【python实战】3 决策树（代码讲解）

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from math import log
import operator

def createDataSet():
dataSet = [[1, 1, 'yes'],
[1, 1, 'yes'],
[1, 0, 'no'],
[0, 1, 'no'],
[0, 1, 'no']]
labels = ['no surfacing','flippers']
#change to discrete values
return dataSet, labels

#计算给定数据集的香农熵
def calcShannonEnt(dataSet):
numEntries = len(dataSet)
labelCounts = {}
for featVec in dataSet: #the the number of unique elements and their occurance
currentLabel = featVec[-1] #数据字典，键值是最后一列的数值。
if currentLabel not in labelCounts.keys(): labelCounts[currentLabel] = 0
labelCounts[currentLabel] += 1 #数个数
shannonEnt = 0.0
for key in labelCounts:
prob = float(labelCounts[key])/numEntries #求概率
shannonEnt -= prob * log(prob,2) #log base 2
return shannonEnt

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

#程序3-3：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) #保留最初的无序度量值的原始香农熵
bestInfoGain = 0.0; bestFeature = -1
for i in range(numFeatures): #遍历所有特征（列）iterate over all the features
featList = [example[i] for example in dataSet] #create a list of这个特征的所有值（每一行的这一列）【http://python.jobbole.com/80823/】
uniqueVals = set(featList) #从列表中创建集合get a set of unique values去掉重复值
newEntropy = 0.0
for value in uniqueVals: #遍历当前特征中的所有唯一属性值，对每一个特征分一次数据集
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

验证：

>>> import trees
>>> myDat,labels=trees.createDataSet()
>>> trees.chooseBestFeatureToSplit(myDat)
0.970951=0.970951-0.000000 #print(infoGain=baseEntropy-newEntropy)
0.970951=0.970951-0.000000
0
>>> exit()

3.1.3 递归构建决策树

以上我们学过的【度量数据集的信息熵】和【如何划分数据集】可看作构建决策树的子功能模块。递归结束的条件是：程序遍历完所有划分数据集的属性，或每个分支下的所有实例都具有相同的分类。 { 其他决策树算法：C4.5 CART暂不考虑。}

目前我们只考虑在算法执行前计算列的数目，查看是否使用了全部属性即可。如果已经处理了全部属性但类标签仍然不是唯一的，则使用【多数表决法】决定该叶子结点的分类。

#投票表决代码，返回出现次数最多的分类名称，该函数使用分类名称的列表，

import operator

def majorityCnt(classList):

classCount={}

for vote in classList:

if vote not in classCount:classCount[vote]=0

classCount+=1

sortedClassCount=sorted(classCount.iteritems(),key=operater.itemgetter[1],reverse=Ture)

return sortClassCount[0][0]

#程序清单3-4：创建树的函数代码

def createTree(dataSet,labels):
classList = [example[-1] for example in dataSet] #数据集的所有类标签(最后一列)
if classList.count(classList[0]) == len(classList): #返回元素在列表中出现的次数。
return classList[0] #stop splitting when all of the classes are equal当剩余元素的类别完全相同
if len(dataSet[0]) == 1: #stop splitting when there are no more features in dataSet仅剩类标签（只有一列）
return majorityCnt(classList) #投票表决函数
bestFeat = chooseBestFeatureToSplit(dataSet) #选择下一个划分的属性返回它的位置i（第几列）
bestFeatLabel = labels[bestFeat]
myTree = {bestFeatLabel:{}} #字典嵌套表示树结构，字典内数据表示叶子{特征：{特征：分类}}
del(labels[bestFeat]) #划分好的属性值，删掉（labels为所有属性标签，即dataset除去最后一列）
featValues = [example[bestFeat] for example in dataSet] #该属性标签下的所有属性值
uniqueVals = set(featValues) #列表变成集合,去掉重复值
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) #递归
return myTree

3.2 matplotlib注解绘制树形图

treePlotter.py

import matplotlib.pyplot as plt

decisionNode = dict(boxstyle="sawtooth", fc="0.8") #锯齿形状
leafNode = dict(boxstyle="round4", fc="0.8") #四个圆角
arrow_args = dict(arrowstyle="<-") #箭头形状

def plotNode(nodeTxt, centerPt, parentPt, nodeType): #绘图功能,绘制树节点
createPlot.ax1.annotate(nodeTxt, xy=parentPt, xycoords='axes fraction', #绘制带箭头的注释
xytext=centerPt, textcoords='axes fraction',
va="center", ha="center", bbox=nodeType, arrowprops=arrow_args )

def createPlot():
fig = plt.figure(1, facecolor='white') #新建绘画窗口
fig.clf() #清空绘图区
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) #绘制结点(节点文字,箭头终点,箭头起点,节点格式形状)
plt.show()

测试：

>>>import treePlotter

>>>treePlotter.createPlot()

3.2.2 构造注解树

# 程序清单3-6 获取叶节点数目和树的层数

def getNumLeafs(myTree):
numLeafs = 0
firstStr = myTree.keys()[0] #树mytree字典的第一个关键字key[0]
secondDict = myTree[firstStr] #从第一个关键字出发，遍历整颗树的所有子节点
for key in secondDict.keys():
if type(secondDict[key]).__name__=='dict': #test to see if the nodes are dictonaires判断子节点是否为字典类型
numLeafs += getNumLeafs(secondDict[key]) #如果是：递归调用；否则：是叶子节点则数目加一
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': #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 #一旦达到叶子节点则从递归调用中返回
return maxDepth

# 下面函数输出预先存储的树信息，避免了每次测试代码时都要从数据中创建树的麻烦

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]

验证：

>>> 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

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