首页 > 程序开发 > 综合编程 > 其他综合 >

机器学习实战03决策树

2017-02-08

机器学习实战03决策树:决策树是一类常见的机器学习方法,决策树是基于树结构来进行决策的。一般的,一颗决策树包含一个根节点,若干个内部节点和若干个叶节点,叶节点对应于决策结果。

机器学习实战03决策树:决策树是一类常见的机器学习方法,决策树是基于树结构来进行决策的。一般的,一颗决策树包含一个根节点,若干个内部节点和若干个叶节点,叶节点对应于决策结果。

其他每个节点则对应于一个属性测试;每个节点包含的样本集合根据属性测试的结果被划分到子节点中;根节点包含样本全集,从根节点到每个叶节点的路径对应了一个判定测试序列。

2、决策树算法
(1)计算信息熵

from math import log
import operator

#计算给定数据集的信息熵
def calcShannonEnt(dataSet):
    numEntries = len(dataSet)
    labelCounts = {}
    for featVec in dataSet: 
        currentLabel = featVec[-1]
        if currentLabel not in labelCounts.keys(): #为所有可能分类创建字典
                labelCounts[currentLabel] = 0  
                labelCounts[currentLabel] += 1
    shannonEnt = 0.0
    for key in labelCounts:
        prob = float(labelCounts[key])/numEntries
        shannonEnt -= prob * log(prob,2)  #以2为底数求对数 
    return shannonEnt  #熵越高,混合的数据也越多

#创建数据集,三列(特征、特征的值,预期结果)
def createDataSet():
    dataSet = [[1, 1, 'yes'],
               [1, 1, 'yes'],
               [1, 0, 'no'],
               [0, 1, 'no'],
               [0, 1, 'no']]
    labels = ['no surfacing','flippers']
    return dataSet, labels

dataSet,labels = createDataSet()
print "dataSet:"
print dataSet
print "\n"

print "信息熵:"
print calcShannonEnt(dataSet)

信息熵的计算结果:

dataSet:
[[1, 1, 'yes'], [1, 1, 'yes'], [1, 0, 'no'], [0, 1, 'no'], [0, 1, 'no']]


信息熵:
0.928771237955

(2)划分数据集

"""
extend和append的区别:
>>> a=[1,2,3]
>>> b=[2,3,4]
>>> a.append(b)
>>> a
[1, 2, 3, [2, 3, 4]]

>>> a=[1,2,3]
>>> b=[2,3,4]
>>> a.extend(b)
>>> a
[1, 2, 3, 2, 3, 4]
"""

from math import log
import operator

def calcShannonEnt(dataSet):#计算信息熵
    numEntries = len(dataSet)
    labelCounts = {}
    for featVec in dataSet: 
        currentLabel = featVec[-1]
        if currentLabel not in labelCounts.keys(): #为所有可能分类创建字典
            labelCounts[currentLabel] = 0
            labelCounts[currentLabel] += 1
    shannonEnt = 0.0
    for key in labelCounts:
        prob = float(labelCounts[key])/numEntries
        shannonEnt -= prob * log(prob,2)  #以2为底数求对数 
    return shannonEnt

def createDataSet():#创建数据集,三列(特征、特征的值,预期结果)
    dataSet = [[1, 1, 'yes'],
               [1, 1, 'yes'],
               [1, 0, 'no'],
               [0, 1, 'no'],
               [0, 1, 'no']]
    labels = ['no surfacing','flippers']
    return dataSet, labels

def splitDataSet(dataSet, axis, value):#划分数据集,dataSet数据集,axis划分数据集的特征,value需要返回的特征的值
    retDataSet = []
    for featVec in dataSet:
        if featVec[axis] == value:
            reducedFeatVec = featVec[:axis]     
            reducedFeatVec.extend(featVec[axis+1:])
            retDataSet.append(reducedFeatVec)
    return retDataSet

dataSet,labels = createDataSet()
print "dataSet:"
print dataSet

retDataSet = splitDataSet(dataSet,1,0)
print "splitDataSet(dataSet,1,0):"
print retDataSet 

retDataSet = splitDataSet(dataSet,1,1)
print "splitDataSet(dataSet,1,1):"
print retDataSet

retDataSet = splitDataSet(dataSet,0,1)
print "splitDataSet(dataSet,0,1):"
print retDataSet

retDataSet = splitDataSet(dataSet,0,0)
print "splitDataSet(dataSet,0,0):"
print retDataSet

划分数据集结果:

dataSet:
[[1, 1, 'yes'], [1, 1, 'yes'], [1, 0, 'no'], [0, 1, 'no'], [0, 1, 'no']]
splitDataSet(dataSet,1,0):
[[1, 'no']]
splitDataSet(dataSet,1,1):
[[1, 'yes'], [1, 'yes'], [0, 'no'], [0, 'no']]
splitDataSet(dataSet,0,1):
[[1, 'yes'], [1, 'yes'], [0, 'no']]
splitDataSet(dataSet,0,0):
[[1, 'no'], [1, 'no']]

