-- coding: utf-8 --

"""
Created on Thu Aug 2 17:09:34 2018
决策树ID3,C4.5的实现
@author: weixw
"""
from math import log
import operator

原始数据

def createDataSet():
dataSet = [[0, 0, 0, 0, 'no'],
[0, 0, 0, 1, 'no'],
[0, 1, 0, 1, 'yes'],
[0, 1, 1, 0, 'yes'],
[0, 0, 0, 0, 'no'],
[1, 0, 0, 0, 'no'],
[1, 0, 0, 1, 'no'],
[1, 1, 1, 1, 'yes'],
[1, 0, 1, 2, 'yes'],
[1, 0, 1, 2, 'yes'],
[2, 0, 1, 2, 'yes'],
[2, 0, 1, 1, 'yes'],
[2, 1, 0, 1, 'yes'],
[2, 1, 0, 2, 'yes'],
[2, 0, 0, 0, 'no']]
# 分类属性
labels = ['age', 'work', 'house', 'loan']
return dataSet, labels

多数表决器

列中相同值数量最多为结果

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

划分数据集

dataSet:原始数据集

axis:进行分割的指定列索引

value:指定列中的值

def splitDataSet(dataSet, axis, value):
retDataSet = []
for featDataVal in dataSet:
if featDataVal[axis] == value:
# 下面两行去除某一项指定列的值，很巧妙有没有
reducedFeatVal = featDataVal[:axis]
reducedFeatVal.extend(featDataVal[axis + 1:])
retDataSet.append(reducedFeatVal)
return retDataSet

计算香农熵

columnIndex = -1表示获取数据集每一项的最后一列的标签值

其他表示获取特征列

def calcShannonEnt(columnIndex, dataSet):
# 数据集总项数
numEntries = len(dataSet)
# 标签计数对象初始化
labelCounts = {}
for featDataVal in dataSet:
# 获取数据集每一项的最后一列的标签值
currentLabel = featDataVal[columnIndex]
# 如果当前标签不在标签存储对象里，则初始化，然后计数
if currentLabel not in labelCounts.keys():
labelCounts[currentLabel] = 0
labelCounts[currentLabel] += 1
# 熵初始化
shannonEnt = 0.0
# 遍历标签对象，求概率，计算熵
for key in labelCounts.keys():
prop = labelCounts[key] / float(numEntries)
shannonEnt -= prop * log(prop, 2)
return shannonEnt

通过信息增益，选出最优特征列索引(ID3)

def chooseBestFeatureToSplit(dataSet):
# 计算特征个数，dataSet最后一列是标签属性，不是特征量
numFeatures = len(dataSet[0]) - 1
# 计算初始数据香农熵
baseEntropy = calcShannonEnt(-1, dataSet)
# 初始化信息增益，最优划分特征列索引
bestInfoGain = 0.0
bestFeatureIndex = -1
for i in range(numFeatures):
# 获取每一列数据
featList = [example[i] for example in dataSet]
# 将每一列数据去重
uniqueVals = set(featList)
newEntropy = 0.0
for value in uniqueVals:
subDataSet = splitDataSet(dataSet, i, value)
# 计算条件概率
prob = len(subDataSet) / float(len(dataSet))
# 计算条件熵
newEntropy += prob * calcShannonEnt(-1, subDataSet)
# 计算信息增益
infoGain = baseEntropy - newEntropy
if (infoGain > bestInfoGain):
bestInfoGain = infoGain
bestFeatureIndex = i
return bestFeatureIndex

通过信息增益率，选出最优特征列索引(C4.5)

