• 决策树基于ID3算法


    
    from math import log
    import operator
    
    """
    使用ID3算法划分数据集,ID3算法可以用于划分标称型数据集
    
    决策树分类器就像带有终止块的流程图,终止块表示分类结果。
    开始处理数据集时,首先需要测量集合中数据的不一致,
    然后寻找最优方案划分数据集,直到数据集中的所有数据属于同一分类
    """
    def createDataSet():
        """ 不浮出水面是否可以生存     是否有脚蹼     属于鱼类 """
        dataSet = [[1, 1, 'yes'],
                   [1, 1, 'yes'],
                   [1, 0, 'no'],
                   [0, 1, 'no'],
                   [0, 1, 'no']]
        labels = ['no surfacing', 'flippers']
        #change to discrete values
        return dataSet, labels
    
    def calcShannonEnt(dataSet):
        """计算信息熵(香农熵)
        H(x) = -∑P(xi)log(2,P(xi)) (i=1,2,..n)
        H 表示信息熵
        P 表示某种语言文字的字符出现的概率
        LOG2是以二为底的对数,用的是二进制,信息熵的单位是比特(BIT,即二进制的0和1)
    
        熵也可以作为一个系统的混乱程度的标准
    
        另一种:基尼不纯度 也可以作为衡量系统混乱程度的标准
            基尼 = 1 − ∑P(xi)^2 (i=1,2,..n)
    
        主要的区别就是基尼中把logP(xi)换成了P(xi),相比于熵,基尼反而有计算量小的优势
        """
        numEntries = len(dataSet)  # 计算数据集中实例的总数
        labelCounts = {}  # 类别次数字典
        for featVec in dataSet:  # the the number of unique elements and their occurance
            currentLabel = featVec[-1]
            if currentLabel not in labelCounts.keys(): labelCounts[currentLabel] = 0
            labelCounts[currentLabel] += 1
        shannonEnt = 0.0   # 信息熵
        for key in labelCounts:
            prob = float(labelCounts[key])/numEntries  # 类别出现的频率
            shannonEnt -= prob * log(prob, 2)  # log base 2
        return shannonEnt
    
    def splitDataSet(dataSet, axis, value):
        """
        按照给定特征划分数据集  第axis个特征是value的数据集
        :param dataSet: 待划分的数据集
        :param axis: 划分数据集的特征索引
        :param value: 需要返回的特征的值
        :return: 符合的数据集
        """
        retDataSet = []
        for featVec in dataSet:
            if featVec[axis] == value:
                reducedFeatVec = featVec[:axis]     #chop out axis used for splitting
                reducedFeatVec.extend(featVec[axis+1:])  # [,,,]
                retDataSet.append(reducedFeatVec)  # [,,[]]
        return retDataSet
    
    def chooseBestFeatureToSplit(dataSet):
        """
        选择最好的数据集划分方式
        选择熵最小的,也就是数据最纯的
        :param dataSet:
        :return: 最好特征划分的索引值
        """
        numFeatures = len(dataSet[0]) - 1  # 特征数,最后一个为标签 #the last column is used for the labels
        baseEntropy = calcShannonEnt(dataSet)
        bestInfoGain = 0.0; bestFeature = -1
        for i in range(numFeatures):        #iterate over all the features
            featList = [example[i] for example in dataSet]#create a list of all the examples of this feature
            uniqueVals = set(featList)  # 当前特征中所有的唯一属性值集合  #get a set of unique values
            newEntropy = 0.0
            for value in uniqueVals:  # 遍历当前特征中所有的唯一属性值,对每个特征划分一次数据集
                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 entropy
            if (infoGain > bestInfoGain):       #compare this to the best gain so far
                bestInfoGain = infoGain         #if better than current best, set to best
                bestFeature = i
        return bestFeature                      #returns an integer
    
    def majorityCnt(classList):
        """出现次数最多的分类名称"""
        classCount={}
        for vote in classList:
            if vote not in classCount.keys(): classCount[vote] = 0
            classCount[vote] += 1
        sortedClassCount = sorted(classCount.items(), key=operator.itemgetter(1), reverse=True)
        return sortedClassCount[0][0]
    
    def createTree(dataSet, labels):
        """
        构建决策数
        :param dataSet: 数据集
        :param labels: 标签列表
        :return: 树
        """
        print(dataSet)
        classList = [example[-1] for example in dataSet]  # 包含数据集的所有类标签
        if classList.count(classList[0]) == len(classList):  # 所有类标签完全相同,直接返回该标签
            return classList[0] #stop splitting when all of the classes are equal
        #
        if len(dataSet[0]) == 1: #stop splitting when there are no more features in dataSet
            return majorityCnt(classList)
        bestFeat = chooseBestFeatureToSplit(dataSet)  # 最好特征划分的索引值
        bestFeatLabel = labels[bestFeat]  # 最好的特征划分标签
        myTree = {bestFeatLabel:{}}
        del(labels[bestFeat])  # 删除已经使用的标签
        featValues = [example[bestFeat] for example in dataSet]
        uniqueVals = set(featValues)
        for value in uniqueVals:
            subLabels = labels[:]  # 拷贝标签     #copy all of labels, so trees don't mess up existing labels
            myTree[bestFeatLabel][value] = createTree(splitDataSet(dataSet, bestFeat, value), subLabels)
        return myTree
    
    
    def classify(inputTree, featLabels, testVec):
        """
        决策树分类
        :param inputTree: 决策树
        :param featLabels: 标签
        :param testVec:
        :return:
        """
        firstStr = list(inputTree)[0]   # 相当于list(inputTree.keys())[0]
        secondDict = inputTree[firstStr]
        featIndex = featLabels.index(firstStr)
        key = testVec[featIndex]
        valueOfFeat = secondDict[key]
        if isinstance(valueOfFeat, dict):
            classLabel = classify(valueOfFeat, featLabels, testVec)
        else: classLabel = valueOfFeat
        return classLabel
    
    """
     存储决策树
    """
    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)
    
    
    if __name__ == '__main__':
        # 熵越高,则混合的数据也越多
        dataSet, labels = createDataSet()
        # dataSet[0][-1] = 'maybe'
        # print(calcShannonEnt(dataSet))
        # print(splitDataSet(dataSet,1,0))
        # print(splitDataSet(dataSet,0,0))
        # print(chooseBestFeatureToSplit(dataSet))  # 第0个特征是最好的用于划分数据集的特征
    
        """ 不浮出水面是否可以生存   是否有脚蹼   属于鱼类
        dataSet = [[1, 1, 'yes'],
                   [1, 1, 'yes'],
                   [1, 0, 'no'],
                   [0, 1, 'no'],
                   [0, 1, 'no']]
        
        按照第一个特征属性划分数据
            特征是1的:两个鱼类,一个不是鱼类
            特征是0的:都是鱼类
        按照第二个特征属性划分数据
            特征是1的:两个鱼类,两个不是鱼类
            特征是0的:都不是鱼类   
        比较得出第一种分组的输出结果较好 
        """
        # print(createTree(dataSet, labels))  # {'no surfacing': {0: 'no', 1: {'flippers': {0: 'no', 1: 'yes'}}}}
    
    
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  • 原文地址:https://www.cnblogs.com/fly-book/p/14210792.html
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