self.dataSet = [] # 数据集 self.labels = [] # 标签集 # 数据导入函数 def loadDataSet(self, path, labels): recordList = [] fp = open(path, "rb") # 读取文件内容 content = fp.read() fp.close() rowList = content.splitlines() # 按行转换为一维表 recordList = [row.split("\t") for row in rowList if row.strip()] # strip()函数删除空格、Tab等 self.dataSet = recordList self.labels = labels # 执行决策树函数 def train(self): labels = copy.deepcopy(self.labels) self.tree = self.buildTree(self.dataSet, labels) # 构件决策树:穿件决策树主程序 def buildTree(self, dataSet, lables): cateList = [data[-1] for data in dataSet] # 抽取源数据集中的决策标签列 # 程序终止条件1:如果classList只有一种决策标签,停止划分,返回这个决策标签 if cateList.count(cateList[0]) == len(cateList): return cateList[0] # 程序终止条件2:如果数据集的第一个决策标签只有一个,返回这个标签 if len(dataSet[0]) == 1: return self.maxCate(cateList) # 核心部分 bestFeat, featValueList= self.getBestFeat(dataSet) # 返回数据集的最优特征轴 bestFeatLabel = lables[bestFeat] tree = {bestFeatLabel: {}} del (lables[bestFeat]) for value in featValueList: # 决策树递归生长 subLables = lables[:] # 将删除后的特征类别集建立子类别集 # 按最优特征列和值分隔数据集 splitDataset = self.splitDataSet(dataSet, bestFeat, value) subTree = self.buildTree(splitDataset, subLables) # 构建子树 tree[bestFeatLabel][value] = subTree return tree # 计算出现次数最多的类别标签 def maxCate(self, cateList): items = dict([(cateList.count(i), i) for i in cateList]) return items[max(items.keys())] # 计算最优特征 def getBestFeat(self, dataSet): Num_Feats = len(dataSet[0][:-1]) totality = len(dataSet) BaseEntropy = self.computeEntropy(dataSet) ConditionEntropy = [] # 初始化条件熵 slpitInfo = [] # for C4.5,caculate gain ratio allFeatVList = [] for f in xrange(Num_Feats): featList = [example[f] for example in dataSet] [splitI, featureValueList] = self.computeSplitInfo(featList) allFeatVList.append(featureValueList) slpitInfo.append(splitI) resultGain = 0.0 for value in featureValueList: subSet = self.splitDataSet(dataSet, f, value) appearNum = float(len(subSet)) subEntropy = self.computeEntropy(subSet) resultGain += (appearNum/totality)*subEntropy ConditionEntropy.append(resultGain) # 总条件熵 infoGainArray = BaseEntropy*ones(Num_Feats)-array(ConditionEntropy) infoGainRatio = infoGainArray/array(slpitInfo) # C4.5信息增益的计算 bestFeatureIndex = argsort(-infoGainRatio)[0] return bestFeatureIndex, allFeatVList[bestFeatureIndex] # 计算划分信息 def computeSplitInfo(self, featureVList): numEntries = len(featureVList) featureVauleSetList = list(set(featureVList)) valueCounts = [featureVList.count(featVec) for featVec in featureVauleSetList] pList = [float(item)/numEntries for item in valueCounts] lList = [item*math.log(item, 2) for item in pList] splitInfo = -sum(lList) return splitInfo, featureVauleSetList # 计算信息熵 # @staticmethod def computeEntropy(self, dataSet): dataLen = float(len(dataSet)) cateList = [data[-1] for data in dataSet] # 从数据集中得到类别标签 # 得到类别为key、 出现次数value的字典 items = dict([(i, cateList.count(i)) for i in cateList]) infoEntropy = 0.0 for key in items: # 香农熵: = -p*log2(p) --infoEntropy = -prob * log(prob, 2) prob = float(items[key]) / dataLen infoEntropy -= prob * math.log(prob, 2) return infoEntropy # 划分数据集: 分割数据集; 删除特征轴所在的数据列,返回剩余的数据集 # dataSet : 数据集; axis: 特征轴; value: 特征轴的取值 def splitDataSet(self, dataSet, axis, value): rtnList = [] for featVec in dataSet: if featVec[axis] == value: rFeatVec = featVec[:axis] # list操作:提取0~(axis-1)的元素 rFeatVec.extend(featVec[axis + 1:]) # 将特征轴之后的元素加回 rtnList.append(rFeatVec) return rtnList # 存取树到文件 def storetree(self, inputTree, filename): fw = open(filename,'w') pickle.dump(inputTree, fw) fw.close() # 从文件抓取树 def grabTree(self, filename): fr = open(filename) return pickle.load(fr)
调用代码
# -*- coding: utf-8 -*- from numpy import * from C45DTree import * dtree = C45DTree() dtree.loadDataSet("dataset.dat",["age", "revenue", "student", "credit"]) dtree.train() dtree.storetree(dtree.tree, "data.tree") mytree = dtree.grabTree("data.tree") print mytree
以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持我们。
赞 (0)