糖逗 发表于 2020-11-18 22:48:46

Python实现FP-growth【链表】【递归】

本帖最后由 糖逗 于 2020-11-19 15:17 编辑

参考书籍:《机器学习实战》


import numpy as np

class treeNode:
    def __init__(self, nameValue, numOccur, parentNode):
      self.name = nameValue
      self.count = numOccur
      self.nodeLink = None
      self.parent = parentNode
      self.children = {}
      
    def inc(self, numOccur):#增加当前节点的值
      self.count += numOccur
   
    def disp(self, ind = 1):#打印树,ind设置成0也可以吗?????
      print(''*ind, self.name, ' ', self.count)
      for child in self.children.values():
            child.disp(ind + 1)#递归


def createTree(dataSet, minSup = 1):#扫描头指针表,删掉哪些出现次数少于minSup的项
    headerTable = {}
    for trans in dataSet:
      for item in trans:
            #https://www.runoob.com/python/att-dictionary-get.html
            headerTable = headerTable.get(item, 0) + dataSet#统计每个字符出现的次数
    for k in list(headerTable.keys()):
      if headerTable < minSup:
            del(headerTable)
    freqItemSet = set(headerTable.keys())#通过minSup筛选的频繁集
    if len(freqItemSet) == 0:#如果每个元素只出现一次,那就直接返回None(没有规律可循)
      return None, None
    for k in headerTable:
      headerTable = , None]#None可能存储指针
    retTree = treeNode('Null Set', 1, None)#根节点设置成空集
    for tranSet, count in dataSet.items():
      localD = {}
      for item in tranSet:
            if item in freqItemSet:#单独出现
                localD = headerTable
      if len(localD) > 0:
            orderedItems = for v in sorted(localD.items(),
                            key = lambda p: p, reverse = True)]#降序排列
            updateTree(orderedItems, retTree, headerTable, count)
    return retTree, headerTable

def updateTree(items, inTree, headerTable, count):
    if items in inTree.children:#如果在节点中,直接更新节点count
      inTree.children].inc(count)
    else:
      inTree.children] = treeNode(items, count, inTree)#如果没有则新增节点
      if headerTable] == None:#当前item不属于频繁集中,更新headerTable
            headerTable] = inTree.children]
      else:
            updateHeader(headerTable], inTree.children])
    if len(items) > 1:
      updateTree(items, inTree.children], headerTable, count)
      
def updateHeader(nodeToTest, targetNode):   
    while (nodeToTest.nodeLink != None):#移动到链表的最后端
      nodeToTest = nodeToTest.nodeLink
    nodeToTest.nodeLink = targetNode#将链表最后段关联当前端      

def ascendTree(leafNode, prefixPath):#从叶子节点递归至根节点
    if leafNode.parent != None:
      prefixPath.append(leafNode.name)
      ascendTree(leafNode.parent, prefixPath)
   
def findPrefixPath(basePat, treeNode): #headerTable中存储的是元素链表的起始指针
    condPats = {}
    while treeNode != None:
      prefixPath = []
      ascendTree(treeNode, prefixPath)
      if len(prefixPath) > 1:
            condPats)] = treeNode.count
      treeNode = treeNode.nodeLink
    return condPats

def loadSimpDat():
    simpDat = [['r', 'z', 'h', 'j', 'p'],
               ['z', 'y', 'x', 'w', 'v', 'u', 't', 's'],
               ['z'],
               ['r', 'x', 'n', 'o', 's'],
               ['y', 'r', 'x', 'z', 'q', 't', 'p'],
               ['y', 'z', 'x', 'e', 'q', 's', 't', 'm']]
    return simpDat

def createInitSet(dataSet):
    retDict = {}
    for trans in dataSet:
      retDict = 1
    return retDict

def mineTree(inTree, headerTable, minSup, preFix, freqItemList):
    bigL = for v in sorted(headerTable.items(), key = lambda p: p)]
    for basePat in bigL:
      newFreqSet = preFix.copy()
      newFreqSet.add(basePat)
      freqItemList.append(newFreqSet)
      condPattBases = findPrefixPath(basePat, headerTable)#提取前缀路径
      myCondTree, myHead = createTree(condPattBases, minSup)#根据前缀路径构建条件FP树
      if myHead != None:      
            mineTree(myCondTree, myHead, minSup, newFreqSet, freqItemList)#从二元扩展到三元,从三元扩展到四元,以此类推递归
    return
            
if __name__ == "__main__":
    simpDat = loadSimpDat()
    initSet = createInitSet(simpDat)
    myFPtree, myHeaderTab = createTree(initSet, 3)
    #myFPtree.disp()
    #res = findPrefixPath('r', myHeaderTab['r'])
    freqItem = []
    mineTree(myFPtree, myHeaderTab, 3, set(), freqItem)
    print(freqItem)

糖逗 发表于 2020-11-18 22:49:17

未完待续,明天接着学习。{:10_259:}
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