机器学习实战---决策树自编程实现（python）

1. ID3 决策树实现

'''
Decision Tree Source Code for Machine Learning in Action Ch. 3
'''
from math import log
import operator

def createDataSet():
    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

"""
func:计算给定数据集的熵

param:
    dataset: 数据集；

return:
    shannonEnt: 给定数据集的熵
"""
def calcShannonEnt(dataSet):
    numEntries = len(dataSet)
    labelCounts = {}
    for featVec in dataSet: #the the number of unique elements and their occurance
        # 当前标签为特征向量的最后一个特征，即类别标签
        currentLabel = featVec[-1]
        # 计算该特征的值对应的出现的次数
        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) #log base 2
    return shannonEnt

# ; 
"""
func:按照给定特征划分数据集

param:
    dataset: 待划分的数据集；
    axis:划分数据集的特征;
    value: 划分数据集的特征的值；

return:
    retDataSet: 划分后的数据集，即特征axis等于给定value，且不包括特征axis列的数据集；
"""
def splitDataSet(dataSet, axis, value):
    # 目的：防止修改原始数据集
    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

"""
func:选择最好的数据集划分方式

param:
    dataset:待划分的数据集

return:
    bestFeature:信息增益最大的特征，注意这里为该特征所在标签列表的索引；
"""
def chooseBestFeatureToSplit(dataSet):
    numFeatures = len(dataSet[0]) - 1      #the last column is used for the labels
    # 计算经验熵 P(D)
    baseEntropy = calcShannonEnt(dataSet)
    bestInfoGain = 0.0; bestFeature = -1
    for i in range(numFeatures):        #iterate over all the features
        # 创建唯一的分类标签列表
        featList = [example[i] for example in dataSet]#create a list of all the examples of this feature
        uniqueVals = set(featList)       #get a set of unique values
        newEntropy = 0.0
        for value in uniqueVals:
            subDataSet = splitDataSet(dataSet, i, value)
            prob = len(subDataSet)/float(len(dataSet))
            # 计算条件经验熵 P(D|Ck)，Ck为某个特征
            newEntropy += prob * calcShannonEnt(subDataSet)     
        # 计算信息增益
        infoGain = baseEntropy - newEntropy     #calculate the info gain; ie reduction in entropy
        # 计算最好的信息增益，获取最有的特征
        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

"""
func:多数表决法，当数据集已经处理完所有属性，
但是类标签依然不是唯一的，依照此方法定义该叶子节点。

param:
    classList: 分类名称的列表

return:
    sortedClassCount[0][0]: 该叶子节点下的所有类标签中出现次数最多的类标签名称
"""
def majorityCnt(classList):
    # 字典：键值为calssList中唯一值，字典对象为每个类标签中出现的次数
    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]

"""
func:创建决策树

param:
    dataset:数据集；
    labels:标签列表；

return:
    myTree: 返回创建好的决策树
"""
def createTree(dataSet,labels):
    # 类别列表
    classList = [example[-1] for example in dataSet]
    # 类别完全相同则停止继续划分
    if classList.count(classList[0]) == len(classList): 
        return classList[0]#stop splitting when all of the classes are equal
    # 遍历完所有特征时返回出现次数最多的；为什么这里可以用len(dataSet[0]) == 1 作为遍历完所有特征的条件? 
    # 
    if len(dataSet[0]) == 1: #stop splitting when there are no more features in dataSet
        return majorityCnt(classList)

    # 划分最优的特征在标签列表上的索引
    bestFeat = chooseBestFeatureToSplit(dataSet)
    # 划分最优的特征标签
    bestFeatLabel = labels[bestFeat]
    myTree = {bestFeatLabel:{}}
    del(labels[bestFeat])

    # 获取列表包含的所有属性值
    featValues = [example[bestFeat] for example in dataSet]
    uniqueVals = set(featValues)
    
    # 在每个数据集划分上递归调用createTree
    for value in uniqueVals:
        # 复制类标签，因为参数时按照引用方式传递的。
        # 若不进行复制，则在递归函数改变了label值后，此处的label也将改变，会造成类标签的混乱
        subLabels = labels[:]       #copy all of labels, so trees don't mess up existing labels
        myTree[bestFeatLabel][value] = createTree(splitDataSet(dataSet, bestFeat, value),subLabels)
    return myTree                            

