Python3 ID3决策树判断申请贷款是否成功的实现代码
1. 定义生成树
# -*- coding: utf-8 -*- #生成树的函数 from numpy import * import numpy as np import pandas as pd from math import log import operator # 计算数据集的信息熵(Information Gain)增益函数(机器学习实战中信息熵叫香农熵) def calcInfoEnt(dataSet):#本题中Label即好or坏瓜 #dataSet每一列是一个属性(列末是Label) numEntries = len(dataSet) #每一行是一个样本 labelCounts = {} #给所有可能的分类创建字典labelCounts for featVec in dataSet: #按行循环:即rowVev取遍了数据集中的每一行 currentLabel = featVec[-1] #故featVec[-1]取遍每行最后一个值即Label if currentLabel not in labelCounts.keys(): #如果当前的Label在字典中还没有 labelCounts[currentLabel] = 0 #则先赋值0来创建这个词 labelCounts[currentLabel] += 1 #计数, 统计每类Label数量(这行不受if限制) InfoEnt = 0.0 for key in labelCounts: #遍历每类Label prob = float(labelCounts[key])/numEntries #各类Label熵累加 InfoEnt -= prob * log(prob,2) #ID3用的信息熵增益公式 return InfoEnt ### 对于离散特征: 取出该特征取值为value的所有样本 def splitDiscreteDataSet(dataSet, axis, value): #dataSet是当前结点(待划分)集合,axis指示划分所依据的属性,value该属性用于划分的取值 retDataSet = [] #为return Data Set分配一个列表用来储存 for featVec in dataSet: if featVec[axis] == value: reducedFeatVec = featVec[:axis] #该特征之前的特征仍保留在样本dataSet中 reducedFeatVec.extend(featVec[axis+1:]) #该特征之后的特征仍保留在样本dataSet中 retDataSet.append(reducedFeatVec) #把这个样本加到list中 return retDataSet ### 对于连续特征: 返回特征取值大于value的所有样本(以value为阈值将集合分成两部分) def splitContinuousDataSet(dataSet, axis, value): retDataSetG = [] #将储存取值大于value的样本 retDataSetL = [] #将储存取值小于value的样本 for featVec in dataSet: if featVec[axis] > value: reducedFeatVecG = featVec[:axis] reducedFeatVecG.extend(featVec[axis+1:]) retDataSetG.append(reducedFeatVecG) else: reducedFeatVecL = featVec[:axis] reducedFeatVecL.extend(featVec[axis+1:]) retDataSetL.append(reducedFeatVecL) return retDataSetG,retDataSetL #返回两个集合, 是含2个元素的tuple形式 ### 根据InfoGain选择当前最好的划分特征(以及对于连续变量还要选择以什么值划分) def chooseBestFeatureToSplit(dataSet,labels): numFeatures = len(dataSet[0])-1 baseEntropy = calcInfoEnt(dataSet) bestInfoGain = 0.0; bestFeature = -1 bestSplitDict = {} for i in range(numFeatures): #遍历所有特征:下面这句是取每一行的第i个, 即得当前集合所有样本第i个feature的值 featList = [example[i] for example in dataSet] #判断是否为离散特征 if not (type(featList[0]).__name__=='float' or type(featList[0]).__name__=='int'): # 对于离散特征:求若以该特征划分的熵增 uniqueVals = set(featList) #从列表中创建集合set(得列表唯一元素值) newEntropy = 0.0 for value in uniqueVals: #遍历该离散特征每个取值 subDataSet = splitDiscreteDataSet(dataSet, i, value)#计算每个取值的信息熵 prob = len(subDataSet)/float(len(dataSet)) newEntropy += prob * calcInfoEnt(subDataSet)#各取值的熵累加 infoGain = baseEntropy - newEntropy #得到以该特征划分的熵增 # 对于连续特征:求若以该特征划分的熵增(区别:n个数据则需添n-1个候选划分点, 并选最佳划分点) else: #产生n-1个候选划分点 sortfeatList=sorted(featList) splitList=[] for j in range(len(sortfeatList)-1): #产生n-1个候选划分点 splitList.append((sortfeatList[j] + sortfeatList[j+1])/2.0) bestSplitEntropy = 10000 #设定一个很大的熵值(之后用) #遍历n-1个候选划分点: 求选第j个候选划分点划分时的熵增, 并选出最佳划分点 for j