决策树分类算法-ID3

 

实现的功能：

 1、对数值型数据和标称型数据 进行分类

 2、可将决策树pickle于txt文件中

但是本人在scikit_learn包中 暂时没成功处理标称型数据

trees.py

'''
Created on 2018年7月27日

@author: hcl
'''
from math import log
import operator
import numpy as np

def createDataSet():
    '''
    产生测试数据
    数据特征：[x1,x2,y] 两个特征 一个输出
    那么label 代表的是特征的名字
    '''
    dataSet = [[1, 1, 'yes'],
               [1, 1, 'yes'],
               [1, 0, 'no'],
               [0, 1, 'no'],
               [0, 1, 'no']]
    feature_names = ['no surfacing','flippers']    
    return dataSet, feature_names

def calcShannonEnt(dataSet):
    '''
    计算给定数据集的香农熵
    '''
    numEntries = len(dataSet)
    labelCounts = {}
    #统计每个类别出现的次数，保存在字典labelCounts中
    for featVec in dataSet: 
        currentLabel = featVec[-1]
        if currentLabel not in labelCounts.keys(): labelCounts[currentLabel] = 0
        labelCounts[currentLabel] += 1 #如果当前键值不存在，则扩展字典并将当前键值加入字典
    shannonEnt = 0.0
    for key in labelCounts:
        #使用所有类标签的发生频率计算类别出现的概率
        prob = float(labelCounts[key])/numEntries
        #用这个概率计算香农熵
        shannonEnt -= prob * log(prob,2) #取2为底的对数
    return shannonEnt

def splitDataSet(dataSet, axis, value):
    '''
    按照给定特征划分数据集
    dataSet：待划分的数据集
    axis：   划分数据集的第axis个特征
    value：  特征的返回值（比较值）
    '''
    retDataSet = []
    #遍历数据集中的每个元素，一旦发现符合要求的值，则将其添加到新创建的列表中
    for featVec in dataSet:
        if featVec[axis] == value:
            
            #下面两句代表去除了featVec[axis]
            reducedFeatVec = featVec[:axis]
            reducedFeatVec.extend(featVec[axis+1:])
            
            #再将去除后的值进行追加至新数据集
            retDataSet.append(reducedFeatVec)
    
    #返回去除 第axis特征值 = value 以后的数据集        
    return retDataSet

def chooseBestFeatureToSplit(dataSet):
    '''
    选择最好的数据集划分方式
    输入：数据集
    输出：最优分类的特征的index
    '''
    #计算特征数量
    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 取出第i列的特征集合 并做去重处理
        featList = [example[i] for example in dataSet]
        uniqueVals = set(featList)

        #计算每种划分方式的信息熵
        newEntropy = 0.0
        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，即infoGain越大划分效果越好
        if (infoGain > bestInfoGain):
            bestInfoGain = infoGain
            bestFeature = i
    return bestFeature

def majorityCnt(classList):
    '''
    投票表决函数
    输入classList:标签集合，本例为：['yes', 'yes', 'no', 'no', 'no']
    输出：得票数最多的分类名称
    '''
    classCount={}
    for vote in classList:
        if vote not in classCount.keys(): classCount[vote] = 0
        classCount[vote] += 1
    #把分类结果进行排序，然后返回得票数最多的分类结果
    sortedClassCount = sorted(classCount.items(), key=operator.itemgetter(1), reverse=True)
    return sortedClassCount[0][0]

def createTree(dataSet,labels):
    '''
    创建树
    输入：数据集和标签列表 ;要对首次传入labels值时进行copy 因为后面出现了del(label)的操作，对原始数据有影响
    输出：树的所有信息
    '''
    
    # classList为数据集的所有类标签
    classList = [example[-1] for example in dataSet]
    # 停止条件1:所有类标签完全相同，直接返回该类标签 如果 第一个标签长的长度 == 总标签的长度  --》 标签完全相同
    if classList.count(classList[0]) == len(classList): 
        return classList[0]
    # 停止条件2:遍历完所有特征时仍不能将数据集划分成仅包含唯一类别的分组，则返回出现次数最多的类标签

    if len(dataSet[0]) == 1:
        return majorityCnt(classList)
    # 选择最优分类特征
    bestFeat = chooseBestFeatureToSplit(dataSet)
    bestFeatName = labels[bestFeat]

    # myTree存储树的所有信息
    myTree = {bestFeatName:{}}
    # 以下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
    for value in uniqueVals:
        subLabels = labels[:]
        myTree[bestFeatName][value] = createTree(splitDataSet(dataSet, bestFeat, value),subLabels)
    return myTree

