Python3 ID3决策树判断申请贷款是否成功的实现代码

# -*- 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)

# 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 

#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) 

#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()

#命令绘决策树的图
createPlot(myTree)