python中是否有计算关闭频繁项集的函数?
我正在搜索一个python算法来计算紧密频繁项集关联,但我只找到了一个java代码?有人能帮我吗?给你python中是否有计算关闭频繁项集的函数?,python,apriori,Python,Apriori,我正在搜索一个python算法来计算紧密频繁项集关联,但我只找到了一个java代码?有人能帮我吗?给你 #Import all basic libray import pandas as pd from mlxtend.preprocessing import TransactionEncoder import time from mlxtend.frequent_patterns import fpgrowth #Task1 : Compute Frequent Item Set using
#Import all basic libray
import pandas as pd
from mlxtend.preprocessing import TransactionEncoder
import time
from mlxtend.frequent_patterns import fpgrowth
#Task1 : Compute Frequent Item Set using mlxtend.frequent_patterns
te = TransactionEncoder()
te_ary = te.fit(dataset).transform(dataset)
df = pd.DataFrame(te_ary, columns=te.columns_)
start_time = time.time()
frequent = fpgrowth(df, min_support=0.001, use_colnames=True)
print('Time to find frequent itemset')
print("--- %s seconds ---" % (time.time() - start_time))
# Task 2&3: Find closed/max frequent itemset using frequent itemset found in task1
su = frequent.support.unique()#all unique support count
#Dictionay storing itemset with same support count key
fredic = {}
for i in range(len(su)):
inset = list(frequent.loc[frequent.support ==su[i]]['itemsets'])
fredic[su[i]] = inset
#Dictionay storing itemset with support count <= key
fredic2 = {}
for i in range(len(su)):
inset2 = list(frequent.loc[frequent.support<=su[i]]['itemsets'])
fredic2[su[i]] = inset2
#Find Closed frequent itemset
start_time = time.time()
cl = []
for index, row in frequent.iterrows():
isclose = True
cli = row['itemsets']
cls = row['support']
checkset = fredic[cls]
for i in checkset:
if (cli!=i):
if(frozenset.issubset(cli,i)):
isclose = False
break
if(isclose):
cl.append(row['itemsets'])
print('Time to find Close frequent itemset')
print("--- %s seconds ---" % (time.time() - start_time))
#Find Max frequent itemset
start_time = time.time()
ml = []
for index, row in frequent.iterrows():
isclose = True
cli = row['itemsets']
cls = row['support']
checkset = fredic2[cls]
for i in checkset:
if (cli!=i):
if(frozenset.issubset(cli,i)):
isclose = False
break
if(isclose):
ml.append(row['itemsets'])
print('Time to find Max frequent itemset')
print("--- %s seconds ---" % (time.time() - start_time))
#导入所有基础库
作为pd进口熊猫
从mlxtend.preprocessing导入事务编码
导入时间
从mlxtend.frequency_模式导入fpgrowth
#任务1:使用mlxtend.Frequent\u模式计算频繁项集
te=TransactionnCoder()
te_ary=te.fit(数据集).transform(数据集)
df=pd.DataFrame(数据元,列=te.columns)
开始时间=time.time()
频繁=fpgrowth(df,最小支持=0.001,使用colnames=True)
打印('查找频繁项集的时间')
打印(“--%s秒--”%(time.time()-start\u time))
#任务2和3:使用在任务1中找到的频繁项集查找已关闭/最大频繁项集
su=频繁.support.unique()#所有唯一支持计数
#使用相同的支持计数键存储条目集
弗雷迪克={}
对于范围内的i(len(su)):
inset=list(frequent.loc[frequent.support==su[i]['itemsets'])
fredic[su[i]]=插图
#词汇使用support count存储项集通过添加更多细节来解释问题我正在寻找一个用python实现的函数来计算close frequent itemets关联规则。即使你得到了一个答案(对你来说很好),你也确实需要继续学习。这将阻止投票被否决或关闭。祝您今天过得愉快