Re:18. 혜진이가 파이썬으로 구현한 expert system (컴퓨터에게 물어보다)

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decision tree expert.py

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# -*- coding: utf-8 -*- """ Created on Wed Apr 26 18:12:18 2017 @author: stu """ #http://clearpal7.blogspot.kr/2016/07/3_10.html  #http://yujuwon.tistory.com/entry/%EC%9D%98%EC%82%AC-%EA%B2%B0%EC%A0%95-%ED%8A%B8%EB%A6%AC   import operator from math import log   def createDataSet():      dataSet=[['M','30','NO','YES','NO','NO'],                 ['F','20','YES','YES','YES','NO'],                 ['F','20','YES','YES','NO','NO'],                 ['F','40','NO','NO','NO','NO'],                 ['F','30','NO','YES','NO','NO'],                 ['F','30','NO','NO','YES','NO'],                 ['F','20','NO','YES','NO','NO'],                 ['F','20','NO','YES','YES','YES'],                 ['F','30','YES','YES','NO','YES'],                 ['M','40','YES','NO','YES','NO'],                 ['M','20','NO','NO','NO','NO'],                 ['M','30','NO','YES','YES','NO'],                 ['M','20','YES','NO','NO','NO'],                 ['F','30','YES','YES','NO','YES'],                 ['M','30','YES','YES','YES','YES'],                 ['F','30','YES','NO','NO','NO'],                 ['F','30','NO','YES','YES','YES'],                 ['M','20','YES','YES','NO','NO'],                 ['M','40','YES','NO','YES','NO'],                 ['M','30','NO','NO','NO','NO'],                 ['F','30','YES','YES','NO','YES'],                 ['M','30','YES','NO','YES','NO'],                 ['F','40','NO','YES','YES','YES'],                 ['M','30','NO','YES','NO','NO'],                 ['F','30','YES','YES','YES','YES'],                 ['F','40','YES','NO','YES','NO'],                 ['M','40','YES','YES','NO','YES'],                 ['F','40','YES','YES','NO','YES']]            labels=['GENDER','AGE','JOB_YN','MARRY_YN','CAR_YN','COUPON_YN']      return dataSet, labels   def calcShannonEnt(dataSet):                #분할 전 엔트로피 구하기     numEntries = len(dataSet)               #row의 개수 28     labelCounts = {}     for feaVec in dataSet:         currentLabel = feaVec[-1]         if currentLabel not in labelCounts:             labelCounts[currentLabel] = 0         labelCounts[currentLabel] += 1      #COUPON_YN labelCounts={'NO': 18, 'YES': 10}     shannonEnt = 0.0     for key in labelCounts:         prob = float(labelCounts[key])/numEntries          shannonEnt -= prob * log(prob, 2)   #분할 전 엔트로피:0.9402859586706309     return shannonEnt   def splitDataSet(dataSet, axis, value):     #변수의 각 속성에 따라 데이터 나누기(그 변수의 값은 제외하고)     retDataSet = []                         #GENDER F  [['20', 'YES', 'YES', 'YES', 'NO'],..,['40', 'YES', 'YES', 'NO', 'YES']     for featVec in dataSet:                 #GENDER M  [['30', 'NO', 'YES', 'NO', 'NO'],..,['40', 'YES', 'YES', 'NO', 'YES']]         if featVec[axis] == value:             reducedFeatVec = featVec[:axis]             reducedFeatVec.extend(featVec[axis+1:])             retDataSet.append(reducedFeatVec)     return retDataSet      def chooseBestFeatureToSplit(dataSet):     #정보이득이 가장 큰 변수 출력     numFeatures = len(dataSet[0]) - 1      #한 row가 갖는 속성의 개수(COUPON_YN 제외) 5개     baseEntropy = calcShannonEnt(dataSet)  #분할 전 엔트로피     bestInfoGain = 0.0     bestFeature = -1     for i in range(numFeatures):           #dataSet의 COUPON_YN변수 제외 속성 리스트         #featList - ['M', 'F', 'F',.., 'F', 'M', 'F'] ['30', '20', '20',.., '40', '40', '40']         featList = [example[i] for example in dataSet]          uniqueVal‎s = set(featList)         #set() : 리스트 요소 중복제거 {'M', 'F'} {'20','30','40'} {'YES','NO'}         