更新完apriori的频繁子项的代码

This commit is contained in:
jiangzhonglian
2017-03-16 17:50:10 +08:00
parent f09919029f
commit a0427c0812
3 changed files with 61 additions and 20 deletions

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@@ -5,8 +5,9 @@
Created on Oct 12, 2010
Update on 2017-02-27
Decision Tree Source Code for Machine Learning in Action Ch. 3
@author: Peter Harrington/jiangzhonglian
@author: Peter Harrington/片刻
'''
print(__doc__)
import operator
from math import log
import decisionTreePlot as dtPlot

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@@ -5,8 +5,9 @@
Created on Feb 4, 2011
Update on 2017-03-02
Tree-Based Regression Methods Source Code for Machine Learning in Action Ch. 9
@author: Peter Harrington/jiangzhonglian
@author: Peter Harrington/片刻
'''
print(__doc__)
from numpy import *

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@@ -3,86 +3,125 @@
'''
Created on Mar 24, 2011
Update on 2017-03-16
Ch 11 code
@author: Peter
@author: Peter/片刻
'''
print(__doc__)
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:
# 遍历所有的元素然后append到C1中
C1.append([item])
# 对数组进行 从小到大 的排序
C1.sort()
return map(frozenset, C1) # use frozen set so we
# can use it as a key in a dict
# frozenset表示冻结的set集合元素无可改变可以把它当字典的key来使用
return map(frozenset, C1)
def scanD(D, Ck, minSupport):
# 临时存放查看Ck每个元素 并 计算元素出现的次数 生成相应的字典
# D用来判断CK中的元素是否存在于原数据D中
ssCnt = {}
for tid in D:
for can in Ck:
# s.issubset(t) 测试是否 s 中的每一个元素都在 t 中
if can.issubset(tid):
if not ssCnt.has_key(can): ssCnt[can]=1
else: ssCnt[can] += 1
if not ssCnt.has_key(can):
ssCnt[can] = 1
else:
ssCnt[can] += 1
# 元素有多少行
numItems = float(len(D))
retList = []
supportData = {}
for key in ssCnt:
# 计算支持度
support = ssCnt[key]/numItems
if support >= minSupport:
# 在retList的首位插入元素只存储支持度满足频繁项集的值
retList.insert(0, key)
# 存储所有的key和对应的support值
supportData[key] = support
return retList, supportData
def aprioriGen(Lk, k): #creates Ck
# creates Ck
def aprioriGen(Lk, k):
"""aprioriGen(循环数据集,然后进行两两合并)
Args:
Lk 频繁项集
k 元素的前k-2相同就进行合并
Returns:
retList 元素两两合并的数据集
"""
retList = []
lenLk = len(Lk)
# 循环Lk这个数组
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
L1 = list(Lk[i])[: k-2]
L2 = list(Lk[j])[: k-2]
# print '-----', Lk, Lk[i], L1
L1.sort()
L2.sort()
# 第一次L1,L2为空元素直接进行合并返回元素两两合并的数据集
# if first k-2 elements are equal
if L1 == L2:
# set union
retList.append(Lk[i] | Lk[j])
return retList
def apriori(dataSet, minSupport = 0.5):
def apriori(dataSet, minSupport=0.5):
# 冻结每一行数据
C1 = createC1(dataSet)
D = map(set, dataSet)
# 计算支持support
# 计算支持support L1表示满足support的key, supportData表示全集的集合
L1, supportData = scanD(D, C1, minSupport)
print("outcome: ", supportData)
# print "L1=", L1, "\n", "outcome: ", supportData
L = [L1]
k = 2
while (len(L[k-2]) > 0):
# print 'L[k-2]=', L[k-2], k
Ck = aprioriGen(L[k-2], k)
Lk, supK = scanD(D, Ck, minSupport)#scan DB to get Lk
# print 'Ck=', Ck
# can DB to get Lk
Lk, supK = scanD(D, Ck, minSupport)
supportData.update(supK)
# L元素在增加
L.append(Lk)
k += 1
# print 'k=', k, len(L[k-2])
return L, supportData
def main():
# project_dir = os.path.dirname(os.path.dirname(os.getcwd()))
# 1.收集并准备数据
# dataMat, labelMat = loadDataSet("%s/resources/Apriori_testdata.txt" % project_dir)
# 1. 加载数据
dataSet = loadDataSet()
print(dataSet)
# 调用 apriori 做购物篮分析
apriori(dataSet, minSupport = 0.7)
L, supportData = apriori(dataSet, minSupport=0.7)
print L, supportData
if __name__=="__main__":
if __name__ == "__main__":
main()