def chooseBestFeatureToSplitOfFurther(dataSet):
# 计算特征个数，dataSet最后一列是标签属性，不是特征量
numFeatures = len(dataSet[0]) - 1
# 计算初始数据香农熵H(Y)
baseEntropy = calcShannonEnt(-1, dataSet)
# 初始化信息增益，最优划分特征列索引
bestInfoGainRatio = 0.0
bestFeatureIndex = -1
for i in range(numFeatures):
# 获取每一特征列香农熵H(X)
featEntropy = calcShannonEnt(i, dataSet)
# 获取每一列数据
featList = [example[i] for example in dataSet]
# 将每一列数据去重
uniqueVals = set(featList)
newEntropy = 0.0
for value in uniqueVals:
subDataSet = splitDataSet(dataSet, i, value)
# 计算条件概率
prob = len(subDataSet) / float(len(dataSet))
# 计算条件熵
newEntropy += prob * calcShannonEnt(-1, subDataSet)
# 计算信息增益
infoGain = baseEntropy - newEntropy
# 计算信息增益率
infoGainRatio = infoGain / float(featEntropy)
if (infoGainRatio > bestInfoGainRatio):
bestInfoGainRatio = infoGainRatio
bestFeatureIndex = i
return bestFeatureIndex

决策树创建

def createTree(dataSet, labels):
# 获取标签属性，dataSet最后一列，区别于labels标签名称
classList = [example[-1] for example in dataSet]
# 树极端终止条件判断
# 标签属性值全部相同，返回标签属性第一项值
if classList.count(classList[0]) == len(classList):
return classList[0]
# 没有特征，只有标签列（1列）
if len(dataSet[0]) == 1:
# 返回实例数最大的类
return majorityCnt(classList)
# #获取最优特征列索引ID3
# bestFeatureIndex = chooseBestFeatureToSplit(dataSet)
# 获取最优特征列索引C4.5
bestFeatureIndex = chooseBestFeatureToSplitOfFurther(dataSet)
# 获取最优索引对应的标签名称
bestFeatureLabel = labels[bestFeatureIndex]
# 创建根节点
myTree = {bestFeatureLabel: {}}
# 去除最优索引对应的标签名，使labels标签能正确遍历
del (labels[bestFeatureIndex])
# 获取最优列
bestFeature = [example[bestFeatureIndex] for example in dataSet]
uniquesVals = set(bestFeature)
for value in uniquesVals:
# 子标签名称集合
subLabels = labels[:]
# 递归
myTree[bestFeatureLabel][value] = createTree(splitDataSet(dataSet, bestFeatureIndex, value), subLabels)
return myTree

获取分类结果

inputTree:决策树字典

featLabels:标签列表

testVec:测试向量例如：简单实例下某一路径 [1,1] => yes（树干值组合，从根结点到叶子节点）

def classify(inputTree, featLabels, testVec):
# 获取根结点名称，将dict转化为list
firstSide = list(inputTree.keys())
# 根结点名称String类型
firstStr = firstSide[0]
# 获取根结点对应的子节点
secondDict = inputTree[firstStr]
# 获取根结点名称在标签列表中对应的索引
featIndex = featLabels.index(firstStr)
# 由索引获取向量表中的对应值
key = testVec[featIndex]
# 获取树干向量后的对象
valueOfFeat = secondDict[key]
# 判断是子结点还是叶子节点：子结点就回调分类函数，叶子结点就是分类结果
# if type(valueOfFeat).__name__=='dict': 等价 if isinstance(valueOfFeat, dict):
if isinstance(valueOfFeat, dict):
classLabel = classify(valueOfFeat, featLabels, testVec)
else:
classLabel = valueOfFeat
return classLabel

将决策树分类器存储在磁盘中，filename一般保存为txt格式

def storeTree(inputTree, filename):
import pickle
fw = open(filename, 'wb+')
pickle.dump(inputTree, fw)
fw.close()

将瓷盘中的对象加载出来，这里的filename就是上面函数中的txt文件

def grabTree(filename):
import pickle
fr = open(filename, 'rb')
return pickle.load(fr)

'''
Created on Oct 14, 2010

@author: Peter Harrington
'''
import matplotlib.pyplot as plt

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

获取树的叶子节点

def getNumLeafs(myTree):
numLeafs = 0
#dict转化为list
firstSides = list(myTree.keys())
firstStr = firstSides[0]
secondDict = myTree[firstStr]
for key in secondDict.keys():
#判断是否是叶子节点（通过类型判断，子类不存在，则类型为str；子类存在，则为dict）
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 numLeafs

获取树的层数

def getTreeDepth(myTree):
maxDepth = 0
#dict转化为list
firstSides = list(myTree.keys())
firstStr = firstSides[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 maxDepth

