目录
前置信息1、决策树2、样本数据
决策树分类算法1、构建数据集2、数据集信息熵3、信息增益4、构造决策树5、实例化构造决策树6、测试样本分类
后置信息：绘制决策树代码
前置信息
1、决策树
决策树是一种十分常用的分类算法，属于监督学习；也就是给出一批样本，每个样本都有一组属性和一个分类结果。算法通过学习这些样本，得到一个决策树，这个决策树能够对新的数据给出合适的分类
2、样本数据
假设现有用户14名，其个人属性及是否购买某一产品的数据如下：
编号年龄收入范围工作性质信用评级购买决策01<30高不稳定较差否02<30高不稳定好否0330-40高不稳定较差是04>40中等不稳定较差是05>40低稳定较差是06>40低稳定好否0730-40低稳定好是08<30中等不稳定较差否09<30低稳定较差是10>40中等稳定较差是11<30中等稳定好是1230-40中等不稳定好是1330-40高稳定较差是14>40中等不稳定好否
决策树分类算法
1、构建数据集
为了方便处理，对模拟数据按以下规则转换为数值型列表数据： 年龄：<30赋值为0；30-40赋值为1；>40赋值为2 收入：低为0；中为1；高为2 工作性质：不稳定为0；稳定为1 信用评级：差为0；好为1
#创建数据集
def createdataset():
dataSet=[[0,2,0,0,'N'],
[0,2,0,1,'N'],
[1,2,0,0,'Y'],
[2,1,0,0,'Y'],
[2,0,1,0,'Y'],
[2,0,1,1,'N'],
[1,0,1,1,'Y'],
[0,1,0,0,'N'],
[0,0,1,0,'Y'],
[2,1,1,0,'Y'],
[0,1,1,1,'Y'],
[1,1,0,1,'Y'],
[1,2,1,0,'Y'],
[2,1,0,1,'N'],]
labels=['age','income','job','credit']
return dataSet,labels
调用函数，可获得数据：
ds1,lab = createdataset()
print(ds1)
print(lab)
[[0, 2, 0, 0, ‘N’], [0, 2, 0, 1, ‘N’], [1, 2, 0, 0, ‘Y’], [2, 1, 0, 0, ‘Y’], [2, 0, 1, 0, ‘Y’], [2, 0, 1, 1, ‘N’], [1, 0, 1, 1, ‘Y’], [0, 1, 0, 0, ‘N’], [0, 0, 1, 0, ‘Y’], [2, 1, 1, 0, ‘Y’], [0, 1, 1, 1, ‘Y’], [1, 1, 0, 1, ‘Y’], [1, 2, 1, 0, ‘Y’], [2, 1, 0, 1, ‘N’]] [‘age’, ‘income’, ‘job’, ‘credit’]
2、数据集信息熵
信息熵也称为香农熵，是随机变量的期望。度量信息的不确定程度。信息的熵越大，信息就越不容易搞清楚。处理信息就是为了把信息搞清楚，就是熵减少的过程。
def calcShannonEnt(dataSet):
numEntries = len(dataSet)
labelCounts = {}
for featVec in dataSet:
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)
return shannonEnt
样本数据信息熵：
shan = calcShannonEnt(ds1)
print(shan)
0.9402859586706309
3、信息增益
信息增益：用于度量属性A降低样本集合X熵的贡献大小。信息增益越大，越适于对X分类。
def chooseBestFeatureToSplit(dataSet):
numFeatures = len(dataSet[0])-1
baseEntropy = calcShannonEnt(dataSet)
bestInfoGain = 0.0;bestFeature = -1
for i in range(numFeatures):
featList = [example[i] for example in dataSet]
uniqueVals = set(featList)
newEntroy = 0.0
for value in uniqueVals:
subDataSet = splitDataSet(dataSet, i, value)
prop = len(subDataSet)/float(len(dataSet))
newEntroy += prop * calcShannonEnt(subDataSet)
infoGain = baseEntropy - newEntroy
if(infoGain > bestInfoGain):
bestInfoGain = infoGain
bestFeature = i
return bestFeature
以上代码实现了基于信息熵增益的ID3决策树学习算法。其核心逻辑原理是：依次选取属性集中的每一个属性，将样本集按照此属性的取值分割为若干个子集；对这些子集计算信息熵，其与样本的信息熵的差，即为按照此属性分割的信息熵增益；找出所有增益中最大的那一个对应的属性，就是用于分割样本集的属性。
计算样本最佳的分割样本属性，结果显示为第0列，即age属性：
col = chooseBestFeatureToSplit(ds1)
col
0
4、构造决策树
def majorityCnt(classList):
classCount = {}
for vote in classList:
if vote not in classCount.keys():classCount[vote] = 0
classCount[vote] += 1
sortedClassCount = sorted(classList.iteritems(),key=operator.itemgetter(1),reverse=True)#利用operator操作键值排序字典