(3)选择最好的数据集划分方式

from math import log
import operator

def calcShannonEnt(dataSet):#计算信息熵
    numEntries = len(dataSet)
    labelCounts = {}
    for featVec in dataSet: 
        currentLabel = featVec[-1]
        if currentLabel not in labelCounts.keys(): #为所有可能分类创建字典
            labelCounts[currentLabel] = 0
            labelCounts[currentLabel] += 1
    shannonEnt = 0.0
    for key in labelCounts:
        prob = float(labelCounts[key])/numEntries
        shannonEnt -= prob * log(prob,2)  #以2为底数求对数 
    return shannonEnt

def createDataSet():#创建数据集,三列(特征、特征的值,预期结果)
    dataSet = [[1, 1, 'yes'],
               [1, 1, 'yes'],
               [1, 0, 'no'],
               [0, 1, 'no'],
               [0, 1, 'no']]
    labels = ['no surfacing','flippers']
    return dataSet, labels

def splitDataSet(dataSet, axis, value):#划分数据集,dataSet数据集,axis划分数据集的特征,value需要返回的特征的值
    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):#遍历整个数据集,循环计算信息熵和splitDataSet()函数,找到最好的特征划分方式
    numFeatures = len(dataSet[0]) - 1      #剩下的特征的个数
    baseEntropy = calcShannonEnt(dataSet)  #计算数据集的熵,放到baseEntropy中
    bestInfoGain = 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: #下面是计算每种划分方式的信息熵,特征i个,每个特征value个值
            subDataSet = splitDataSet(dataSet, i, value)
            prob = len(subDataSet)/float(len(dataSet))
            newEntropy += prob * calcShannonEnt(subDataSet)     
        infoGain = baseEntropy - newEntropy     #计算i个特征的信息熵
        print i  #输出特征值
        print infoGain #输出特征值的信息熵
        if (infoGain > bestInfoGain):      
            bestInfoGain = infoGain        
            bestFeature = i
    return bestFeature                      

dataSet,labels = createDataSet()
print "dataSet:"
print dataSet

bestFeature = chooseBestFeatureToSplit(dataSet)
print "bestFeature:"
print bestFeature

最好的划分方式:

dataSet:
[[1, 1, 'yes'], [1, 1, 'yes'], [1, 0, 'no'], [0, 1, 'no'], [0, 1, 'no']]
0
0.0947862376665
1
0.128771237955
bestFeature:
1

(4)递归构建决策树

得到原始的数据集,然后基于最好的属性值划分数据集,由于特征值可能多于两个,因此可能存在大于两个分支的数据集划分。
第一次划分之后,数据将被向下传递到树分支的下一个节点,在这个节点上,可以再次划分数据,采用递归的原则处理数据集。

递归结束的条件:
程序遍历完所有划分数据集的属性,或者每个分支下的所有实例都具有相同的分类。
如果所有实例具有相同的分类,则得到一个叶子节点或者终止块,任何到达叶子节点的数据必然属于叶子节点的分类。

(5)创建树的函数代码

#创建树,两个参数,数据集和标签列表
def createTree(dataSet,labels):
    classList = [example[-1] for example in dataSet] #classList变量包含了数据集的所有类标签
    if classList.count(classList[0]) == len(classList): 
        return classList[0] #类标签完全相同就停止继续划分
    if len(dataSet[0]) == 1: #遍历完所有特征时返回出现次数最多的
        return majorityCnt(classList)
    bestFeat = chooseBestFeatureToSplit(dataSet) #选取数据集中的最好特征存储在bestFeat中
    bestFeatLabel = labels[bestFeat]
    myTree = {bestFeatLabel:{}} #存储树的所有信息
    del(labels[bestFeat]) 
    featValues = [example[bestFeat] for example in dataSet]
    uniqueVals = set(featValues)
    for value in uniqueVals:
        subLabels = labels[:]       #复制类标签,并将其存储在新列表变量subLabels中
            #这样做的原因是:python语言中函数参数是列表类型时,参数是按照引用方式传递的,
            #为了保证每次调用函数createTree()时不改变原始列表的内容,使用新变量subLabels代替原始列表。
        myTree[bestFeatLabel][value] = createTree(splitDataSet(dataSet, bestFeat, value),subLabels)
        #递归调用createTree()函数,得到的返回值插入到字典变量myTree中
    return myTree  