"""
func: 对测试数据进行分类

param:
    inputTree:训练好的决策树；
    featLabels:标签列表；
    testVec:待测试的数据

return:
    classLabel: 类标签，即该测试数据所属的类别
"""
def classify(inputTree,featLabels,testVec):
    # 获取决策树的第一个特征标签
    firstStr = list(inputTree)[0]
    # 获取第一个特征标签对应的字典值，该值为一个字典, 键值为第一个特征标签的属性。
    # 如{0: 'no', 1: {'flippers': {0: 'no', 1: 'yes'}}}
    secondDict = inputTree[firstStr]
    # 将标签字符串转换为索引
    featIndex = featLabels.index(firstStr)
    # 获取测试数据该索引对应的属性。
    key = testVec[featIndex]
    # 获取特征标签对应的值
    valueOfFeat = secondDict[key]
    # 判断该值是否属于字典，若属于，则递归调用classify，若不属于，则返回对应的分类标签
    if isinstance(valueOfFeat, dict): 
        classLabel = classify(valueOfFeat, featLabels, testVec)
    else: classLabel = valueOfFeat
    return classLabel

"""
func:存储创建好的决策树

param:
    inputTree:创建好的决策树；
    filename:文件名

return:
    None
"""
def storeTree(inputTree,filename):
    import pickle
    fw = open(filename,'w')
    # 序列化
    pickle.dump(inputTree,fw)
    fw.close()

"""
func:打开创建好的决策树

param:
    filename:文件名

return:
    决策树
"""
def grabTree(filename):
    import pickle
    fr = open(filename)
    return pickle.load(fr)
    
"""
fr = open("lenses.txt")
lenses = [inst.strip().split('\t') for inst in fr.readlines()]
lensesLabels = ['age', 'prescript','astigmatic','tearRate']
lensesTree = trees.createTree(lenses,lensesLabels)
lensesTree
treePlotter.createPlot(lensesTree)
"""

2. 决策树可视化

'''
Created on Oct 14, 2010

@author: Peter Harrington

绘制决策树
'''

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="<-")

"""
func:获取叶节点的数目

param:
    myTree: 决策树

return:
    numLeafs: 叶节点的数目
"""
def getNumLeafs(myTree):
    numLeafs = 0
    firstStr = list(myTree)[0]
    secondDict = myTree[firstStr]
    for key in secondDict.keys():
        # 测试节点的数据类型是否是字典，若为字典则递归调用函数getNumLeafs
        if type(secondDict[key]).__name__=='dict':#test to see if the nodes are dictonaires, if not they are leaf nodes
            numLeafs += getNumLeafs(secondDict[key])
        else:   numLeafs +=1
    return numLeafs

"""
func:获取树的层数

param:
    myTree: 决策树

return:
    maxDepth: 树的层数

"""
def getTreeDepth(myTree):
    maxDepth = 0
    firstStr = list(myTree)[0]
    secondDict = myTree[firstStr]
    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
            thisDepth = 1 + getTreeDepth(secondDict[key])
        else:   thisDepth = 1
        if thisDepth > maxDepth: maxDepth = thisDepth
    return maxDepth

"""
func:绘制带箭头的注解

param:
    nodeTxt: 节点注释
    centerPt：箭头位置
    panrentPt: 键尾位置
    nodeType: 节点类型

return:
    None
"""
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 )

"""
func: 在父子节点间填充文本信息

param:
    cntrPt: 
    parentPt: 箭尾坐标
    txtString: 文本信息

return:
    None
"""    
def plotMidText(cntrPt, parentPt, txtString):
    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)

"""
func: 绘制决策树

param:
    myTree: 决策树
    parentPt: 箭尾坐标
    nodeTxt: 节点中的文本信息

return:
    None
""" 
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)
    firstStr = list(myTree)[0]     #the text label for this node should be this
    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
#if you do get a dictonary you know it's a tree, and the first element will be another dict

"""
func: 创建并显示图片

param:
    inTree：创建好的决策树

return:
    None
""" 
def createPlot(inTree):
    fig = plt.figure(1, facecolor='white')
    fig.clf()
    axprops = dict(xticks=[], yticks=[])
    createPlot.ax1 = plt.subplot(111, frameon=False, **axprops)    #no ticks
    #createPlot.ax1 = plt.subplot(111, frameon=False) #ticks for demo puropses 
    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()

#def createPlot():
#    fig = plt.figure(1, facecolor='white')
#    fig.clf()
#    createPlot.ax1 = plt.subplot(111, frameon=False) #ticks for demo puropses 
#    plotNode('a decision node', (0.5, 0.1), (0.1, 0.5), decisionNode)
#    plotNode('a leaf node', (0.8, 0.1), (0.3, 0.8), leafNode)
#    plt.show()

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]

#createPlot(thisTree)