一个很重要的思想:
如果某元素不是频繁的,那么包含该元素的超集也是不频繁的。
发现dataSet的频繁集:
import numpy as np import pandas as pd def loadDataSet(): return [[1, 3, 4], [2, 3, 5], [1, 2, 3, 5], [2, 5]] #提取数据集中所有的单独数据,参数是数据集 def createC1(dataSet): C1 = [] for transaction in dataSet: for item in transaction: if not [item] in C1: C1.append([item]) C1.sort() return map(frozenset, C1)#use frozen set so we #can use it as a key in a dict #找出ck字典中所有的序列在D数据集中出现的次数以及由此 计算出的频繁度 def scanD(D, Ck, minSupport): ssCnt = {} #找出ck字典中所有的序列在D数据集中出现的次数 for tid in D: for can in Ck: if can.issubset(tid): if not ssCnt.has_key(can): ssCnt[can]=1 else: ssCnt[can] += 1 numItems = float(len(D)) retList = [] supportData = {} #计算每个序列的频繁度 for key in ssCnt: support = ssCnt[key]/numItems if support >= minSupport: retList.insert(0,key) supportData[key] = support #返回合格的序列和对于的支持度 return retList, supportData #Lk是k-1层的频繁度序列,这里是拼接处k层序列,在用上面的scanD函数计算出合格的序列 def aprioriGen(Lk, k): #creates Ck retList = [] lenLk = len(Lk) for i in range(lenLk): for j in range(i+1, lenLk): L1 = list(Lk[i])[:k-2]; L2 = list(Lk[j])[:k-2] L1.sort(); L2.sort() if L1==L2: #if first k-2 elements are equal retList.append(Lk[i] | Lk[j]) #set union return retList # Apriori算法的主函数,用到以上的所有函数 def apriori(dataSet, minSupport = 0.5): C1 = createC1(dataSet) D = map(set, dataSet) L1, supportData = scanD(D, C1, minSupport) L = [L1] k = 2 while (len(L[k-2]) > 0): Ck = aprioriGen(L[k-2], k) Lk, supK = scanD(D, Ck, minSupport) supportData.update(supK) L.append(Lk) k += 1 #L是所有的合格的频繁度序列,supportData是一个字典,键是序列,值是支持度 return L, supportData
从找出的频繁集中找关联规则:
def generateRules(L, supportData, minConf=0.7): #supportData is a dict coming from scanD bigRuleList = [] for i in range(1, len(L)):#only get the sets with two or more items for freqSet in L[i]: H1 = [frozenset([item]) for item in freqSet] if (i > 1): rulesFromConseq(freqSet, H1, supportData, bigRuleList, minConf) else: calcConf(freqSet, H1, supportData, bigRuleList, minConf) return bigRuleList def calcConf(freqSet, H, supportData, brl, minConf=0.7): prunedH = [] #create new list to return for conseq in H: conf = supportData[freqSet]/supportData[freqSet-conseq] #calc confidence if conf >= minConf: print freqSet-conseq,'-->',conseq,'conf:',conf brl.append((freqSet-conseq, conseq, conf)) prunedH.append(conseq) return prunedH def rulesFromConseq(freqSet, H, supportData, brl, minConf=0.7): m = len(H[0]) if (len(freqSet) > (m + 1)): #try further merging Hmp1 = aprioriGen(H, m+1)#create Hm+1 new candidates Hmp1 = calcConf(freqSet, Hmp1, supportData, brl, minConf) if (len(Hmp1) > 1): #need at least two sets to merge rulesFromConseq(freqSet, Hmp1, supportData, brl, minConf)
免责声明:本站发布的内容(图片、视频和文字)以原创、转载和分享为主,文章观点不代表本网站立场,如果涉及侵权请联系站长邮箱:is@yisu.com进行举报,并提供相关证据,一经查实,将立刻删除涉嫌侵权内容。