机器学习实战

机器学习实战-决策树

2017-04-18  本文已影响0人  mov觉得高数好难

第二章介绍的k-近邻算法可以完成很多分类任务,但是它最大的缺点就是无法给出数据内在的含义,决策树的主要优势在于数据形式非常容易理解。

决策树
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的问题
适用数据类型:数值型和标称型

先计算给定数据集的香农熵

在CSDN的Markdown中,第一行居然不居中,简书的是默认居中了

序号 不浮出水面是否可以生存 是否有脚蹼 属于鱼类
1 1 1 1
2 1 1 1
3 1 0 0
3 0 1 0
3 0 1 0
from math import log

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

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

测试数据集的熵

import trees
reload(trees)
myDat,labels = trees.createDataSet()
>>> myDat
[[1, 1, 'yes'], [1, 1, 'yes'], [1, 0, 'no'], [0, 1, 'no'], [0, 1, 'no']]
>>> trees.calcShannonEnt(myDat)
0.9709505944546686

熵越高,则混合的数据也越多
增加一个测试分类来测试熵的变化:

>>> myDat[0][-1]='maybe'
>>> myDat
[[1, 1, 'maybe'], [1, 1, 'yes'], [1, 0, 'no'], [0, 1, 'no'], [0, 1, 'no']]
>>> trees.calcShannonEnt(myDat)
1.3709505944546687

下面,开始划分数据集:

def splitDataSet(dataSet, axis, value):
    retDataSet = []
    for featVec in dataSet:
        #将符合要求的数据集添加到列表中
        if featVec[axis] == value:
            reducedFeatVec = featVec[:axis]
            reducedFeatVec.extend(featVec[axis+1:])
            retDataSet.append(reducedFeatVec)
    return retDataSet

本段代码使用了三个输入参数:待划分的数据集,划分数据集的特征,需要返回的特征的值。
然后在Python命令提示符内输入下述命令:

>>> myDat
[[1, 1, 'yes'], [1, 1, 'yes'], [1, 0, 'no'], [0, 1, 'no'], [0, 1, 'no']]
>>> trees.splitDataSet(myDat,0,1)
[[1, 'yes'], [1, 'yes'], [0, 'no']]
>>> trees.splitDataSet(myDat,0,0)
[[1, 'no'], [1, 'no']]
>>> 

接下来我们开始遍历整个数据集,循环计算香农熵和splitDataSet()函数,找到最好的特征划分方式。

def chooseBestFeatureToSplit(dataSet):
    numFeatures = len(dataSet[0]) - 1
    baseEntropy = calcShannonEnt(dataSet)#计算原始香农熵
    bastInfoGain = 0.0;baseFeature = -1
    for i in range(numFeatures):
        featList = [example[i] for example in dataSet]
        uniqueVals = set(featList)#创建分类标签集合
        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
        if (infoGain > bestInfogain):
            bestInfoGain = infoGain
            bestFeature = i
        return bestFeature

从列表中新建集合是Python语言得到列表中唯一元素的最快方法

下面开始测试上面代码的实际输出结果

>>> trees.chooseBestFeatureToSplit(myDat)
0
>>> myDat
[[1, 1, 'yes'], [1, 1, 'yes'], [1, 0, 'no'], [0, 1, 'no'], [0, 1, 'no']]

代码告诉我们,第0个特征是最好的用于划分数据集的特征

下面我们开始采用递归的方式处理数据集

如果数据集已经处理了所有属性,但是类标签依然不是唯一的,此时我们需要决定如何定义该叶子节点,在这种情况下,我们通常会采用多数表决的方法决定该叶子节点的分类

import operator
def majorityCnt(classList):
    classCount = {}
    for vote in classList:#统计数据字典classList每一个类标签出现的频率
        if vote not in classCount.keys(): classCount[vote] = 0
        classCount[vote] += 1
    sortedClassCount = sorted(classCount.iteritems(), key=operator.itemgette(1), reverse=True)#排序
    return sortedClassCount[0][0]#返回次数最多的

下面开始加入创建树的函数代码

def createTree(dataSet, labels):#数据集,标签列表
    classList = [example[-1] for example in dataSet]#最后一个属性加入列表变量classList
    #类别完全相同则停止划分
    if classList.count(classList[0]) == len(classList):
        return classList[0]
    if len(dataSet[0]) == 1:#遍历完所有特征,返回次数最多的类别
        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[:]
        myTree[bestFeatLabel][value] = createTree(splitDataSet(dataSet, bestFeat, value),subLabels)
    return myTree

subLabels = labels[:],这行代码复制了类标签,并将其存储在新列表变量subLabels中。之所以这样做,是因为在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。

下面是测试实际输出结果

#trees-1.py
import trees
reload(trees)
myDat,labels = trees.createDataSet()
myTree = trees.createTree(myDat,labels)
>>> myTree
{'no surfacing': {0: 'no', 1: {'flippers': {0: 'no', 1: 'yes'}}}}

