《机器学习实战》（三）决策树(decision trees)

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决策树的构造

优点: 计算复杂度不高，输出结果易于理解，对中间值的缺失不敏感，可以处理不相关特征数据。
缺点: 可能会产生过度匹配(overfitting)问题。
试用数据范围: 数值型和标称型。

信息增益：在划分数据集之前之后信息发生的变化。

符号xi的信息定义为: $l (x i) = - l o g 2 p (x i)$
其中p(xi)是选择该分类的概率
集合信息的度量方式称为香农熵或者简称为熵: $H = - \sum i = 1 n p (x i) l o g 2 p (x i)$

计算给定数据集的香农熵

from math import logdef calcShannonEnt(dataSet):    numEntries = len(dataSet)    labelCounts = {}    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    for key in labelCounts:        prob = float(labelCounts[key])/numEntries        shannonEnt -= prob * log(prob, 2)  # log base 2    return shannonEnt

熵越高，则混合的数据越多

按照给定特征划分数据集

def splitDataSet(dataSet, axis, value):    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

选择最好的数据集划分方式

def chooseBestFeatureToSplit(dataSet):    numFeatures = len(dataSet[0]) - 1      # 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)       # get a set of unique values        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     # 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

递归构建决策树

ID3算法

def createTree(dataSet, labels):    classList = [example[-1] for example in dataSet]    if classList.count(classList[0]) == len(classList):        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)    bestFeatLabel = labels[bestFeat]    myTree = {bestFeatLabel: {}}    del(labels[bestFeat])    featValues = [example[bestFeat] for example in dataSet]    uniqueVals = set(featValues)    for value in uniqueVals:        subLabels = labels[:]       # copy all of labels, so trees don't mess up existing labels        myTree[bestFeatLabel][value] = createTree(splitDataSet(dataSet, bestFeat, value), subLabels)    return myTreedef 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]

用Matplotlib注解绘制树形图

import matplotlib.pyplot as pltdecisionNode = dict(boxstyle="sawtooth", fc="0.8")leafNode = dict(boxstyle="round4", fc="0.8")arrow_args = dict(arrowstyle="<-")def getNumLeafs(myTree):    numLeafs = 0    firstStr = myTree.keys()[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            numLeafs += getNumLeafs(secondDict[key])        else:            numLeafs += 1    return numLeafsdef getTreeDepth(myTree):    maxDepth = 0    firstStr = myTree.keys()[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 maxDepthdef 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 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)def plotTree(myTree, parentPt, nodeTxt):  # if the first key tells you what feat was split on    numLeafs = getNumLeafs(myTree)  # this determines the x width of this tree    depth = getTreeDepth(myTree)    firstStr = myTree.keys()[0]     # the text label for this node should be this    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__ == '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 dictdef 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 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]

使用决策树的分类算法

def classify(inputTree, featLabels, testVec):    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

决策树的存储

在python中，一般可以使用pickle类来进行python对象的序列化，而cPickle提供了一个更快速简单的接口，如python文档所说的：“cPickle – A faster pickle”。
cPickle可以对任意一种类型的python对象进行序列化操作，比如list，dict，甚至是一个类的对象等。而所谓的序列化，我的粗浅的理解就是为了能够完整的保存并能够完全可逆的恢复。在cPickle中，主要有四个函数可以做这一工作，下面使用例子来介绍。
1 dump：将python对象序列化保存到本地的文件。

 import cPickle data = range(1000)cPickle.dump(data,open("test\\data.pkl","wb"))

dump函数需要指定两个参数，第一个是需要序列化的python对象名称，第二个是本地的文件，需要注意的是，在这里需要使用open函数打开一个文件，并指定“写”操作。

2 load：载入本地文件，恢复python对象
data = cPickle.load(open("test\\data.pkl","rb"))
同dump一样，这里需要使用open函数打开本地的一个文件，并指定“读”操作

3 dumps：将python对象序列化保存到一个字符串变量中。
data_string = cPickle.dumps(data)

4 loads：从字符串变量中载入python对象
代码如下:
data = cPickle.loads(data_string)

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