return sortedClassCount[0][0]
#创建树的函数
def createTree(dataSet,labels):
classList = [example[-1] for example in dataSet]
if classList.count(classList[0]) == len(classList):
return classList[0]
if len(dataSet[0]) == 1:
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
majorityCnt函数用于处理一下情况：最终的理想决策树应该沿着决策分支到达最底端时，所有的样本应该都是相同的分类结果。但是真实样本中难免会出现所有属性一致但分类结果不一样的情况，此时majorityCnt将这类样本的分类标签都调整为出现次数最多的那一个分类结果。
createTree是核心任务函数，它对所有的属性依次调用ID3信息熵增益算法进行计算处理，最终生成决策树。
5、实例化构造决策树
利用样本数据构造决策树：
Tree = createTree(ds1, lab)
print("样本数据决策树：")
print(Tree)
样本数据决策树： {‘age’: {0: {‘job’: {0: ‘N’, 1: ‘Y’}}, 1: ‘Y’, 2: {‘credit’: {0: ‘Y’, 1: ‘N’}}}}
6、测试样本分类
给出一个新的用户信息，判断ta是否购买某一产品：
年龄收入范围工作性质信用评级<30低稳定好<30高不稳定好
def classify(inputtree,featlabels,testvec):
firststr = list(inputtree.keys())[0]
seconddict = inputtree[firststr]
featindex = featlabels.index(firststr)
for key in seconddict.keys():
if testvec[featindex]==key:
if type(seconddict[key]).__name__=='dict':
classlabel=classify(seconddict[key],featlabels,testvec)
else:
classlabel=seconddict[key]
return classlabel
labels=['age','income','job','credit']
tsvec=[0,0,1,1]
print('result:',classify(Tree,labels,tsvec))
tsvec1=[0,2,0,1]
print('result1:',classify(Tree,labels,tsvec1))
result: Y result1: N
后置信息：绘制决策树代码
以下代码用于绘制决策树图形，非决策树算法重点，有兴趣可参考学习
import matplotlib.pyplot as plt
decisionNode = dict(boxstyle="sawtooth", fc="0.8")
leafNode = dict(boxstyle="round4", fc="0.8")
arrow_args = dict(arrowstyle="<-")
#获取叶节点的数目
def getNumLeafs(myTree):
numLeafs = 0
firstStr = list(myTree.keys())[0]
secondDict = myTree[firstStr]
for key in secondDict.keys():
if type(secondDict[key]).__name__=='dict':#测试节点的数据是否为字典，以此判断是否为叶节点
numLeafs += getNumLeafs(secondDict[key])
else: numLeafs +=1
return numLeafs
#获取树的层数
def getTreeDepth(myTree):
maxDepth = 0
firstStr = list(myTree.keys())[0]
secondDict = myTree[firstStr]
for key in secondDict.keys():
if type(secondDict[key]).__name__=='dict':#测试节点的数据是否为字典，以此判断是否为叶节点
thisDepth = 1 + getTreeDepth(secondDict[key])
else: thisDepth = 1
if thisDepth > maxDepth: maxDepth = thisDepth
return maxDepth
#绘制节点
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 )
#绘制连接线
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)
#绘制树结构
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 = list(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
#创建决策树图形
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.savefig('决策树.png',dpi=300,bbox_inches='tight')
plt.show()
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