下面开始学习使用Matplotlib画图

#-*- coding=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 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():
    fig = plt.figure(1,facecolor='white')
    fig.clf()
    createPlot.ax1 = plt.subplot(111,frameon=False)

    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()
>>> import treePlotter;treePlotter.createPlot()
函数plotNode例图

下面开始获取叶节点的数目和树的层数

def getNumLeafs(myTree):
    numLeafs = 0
    firstStr = myTree.keys()[0]
    secondDict = myTree[firstStr]
    for key in secondDict.keys():
        if type(secondDict[key]).__name__=='dict':
             numLeafs += getNumLeafs(secondDict[key])
        else:   numLeafs +=1
    return numLeafs

def getTreeDepth(myTree):
    maxDepth = 0
    firstStr = myTree.keys()[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 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]
>>> import treePlotter
>>> treePlotter.retrieveTree(1)
{'no surfacing': {0: 'no', 1: {'flippers': {0: {'head': {0: 'no', 1: 'yes'}}, 1: 'no'}}}}
>>> myTree = treePlotter.retrieveTree(0)
>>> treePlotter.getNumLeafs(myTree)
3
>>> treePlotter.getTreeDepth(myTree)
2

更新部分代码,开始尝试画图

def plotMidText(cntrPt, parentPt, txtString):#计算父节点和子节点的中间位置
    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):
    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__=='dict':
           plotTree(secondDict[key],cntrPt,str(key))
        else:
            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

并更新creaePlot()部分的代码

def createPlot(inTree):
    fig = plt.figure(1,facecolor='white')
    fig.clf()
    axprops = dict(xticks=[], yticks=[])
    createPlot.ax1 = plt.subplot(111,frameon=False, **axprops)
    plotTree.totalW = float(getNumLeafs(inTree))#宽度
    plotTree.totalD = float(getTreeDepth(inTree))#高度
    plotTree.xOff = -0.5/plotTree.totalW;plotTree.yOff = 1.0;
    plotTree(inTree, (0.5,1.0),'')
    plt.show()

开始画图

#treePlotter-1.py
import treePlotter
myTree=treePlotter.retrieveTree(0)
treePlotter.createPlot(myTree)
图3-6

接着添加字典的内容,重新绘制图片

>>> myTree['no surfacing'][3] = 'maybe'
>>> myTree
{'no surfacing': {0: 'no', 1: {'flippers': {0: 'no', 1: 'yes'}}, 3: 'maybe'}}
>>> treePlotter.createPlot(myTree)
图3-7

下面开始重点讲如何利用决策树执行数据分类
在执行数据分类时,需要使用决策树以及用于构造决策树的标签向量。然后,程序比较测试数据与决策树上的数值,递归执行该过程直到进入叶子节点;最后将测试数据定义为叶子节点所属的类型。

#使用决策树的分类函数
#添加到trees.py
def classify(inputTree, featLabels, testVec):
    firstStr = inputTree.keys()[0]
    secondDict = inputTree[firstStr]
    featIndex = featLabels.index(firstStr)#将标签字符串转化为索引
    for key in secondDict.keys():
        if testVec[featIndex] == key:
            if type(secondDict[key]).__name__=='dict':#递归遍历
                classLabel = classify(secondDict[key],featLabels,testVec)
            else: classLabel = secondDict[key]
    #key = testVec[featIndex]
    #valueOfFeat = secondDict[key]
    #if isinstance(valueOfFeat, dict):#是否是实例
    #    classLabel = classify(valueOfFeat, featLabels, testVec)
    #else: classLabel = valueOfFeat
    return classLabel
#trees-1
import trees
import treePlotter
myDat,labels = trees.createDataSet()
myTree=treePlotter.retrieveTree(0)
>>> labels
['no surfacing', 'flippers']
>>> myTree
{'no surfacing': {0: 'no', 1: {'flippers': {0: 'no', 1: 'yes'}}}}
>>> trees.classify(myTree,labels,[1,0])
'no'
>>> trees.classify(myTree,labels,[1,1])
'yes'

下面开始在硬盘上存储决策树分类器

#trees.py
def storeTree(inputTree,filename):
    import pickle#重点
    fw = open(filename,'w')
    pickle.dump(inputTree,fw)
    fw.close()
    
def grabTree(filename):
    import pickle
    fr = open(filename)
    return pickle.load(fr)
#tree-1.py
import trees
import treePlotter
myDat,labels = trees.createDataSet()
myTree = trees.createTree(myDat,labels)
trees.storeTree(myTree,'classifierStorage.txt')
>>> trees.grabTree('classifierStorage.txt')
{'no surfacing': {0: 'no', 1: {'flippers': {0: 'no', 1: 'yes'}}}}
import trees
import treePlotter
myDat,labels = trees.createDataSet()
myTree = trees.createTree(myDat,labels)

fr=open('lenses.txt')
lenses=[inst.strip().split('\t') for inst in fr.readlines()]
lensesLabels=['age','prescript','astigmatic','tearRate']
lensesTree = trees.createTree(lenses,lensesLabels)
>>> treePlotter.createPlot(lensesTree)
ID3算法产生的决策树
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