def classify(inputTree,feature_names,testVec):
    '''
    决策树的分类函数
    inputTree:训练好的树信息
    feature_names:标签列表
    testVec:测试向量
    '''
    # 在2.7中，找到key所对应的第一个元素为：firstStr = myTree.keys()[0]，
    # 这在3.4中运行会报错：‘dict_keys‘ object does not support indexing，这是因为python3改变了dict.keys,
    # 返回的是dict_keys对象,支持iterable 但不支持indexable，
    # 我们可以将其明确的转化成list，则此项功能在3中应这样实现：
    firstSides = list(inputTree.keys()) #['no surfacing'] ; ['flippers']
    firstStr = firstSides[0]            #no surfacing     ; flippers
    secondDict = inputTree[firstStr]    #{0: 'no', 1: {'flippers': {0: 'no', 1: 'yes'}}}  ; {0: 'no', 1: 'yes'}
    # 将标签字符串转换成索引
    featIndex = feature_names.index(firstStr)  # 0； 1
    key = testVec[featIndex]                  # 1； 0
    valueOfFeat = secondDict[key]            # {'flippers': {0: 'no', 1: 'yes'}}；no     
    # 递归遍历整棵树，比较testVec变量中的值与树节点的值，如果到达叶子节点，则返回当前节点的分类标签
    # isinstance用于判断对象的类型
    if isinstance(valueOfFeat, dict): 
        classLabel = classify(valueOfFeat, feature_names, testVec)
    else: 
        classLabel = valueOfFeat
    return classLabel

def storeTree(inputTree,filename):
    '''
    使用pickle模块存储决策树
    '''
    import pickle
    fw = open(filename,'wb+')
    pickle.dump(inputTree,fw)
    fw.close()
    
def grabTree(filename):
    '''
    导入决策树模型
    '''
    import pickle
    fr = open(filename,'rb')
    return pickle.load(fr)

if __name__== "__main__":  
    
    dataSet,feature_names = createDataSet()
    myTree = createTree(dataSet, feature_names.copy())
    ans = classify(myTree,feature_names,[1,0])
    print(ans)
#     storeTree(myTree,'mt.txt')
#     myTree2 = grabTree('mt.txt')
#     print(myTree2)

通过matlibplot 绘制决策树

trees_plot.py

'''
Created on 2018年7月27日

@author: hcl
'''
import matplotlib.pyplot as plt

# 定义文本框和箭头格式
decisionNode = dict(boxstyle="sawtooth", fc="0.8")
leafNode = dict(boxstyle="round4", fc="0.8")
arrow_args = dict(arrowstyle="<-")

# 绘制带箭头的注释
def plotNode(nodeTxt, centerPt, parentPt, nodeType):
    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(inTree):
    '''
    绘树主函数
    '''
    fig = plt.figure(1, facecolor='white')
    fig.clf()
    # 设置坐标轴数据
    axprops = dict(xticks=[], yticks=[])
    # 无坐标轴
    createPlot.ax1 = plt.subplot(111, frameon=False, **axprops)
    # 带坐标轴
#    createPlot.ax1 = plt.subplot(111, frameon=False)
    plotTree.totalW = float(getNumLeafs(inTree))
    plotTree.totalD = float(getTreeDepth(inTree))
    # 两个全局变量plotTree.xOff和plotTree.yOff追踪已经绘制的节点位置，
    # 以及放置下一个节点的恰当位置
    plotTree.xOff = -0.5/plotTree.totalW; plotTree.yOff = 1.0;
    plotTree(inTree, (0.5,1.0), '')
    plt.show()

def getNumLeafs(myTree):
    '''
    获取叶节点的数目
    '''
    numLeafs = 0
    firstSides = list(myTree.keys())
    firstStr = firstSides[0]
    secondDict = myTree[firstStr]
    for key in secondDict.keys():
        # 判断节点是否为字典来以此判断是否为叶子节点
        if type(secondDict[key]).__name__=='dict':
            numLeafs += getNumLeafs(secondDict[key])
        else:   numLeafs +=1
    return numLeafs

def getTreeDepth(myTree):
    '''
    获取树的层数
    '''
    maxDepth = 0
    firstSides = list(myTree.keys())
    firstStr = firstSides[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 plotMidText(cntrPt, parentPt, txtString):
    '''
    计算父节点和子节点的中间位置，并在此处添加简单的文本标签信息
    '''
    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 plotTree(myTree, parentPt, nodeTxt):#if the first key tells you what feat was split on
    # 计算宽与高
    numLeafs = getNumLeafs(myTree)  #this determines the x width of this tree
    depth = getTreeDepth(myTree)
    firstSides = list(myTree.keys())
    firstStr = firstSides[0]
    cntrPt = (plotTree.xOff + (1.0 + float(numLeafs))/2.0/plotTree.totalW, plotTree.yOff)
    # 标记子节点属性值    
    plotMidText(cntrPt, parentPt, nodeTxt)
    plotNode(firstStr, cntrPt, parentPt, decisionNode)
    secondDict = myTree[firstStr]
    # 减少y偏移
    plotTree.yOff = plotTree.yOff - 1.0/plotTree.totalD
    for key in secondDict.keys():
        if type(secondDict[key]).__name__=='dict':#test to see if the nodes are dictonaires, if not they are leaf nodes   
            plotTree(secondDict[key],cntrPt,str(key))        #recursion
        else:   #it's a leaf node print the leaf node
            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 retrieveTree(i):
    '''
    保存了树的测试数据
    '''
    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]

if __name__== "__main__":  
    '''
    绘制树
    '''
    createPlot(retrieveTree(0))

输出：

预测隐形眼睛分类

数据集：

输入   test.py

'''
Created on 2018年7月26日

@author: hcl
'''
import trees
import trees_plot

fr = open('lenses.txt')
lenses = [inst.strip().split('\t') for inst in fr.readlines()]
lensesLabels=['age','prescript','astigmatic','tearRate']
lensesTree = trees.createTree(lenses,lensesLabels)
trees_plot.createPlot(lensesTree)

输出：