Python实现决策树C4.5算法的示例

为什么要改进成C4.5算法

原理

C4.5算法是在ID3算法上的一种改进，它与ID3算法最大的区别就是特征选择上有所不同，一个是基于信息增益比，一个是基于信息增益。

之所以这样做是因为信息增益倾向于选择取值比较多的特征(特征越多，条件熵(特征划分后的类别变量的熵)越小，信息增益就越大)；因此在信息增益下面加一个分母，该分母是当前所选特征的熵，注意：这里而不是类别变量的熵了。

这样就构成了新的特征选择准则，叫做信息增益比。为什么加了这样一个分母就会消除ID3算法倾向于选择取值较多的特征呢？

因为特征取值越多，该特征的熵就越大，分母也就越大，所以信息增益比就会减小，而不是像信息增益那样增大了，一定程度消除了算法对特征取值范围的影响。

实现

在算法实现上，C4.5算法只是修改了信息增益计算的函数calcShannonEntOfFeature和最优特征选择函数chooseBestFeatureToSplit。

calcShannonEntOfFeature在ID3的calcShannonEnt函数上加了个参数feat，ID3中该函数只用计算类别变量的熵，而calcShannonEntOfFeature可以计算指定特征或者类别变量的熵。

chooseBestFeatureToSplit函数在计算好信息增益后，同时计算了当前特征的熵IV，然后相除得到信息增益比，以最大信息增益比作为最优特征。

在划分数据的时候，有可能出现特征取同一个值，那么该特征的熵为0，同时信息增益也为0(类别变量划分前后一样，因为特征只有一个取值)，0/0没有意义，可以跳过该特征。

#coding=utf-8import operatorfrom math import logimport timeimport os, sysimport stringdef createDataSet(trainDataFile): print trainDataFile dataSet = [] try: fin = open(trainDataFile) for line in fin:  line = line.strip()  cols = line.split('/t')  row = [cols[1], cols[2], cols[3], cols[4], cols[5], cols[6], cols[7], cols[8], cols[9], cols[10], cols[0]]  dataSet.append(row)  #print row except: print 'Usage xxx.py trainDataFilePath' sys.exit() labels = ['cip1', 'cip2', 'cip3', 'cip4', 'sip1', 'sip2', 'sip3', 'sip4', 'sport', 'domain'] print 'dataSetlen', len(dataSet) return dataSet, labels#calc shannon entropy of label or featuredef calcShannonEntOfFeature(dataSet, feat): numEntries = len(dataSet) labelCounts = {} for feaVec in dataSet: currentLabel = feaVec[feat] if currentLabel not in labelCounts:  labelCounts[currentLabel] = 0 labelCounts[currentLabel] += 1 shannonEnt = 0.0 for key in labelCounts: prob = float(labelCounts[key])/numEntries shannonEnt -= prob * log(prob, 2) return shannonEntdef splitDataSet(dataSet, axis, value): retDataSet = [] for featVec in dataSet: if featVec[axis] == value:  reducedFeatVec = featVec[:axis]  reducedFeatVec.extend(featVec[axis+1:])  retDataSet.append(reducedFeatVec) return retDataSet def chooseBestFeatureToSplit(dataSet): numFeatures = len(dataSet[0]) - 1 #last col is label baseEntropy = calcShannonEntOfFeature(dataSet, -1) bestInfoGainRate = 0.0 bestFeature = -1 for i in range(numFeatures): featList = [example[i] for example in dataSet] uniqueVals = set(featList) newEntropy = 0.0 for value in uniqueVals:  subDataSet = splitDataSet(dataSet, i, value)  prob = len(subDataSet) / float(len(dataSet))  newEntropy += prob *calcShannonEntOfFeature(subDataSet, -1) #calc conditional entropy infoGain = baseEntropy - newEntropy 　　 iv = calcShannonEntOfFeature(dataSet, i) if(iv == 0): #value of the feature is all same,infoGain and iv all equal 0, skip the feature continue 　　 infoGainRate = infoGain / iv if infoGainRate > bestInfoGainRate:  bestInfoGainRate = infoGainRate  bestFeature = i return bestFeature  #feature is exhaustive, reture what you want labeldef majorityCnt(classList): classCount = {} for vote in classList: if vote not in classCount.keys():  classCount[vote] = 0 classCount[vote] += 1 return max(classCount)   def createTree(dataSet, labels): classList = [example[-1] for example in dataSet] if classList.count(classList[0]) ==len(classList): #all data is the same label return classList[0] if len(dataSet[0]) == 1: #all feature is exhaustive return majorityCnt(classList) bestFeat = chooseBestFeatureToSplit(dataSet) bestFeatLabel = labels[bestFeat] if(bestFeat == -1): #特征一样，但类别不一样，即类别与特征不相关，随机选第一个类别做分类结果 return classList[0]  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 def main(): if(len(sys.argv) < 3): print 'Usage xxx.py trainSet outputTreeFile' sys.exit() data,label = createDataSet(sys.argv[1]) t1 = time.clock() myTree = createTree(data,label) t2 = time.clock() fout = open(sys.argv[2], 'w') fout.write(str(myTree)) fout.close() print 'execute for ',t2-t1if __name__=='__main__': main()