其中当所有的特征都用完时,采用多数表决的方法来决定该叶子节点的分类,即该叶节点中属于某一类最多的样本数,那么我们就说该叶节点属于那一类。

def majorityCnt(classList):
    classCount={}
    for vote in classList:
        if vote not in classCount.keys(): classCount[vote] = 0
        classCount[vote] += 1
    sortedClassCount = sorted(classCount.iteritems(), key=operator.itemgetter(1), reverse=True)
    return sortedClassCount[0][0]

(6)使用决策树进行分类

def classify(inputTree,featLabels,testVec):#决策树的分类函数
    firstStr = 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

(7)决策树的存储

def storeTree(inputTree,filename): #存储决策树
    import pickle
    fw = open(filename,'w')
    pickle.dump(inputTree,fw) #pickle序列化对象,可以在磁盘上保存对象
    fw.close()

def grabTree(filename): #并在需要的时候将其读取出来  
    import pickle
    fr = open(filename)
    return pickle.load(fr)

3、Matplotlib注解
Matplotlib提供了一个注解工具annotations,它可以在数据图形上添加文本注释。

#encoding:utf-8
import matplotlib.pyplot as plt 

decisionNode = dict(boxstyle="sawtooth", fc="0.8")
leafNode = dict(boxstyle="round4", fc="0.8")
arrow_args = dict(arrowstyle="<-")

def getNumLeafs(myTree):#递归计算获取叶节点的数目
    numLeafs = 0
    firstStr = myTree.keys()[0]
    secondDict = myTree[firstStr]
    for key in secondDict.keys():
        if type(secondDict[key]).__name__==&#39;dict&#39;:#test to see if the nodes are dictonaires, if not they are leaf nodes
            numLeafs += getNumLeafs(secondDict[key])
        else:   numLeafs +=1
    return numLeafs

def getTreeDepth(myTree):#递归计算获取树的深度
    maxDepth = 0
    firstStr = myTree.keys()[0]
    secondDict = myTree[firstStr]
    for key in secondDict.keys():
        if type(secondDict[key]).__name__==&#39;dict&#39;:#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

def plotNode(nodeTxt, centerPt, parentPt, nodeType):#绘制带箭头的注解
    createPlot.ax1.annotate(nodeTxt, xy=parentPt,  xycoords=&#39;axes fraction&#39;,
             xytext=centerPt, textcoords=&#39;axes fraction&#39;,
             va="center", ha="center", bbox=nodeType, arrowprops=arrow_args )

def plotMidText(cntrPt, parentPt, txtString):#在父子节点间填充信息,绘制线上的标注
    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):#计算宽和高,递归,决定整个树图的绘制
    numLeafs = getNumLeafs(myTree)  
    depth = getTreeDepth(myTree)
    firstStr = myTree.keys()[0]     
    cntrPt = (plotTree.xOff + (1.0 + float(numLeafs))/2.0/plotTree.totalW, plotTree.yOff)
    plotMidText(cntrPt, parentPt, nodeTxt)
    plotNode(firstStr, cntrPt, parentPt, decisionNode)
    secondDict = myTree[firstStr]
    plotTree.yOff = plotTree.yOff - 1.0/plotTree.totalD
    for key in secondDict.keys():
        if type(secondDict[key]).__name__==&#39;dict&#39;:#test to see if the nodes are dictonaires, if not they are leaf nodes   
            plotTree(secondDict[key],cntrPt,str(key))        #recursion
        else:   #it&#39;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&#39;s a tree, and the first element will be another dict

def createPlot(inTree):#这里是真正的绘制,上面是逻辑绘制
    fig = plt.figure(1, facecolor=&#39;white&#39;)
    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), &#39;&#39;)
    plt.show()

"""
def createPlot():#简单绘制图形的节点和箭头
    fig = plt.figure(1, facecolor=&#39;white&#39;)
    fig.clf()
    createPlot.ax1 = plt.subplot(111, frameon=False) #ticks for demo puropses 
    plotNode(&#39;a decision node&#39;, (0.5, 0.1), (0.1, 0.5), decisionNode)
    plotNode(&#39;a leaf node&#39;, (0.8, 0.1), (0.3, 0.8), leafNode)
    plt.show()
"""

def retrieveTree(i):#预先存储树的信息
    listOfTrees =[{&#39;no surfacing&#39;: {0: &#39;no&#39;, 1: {&#39;flippers&#39;: {0: &#39;no&#39;, 1: &#39;yes&#39;}}}},
                  {&#39;no surfacing&#39;: {0: &#39;no&#39;, 1: {&#39;flippers&#39;: {0: {&#39;head&#39;: {0: &#39;no&#39;, 1: &#39;yes&#39;}}, 1: &#39;no&#39;}}}}
                  ]
    return listOfTrees[i]

createPlot(retrieveTree(1))

相关文章
最新文章
热点推荐