newEntropy = 0.0         for value in uniqueVal‎s:             subDataSet = splitDataSet(dataSet, i, value)             prob = len(subDataSet) / float(len(dataSet))             newEntropy += prob * calcShannonEnt(subDataSet) #분할 후 엔트로피         infoGain = baseEntropy -newEntropy #정보이득         if infoGain > bestInfoGain:             bestInfoGain = infoGain             bestFeature = i     return bestFeature                     #가장 큰 정보이득을 갖는 변수의 위치                def majorityCnt(classList):                #각 변수의 요소 중 개수가 가장 많은 것을 리턴     classCount = {}     for vote in classList:                 #AGE ['30', '20', '20',.., '40', '40', '40']         if vote not in classCount.keys():             classCount[vote] = 0         classCount[vote] += 1              #COUPON_YN classCount {20':7,'30':13,'40':8}     #[('20', 7), ('30', 13), ('40', 8)]     sortedClassCount = sorted(classCount.items(), key=operator.itemgetter(1), reverse=True)     return sortedClassCount[0][0]          #내림차순으로 정렬, 숫자가 가장 큰 '30'을 리턴         def createTree(dataSet, labels):     #COUPON_YN 데이터 ['NO', 'NO', 'NO',.., 'NO', 'YES', 'YES']     classList = [example[-1] for example in dataSet]     if classList.count(classList[0]) == len(classList): #종료조건1 : 모두 'yes'이거나 모두 'no'이면 종료하고 그 항목 리턴         return classList[0]     if len(dataSet[0]) == 1:                            #종료조건2 : [1,0,'yes']->[1,0]->[1] 데이터셋의 변수가 1개가 됐을 때 majorityCnt리턴         return majorityCnt(classList)                   #둘 다 만족하지 못하면 정보이득이 가장 큰 조건으로     bestFeat = chooseBestFeatureToSplit(dataSet)        #정보이득이 가장 큰 속성의 위치(ex.0~4)     bestFeatLabel = labels[bestFeat]                    #정보이득이 가장 큰 속성(ex.'GENDER','AGE','JOB_YN','MARRY_YN','CAR_YN')     myTree = {bestFeatLabel:{}}     del(labels[bestFeat])                               #정보이득이 가장 큰 속성 이름 삭제     featValues = [example[bestFeat] for example in dataSet] #정보이득이 가장 큰 속성 값들의 리스트     uniqueVal‎s = set(featValues)                        #그 속성의 중복값 제거 ex.{'M', 'F'} {'20','30','40'} {'YES','NO'}     for value in uniqueVal‎s:         subLabels = labels[:]                           #subLabels : 그 속성을 제외한 label 리스트         myTree[bestFeatLabel][value] = createTree(splitDataSet(dataSet,                                          bestFeat, value),subLabels)                                                          #ex. myTree={'MARRY_YN':{'YES':'aaa'}}                                                         #myTree['MARRY_YN']['NO']='bbb' {'MARRY_YN': {'YES': 'aaa', 'NO': 'bbb'}}     return myTree       def classify(inputTree,featLabels,testVec):     firstStr=list(inputTree.keys())[0]      #myTree의 맨 처음 key : MARRY_YN     secondDict=inputTree[firstStr]          #MARRY_YN의 값 : {'YES': {'AGE':...}, 'NO': 'NO'}     featIndex=featLabels.index(firstStr)    #속성 리스트에서 MARRY_YN의 위치 3     for key in secondDict.keys():           #MARRY_YN의 값의 key : ['YES','NO']         if testVec[featIndex]==key:         #MARRY_YN에 대한 대답과 key가 같다면            if type(secondDict[key]).__name__=='dict': #그 key의 값이 딕셔너리 타입이면 classify 재실행               classLabel=classify(secondDict[key], featLabels, testVec)            else: classLabel=secondDict[key] #아니면, classLabel은 그 key의 값을 리턴     return classLabel     def main():     data,label = createDataSet()     myTree = createTree(data,label)     featlabel=['GENDER','AGE','JOB_YN','MARRY_YN','CAR_YN']     a=[]     for i in featlabel:            a.append(input('{0}?'.format(i)).upper())     answer=classify(myTree,featlabel,a)     print (myTree)     print('\n'+'쿠폰 반응여부 : '+answer)   if __name__=='__main__':     main()   Colored by Color Scripter

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오우 정말 잘했다