in range(len(splitList)): value = splitList[j] newEntropy = 0.0 DataSet = splitContinuousDataSet(dataSet, i, value) subDataSetG = DataSet[0] subDataSetL = DataSet[1] probG = len(subDataSetG) / float(len(dataSet)) newEntropy += probG * calcInfoEnt(subDataSetG) probL = len(subDataSetL) / float(len(dataSet)) newEntropy += probL * calcInfoEnt(subDataSetL) if newEntropy < bestSplitEntropy: bestSplitEntropy = newEntropy bestSplit = j bestSplitDict[labels[i]] = splitList[bestSplit]#字典记录当前连续属性的最佳划分点 infoGain = baseEntropy - bestSplitEntropy #计算以该节点划分的熵增 # 在所有属性(包括连续和离散)中选择可以获得最大熵增的属性 if infoGain > bestInfoGain: bestInfoGain = infoGain bestFeature = i #若当前节点的最佳划分特征为连续特征,则需根据“是否小于等于其最佳划分点”进行二值化处理 #即将该特征改为“是否小于等于bestSplitValue”, 例如将“密度”变为“密度<=0.3815” #注意:以下这段直接操作了原dataSet数据, 之前的那些float型的值相应变为0和1 #【为何这样做?】在函数createTree()末尾将看到解释 if type(dataSet[0][bestFeature]).__name__=='float' or type(dataSet[0][bestFeature]).__name__=='int': bestSplitValue = bestSplitDict[labels[bestFeature]] labels[bestFeature] = labels[bestFeature] + '<=' + str(bestSplitValue) for i in range(shape(dataSet)[0]): if dataSet[i][bestFeature] <= bestSplitValue: dataSet[i][bestFeature] = 1 else: dataSet[i][bestFeature] = 0 return bestFeature # 若特征已经划分完,节点下的样本还没有统一取值,则需要进行投票:计算每类Label个数, 取max者 def majorityCnt(classList): classCount = {} #将创建键值为Label类型的字典 for vote in classList: if vote not in classCount.keys(): classCount[vote] = 0 #第一次出现的Label加入字典 classCount[vote] += 1 #计数 return max(classCount)
2. 递归产生决策树
# 主程序:递归产生决策树 # dataSet:当前用于构建树的数据集, 最开始就是data_full,然后随着划分的进行越来越小。这是因为进行到到树分叉点上了. 第一次划分之前17个瓜的数据在根节点,然后选择第一个bestFeat是纹理. 纹理的取值有清晰、模糊、稍糊三种;将瓜分成了清晰(9个),稍糊(5个),模糊(3个),这时应该将划分的类别减少1以便于下次划分。 # labels:当前数据集中有的用于划分的类别(这是因为有些Label当前数据集没了, 比如假如到某个点上西瓜都是浅白没有深绿了) # data_full:全部的数据 # label_full:全部的类别 numLine = numColumn = 2 #这句是因为之后要用global numLine……至于为什么我一定要用global # 我也不完全理解。如果我只定义local变量总报错,我只好在那里的if里用global变量了。求解。 def createTree(dataSet,labels,data_full,labels_full): classList = [example[-1] for example in dataSet] #递归停止条件1:当前节点所有样本属于同一类;(注:count()方法统计某元素在列表中出现的次数) if classList.count(classList[0]) == len(classList): return classList[0] #递归停止条件2:当前节点上样本集合为空集(即特征的某个取值上已经没有样本了): global numLine,numColumn (numLine,numColumn) = shape(dataSet) if float(numLine) == 0: return 'empty' #递归停止条件3:所有可用于划分的特征均使用过了,则调用majorityCnt()投票定Label; if float(numColumn) == 1: return majorityCnt(classList) #不停止时继续划分: bestFeat = chooseBestFeatureToSplit(dataSet,labels)#调用函数找出当前最佳划分特征是第几个 bestFeatLabel = labels[bestFeat] #当前最佳划分特征 myTree = {bestFeatLabel:{}} featValues = [example[bestFeat] for example in dataSet] uniqueVals = set(featValues) if type(dataSet[0][bestFeat]).__name__=='str': currentlabel = labels_full.index(labels[bestFeat]) featValuesFull = [example[currentlabel] for example in data_full] uniqueValsFull = set(featValuesFull) del(labels[bestFeat]) #划分完后, 即当前特征已经使用过了, 故将其从“待划分特征集”中删去 #【递归调用】针对当前用于划分的特征(beatFeat)的每个取值,划分出一个子树。 