Question

问题：

apriori包的arules功能从输入的交易中推断出关联规则并报告支持，置信度和解除每条规则的。关联规则源自频繁项集。我想在输入事务中获得最频繁的项目集。具体来说，我想获得具有给定最小支持的所有项目集。 itemset的支持是包含itemset的事务数与事务总数的比率。

要求：

我非常希望从apriori函数的中间结果中找到最常用的项集。也就是说，我宁愿不从头开始编写程序来计算最频繁的项集，因为apriori函数已经将它计算为中间步骤。尽管如此，如果确实没有合理的方式来访问apriori函数的中间结果，我会对其他解决方案持开放态度。
我宁愿不对apriori函数的结果进行字符串操作，因为这种方法过于依赖于apriori函数结果的字符串表示。同样，如果事实证明没有更好的选择，我可以采用这种方法。
我知道arules包提供的itemFrequency功能。不幸的是，此功能仅使用单个项目报告项目集。我对所有长度的项目集感兴趣，并且支持率最低。
我希望输出按数字顺序排序，然后按字典顺序排序。

示例输入：

a,b
a,b,c

程序：

# The following is how I'm using apriori to infer the association rules.
library(package = "arules")
transactions = read.transactions(file = file("stdin"), format = "basket", sep = ",")
rules = apriori(transactions, parameter = list(minlen=1, sup = 0.001, conf = 0.001))
WRITE(rules, file = "", sep = ",", quote = TRUE, col.names = NA)

当前输出：

"","rules","support","confidence","lift"
"1","{} => {c}",0.5,0.5,1
"2","{} => {b}",1,1,1
"3","{} => {a}",1,1,1
"4","{c} => {b}",0.5,1,1
"5","{b} => {c}",0.5,0.5,1
"6","{c} => {a}",0.5,1,1
"7","{a} => {c}",0.5,0.5,1
"8","{b} => {a}",1,1,1
"9","{a} => {b}",1,1,1
"10","{b,c} => {a}",0.5,1,1
"11","{a,c} => {b}",0.5,1,1
"12","{a,b} => {c}",0.5,0.5,1

期望输出：

"itemset","support"
"{a}",1
"{a,b}",1
"{b}",1
"{a,b,c}",0.5
"{a,c}",0.5
"{b,c}",0.5
"{c}",0.5

Answer 1

我在arules包的引用manual中找到了generatingItemsets函数。

library(package = "arules")
transactions = read.transactions(file = file("stdin"), format = "basket", sep = ",")
rules = apriori(transactions, parameter = list(minlen=1, sup = 0.001, conf = 0.001))
itemsets <- unique(generatingItemsets(rules))
itemsets.df <- as(itemsets, "data.frame")
frequentItemsets <- itemsets.df[with(itemsets.df, order(-support,items)),]
names(frequentItemsets)[1] <- "itemset"
write.table(frequentItemsets, file = "", sep = ",", row.names = FALSE)

如何从R中的apriori调用中获取常用项集的频率？

1 个答案: