在先验算法中创建项目集

Question

我在读关于联想分析的书，题为“机器学习中的行动”。以下代码在书中给出

k-2的事情可能有点混乱。让我们再看一看。当您从{0}、{1}、{2}创建{0,1} {0,2}时，您只是组合了项。现在，如果要使用{0,1} {0,2}，{1,2}创建三项集，怎么办？如果您对每个集合进行合并，则得到{0，1，2}，{0，1，2}，{0，1，2}。没错。三次都是一样的。现在，您必须扫描三项集的列表，才能得到唯一的值。你试图将你浏览列表的次数保持在最低限度。现在，如果将第一个元素{0,1} {0, 2}，{ 1，2}进行比较，并且只接受具有相同第一项的元素的合并，那么您会得到什么呢？{0，1，2}只需一次。现在，您不必查看列表，寻找唯一的值。

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] # Join sets if first k-2 items are equal
            L1.sort(); L2.sort()
            if L1==L2:
                retList.append(Lk[i] | Lk[j])
    return retLis

假设我正在调用上面的函数

Lk = [frozenset({2, 3}), frozenset({3, 5}), frozenset({2, 5}), frozenset({1, 3})]

k = 3

aprioriGen(Lk,3)

我得到了以下输出

[frozenset({2, 3, 5})]

我认为上述逻辑中存在缺陷，因为我们缺少了其他组合，如{1,2,3}，{1,3,5}。难到不是么？我的理解对吗？

Answer 1

我认为您正在遵循下面的链接，输出集取决于我们传递的minSupport。

http://adataanalyst.com/machine-learning/apriori-algorithm-python-3-0/

如果我们将minSupport值降到0.2，就会得到所有的集合。

下面是完整的代码

# -*- coding: utf-8 -*-
"""
Created on Mon Dec 31 16:57:26 2018

@author: rponnurx
"""

from numpy import *

def loadDataSet():
    return [[1, 3, 4], [2, 3, 5], [1, 2, 3, 5], [2, 5]]

def createC1(dataSet):
    C1 = []
    for transaction in dataSet:
        for item in transaction:
            if not [item] in C1:
                C1.append([item])

    C1.sort()
    return list(map(frozenset, C1))#use frozen set so we
                            #can use it as a key in a dict  

def scanD(D, Ck, minSupport):
    ssCnt = {}
    for tid in D:
        for can in Ck:
            if can.issubset(tid):
                if not can in ssCnt: ssCnt[can]=1
                else: ssCnt[can] += 1
    numItems = float(len(D))
    retList = []
    supportData = {}
    for key in ssCnt:
        support = ssCnt[key]/numItems
        if support >= minSupport:
            retList.insert(0,key)
        supportData[key] = support
    return retList, supportData

dataSet = loadDataSet()
print(dataSet)

C1 = createC1(dataSet)

print(C1)

#D is a dataset in the setform.

D = list(map(set,dataSet))
print(D)

L1,suppDat0 = scanD(D,C1,0.5)
print(L1)

def aprioriGen(Lk, k): #creates Ck
    retList = []
    print("Lk")
    print(Lk)
    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

def apriori(dataSet, minSupport = 0.5):
    C1 = createC1(dataSet)
    D = list(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)#scan DB to get Lk
        supportData.update(supK)
        L.append(Lk)
        k += 1
    return L, supportData

L,suppData = apriori(dataSet,0.2)

print(L)

输出：[frozenset({5})，frozenset({2})，frozenset({4})，frozenset({3})，frozenset({1，4})，frozenset({1，2})，frozenset({1，5})，frozenset({2，3})，frozenset({3，5})，frozenset({2，5})，frozenset({1，3})，frozenset({1，4})，frozenset({3，4})，frozenset({1，3，5})，frozenset({1，3，5}))，冻结({1，2，5})，冻结({ 2，3，5})，冻结({1，3，4})，冻结({1，2，3，5})，[]

谢谢，拉杰斯瓦里·庞努鲁