for value in uniqueVals: #遍历该特征【现存的】取值 subLabels = labels[:] if type(dataSet[0][bestFeat]).__name__=='str': uniqueValsFull.remove(value) #划分后删去(从uniqueValsFull中删!) myTree[bestFeatLabel][value] = createTree(splitDiscreteDataSet(dataSet,bestFeat,value),subLabels,data_full,labels_full)#用splitDiscreteDataSet() #是由于, 所有的连续特征在划分后都被我们定义的chooseBestFeatureToSplit()处理成离散取值了。 if type(dataSet[0][bestFeat]).__name__=='str': #若该特征离散【更详见后注】 for value in uniqueValsFull:#则可能有些取值已经不在【现存的】取值中了 #这就是上面为何从“uniqueValsFull”中删去 #因为那些现有数据集中没取到的该特征的值,保留在了其中 myTree[bestFeatLabel][value] = majorityCnt(classList) return myTree
3. 调用生成树
#生成树调用的语句 df = pd.read_excel(r'E:\BaiduNetdiskDownload\spss\数据\实验data\银行贷款.xlsx') data = df.values[:,1:].tolist() data_full = data[:] labels = df.columns.values[1:-1].tolist() labels_full = labels[:] myTree = createTree(data,labels,data_full,labels_full)
查看数据
data
labels
4. 绘制决策树
#绘决策树的函数 import matplotlib.pyplot as plt decisionNode = dict(boxstyle = "sawtooth",fc = "0.8") #定义分支点的样式 leafNode = dict(boxstyle = "round4",fc = "0.8") #定义叶节点的样式 arrow_args = dict(arrowstyle = "<-") #定义箭头标识样式 # 计算树的叶子节点数量 def getNumLeafs(myTree): numLeafs = 0 firstStr = list(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 # 计算树的最大深度 def getTreeDepth(myTree): maxDepth = 0 firstStr = list(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 # 画出节点 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) # 标箭头上的文字 def plotMidText(cntrPt,parentPt,txtString): lens = len(txtString) xMid = (parentPt[0] + cntrPt[0]) / 2.0 - lens*0.002 yMid = (parentPt[1] + cntrPt[1]) / 2.0 createPlot.ax1.text(xMid,yMid,txtString) def plotTree(myTree,parentPt,nodeTxt): numLeafs = getNumLeafs(myTree) depth = getTreeDepth(myTree) firstStr = list(myTree.keys())[0] cntrPt = (plotTree.x0ff + (1.0 + float(numLeafs))/2.0/plotTree.totalW,plotTree.y0ff) plotMidText(cntrPt,parentPt,nodeTxt) plotNode(firstStr,cntrPt,parentPt,decisionNode) secondDict = myTree[firstStr] plotTree.y0ff = plotTree.y0ff - 1.0/plotTree.totalD for key in secondDict.keys(): if type(secondDict[key]).__name__=='dict': plotTree(secondDict[key],cntrPt,str(key)) else: plotTree.x0ff = plotTree.x0ff + 1.0/plotTree.totalW plotNode(secondDict[key],(plotTree.x0ff,plotTree.y0ff),cntrPt,leafNode) plotMidText((plotTree.x0ff,plotTree.y0ff),cntrPt,str(key)) plotTree.y0ff = plotTree.y0ff + 1.0/plotTree.totalD 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)) plotTree.totalD = float(getTreeDepth(inTree)) plotTree.x0ff = -0.5/plotTree.totalW plotTree.y0ff = 1.0 plotTree(inTree,(0.5,1.0),'') plt.show()
5. 调用函数
#命令绘决策树的图 createPlot(myTree)
myTree
总结
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