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