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 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)
firstSides = list(myTree.keys())
firstStr = firstSides[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 dict

绘制决策树

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.show()

绘制树的根节点和叶子节点（根节点形状：长方形，叶子节点：椭圆形）

def createPlot():

fig = plt.figure(1, facecolor='white')

fig.clf()

createPlot.ax1 = plt.subplot(111, frameon=False) #ticks for demo puropses

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()

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]

thisTree = retrieveTree(0)

createPlot(thisTree)

createPlot()

myTree = retrieveTree(0)

numLeafs =getNumLeafs(myTree)

treeDepth =getTreeDepth(myTree)

print(u"叶子节点数目：%d"% numLeafs)

print(u"树深度：%d"%treeDepth)

-- coding: utf-8 --

"""
Created on Fri Aug 3 19:52:10 2018

@author: weixw
"""
import Demo_1.myTrees as mt
import Demo_1.treePlotter as tp

测试

dataSet, labels = mt.createDataSet()

copy函数：新开辟一块内存，然后将list的所有值复制到新开辟的内存中

labels1 = labels.copy()

createTree函数中将labels1的值改变了，所以在分类测试时不能用labels1

myTree = mt.createTree(dataSet,labels1)

保存树到本地

mt.storeTree(myTree,'myTree.txt')

在本地磁盘获取树

myTree = mt.grabTree('myTree.txt')
print (u"决策树结构：%s"%myTree)

绘制决策树

print(u"绘制决策树：")
tp.createPlot(myTree)
numLeafs =tp.getNumLeafs(myTree)
treeDepth =tp.getTreeDepth(myTree)
print(u"叶子节点数目：%d"% numLeafs)
print(u"树深度：%d"%treeDepth)

测试分类简单样本数据3列

labelResult =mt.classify(myTree,labels,[1,1,1,0])
print(u"[1,1] 测试结果为：%s"%labelResult)
labelResult =mt.classify(myTree,labels,[1,0,0,0])
print(u"[1,0] 测试结果为：%s"%labelResult)

巴特西

决策树——C4.5

-- coding: utf-8 --

原始数据

多数表决器

列中相同值数量最多为结果

划分数据集

dataSet:原始数据集

axis:进行分割的指定列索引

value:指定列中的值

计算香农熵

columnIndex = -1表示获取数据集每一项的最后一列的标签值

其他表示获取特征列

通过信息增益，选出最优特征列索引(ID3)

通过信息增益率，选出最优特征列索引(C4.5)

决策树创建

获取分类结果

inputTree:决策树字典

featLabels:标签列表

testVec:测试向量 例如：简单实例下某一路径 [1,1] => yes（树干值组合，从根结点到叶子节点）

将决策树分类器存储在磁盘中，filename一般保存为txt格式

将瓷盘中的对象加载出来，这里的filename就是上面函数中的txt文件

获取树的叶子节点

获取树的层数

if you do get a dictonary you know it's a tree, and the first element will be another dict

绘制决策树

绘制树的根节点和叶子节点（根节点形状：长方形，叶子节点：椭圆形）

def createPlot():

fig = plt.figure(1, facecolor='white')

fig.clf()

createPlot.ax1 = plt.subplot(111, frameon=False) #ticks for demo puropses

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()

thisTree = retrieveTree(0)

createPlot(thisTree)

createPlot()

myTree = retrieveTree(0)

numLeafs =getNumLeafs(myTree)

treeDepth =getTreeDepth(myTree)

print(u"叶子节点数目：%d"% numLeafs)

print(u"树深度：%d"%treeDepth)

-- coding: utf-8 --

测试

copy函数：新开辟一块内存，然后将list的所有值复制到新开辟的内存中

createTree函数中将labels1的值改变了，所以在分类测试时不能用labels1

保存树到本地

在本地磁盘获取树

绘制决策树

测试分类 简单样本数据3列

最新文章

热门文章

testVec:测试向量例如：简单实例下某一路径 [1,1] => yes（树干值组合，从根结点到叶子节点）

测试分类简单样本数据3列