arules包.错误:为产生建议下标

Question

我正在尝试用arules包做推荐

我有这些数据

Data
      Client product  N       Date
    1      A  Banana  1 01/01/2016
    2      A  Tomato  1 01/01/2016
    3      A    Tuna  1 01/01/2016
    4      B  Orange  2 01/01/2016
    5      B  Tomato  3 02/01/2016
    6      C    Kiwi 11 08/01/2016

接下来，我使用了以下代码

trans = as(split(Data$product, Data$Client), "transactions")

Sales<- as(trans, "data.frame")

rules = apriori(trans, parameter = list(support = 0.001, confidence = 0.005))
rules.sorted <- sort(rules, by="lift")

# find redundant rules
subset.matrix <- is.subset(rules.sorted, rules.sorted)
subset.matrix[lower.tri(subset.matrix, diag=T)] <- NA
redundant <- colSums(subset.matrix, na.rm=T) >= 1
which(redundant)
rules.pruned <- rules.sorted[!redundant]
inspect(rules.pruned)
rules = rules.pruned

我知道这些规则：

lhs         rhs        support confidence lift
1 {Tuna}   => {Banana} 0.3333333  1.0000000  3.0
2 {Orange} => {Tomato} 0.3333333  1.0000000  1.5
3 {Tuna}   => {Tomato} 0.3333333  1.0000000  1.5
4 {Banana} => {Tomato} 0.3333333  1.0000000  1.5
5 {}       => {Kiwi}   0.3333333  0.3333333  1.0
6 {}       => {Orange} 0.3333333  0.3333333  1.0
7 {}       => {Tuna}   0.3333333  0.3333333  1.0
8 {}       => {Banana} 0.3333333  0.3333333  1.0
9 {}       => {Tomato} 0.6666667  0.6666667  1.0

但是现在，对于所有的客户，我想推荐3种产品：

for (i in 1:3) {

        reco=function(x){
                rulesMatchLHS = is.subset(rules@lhs,x)
                suitableRules =  rulesMatchLHS & !(is.subset(rules@rhs,x))
                order.rules = sort(rules[suitableRules], by = "lift")
                LIST(order.rules@rhs)[[i]]


        }

        NewS <- sapply(1:length(trans), function(x) reco(trans[x]))
        NewS <- as.data.frame(NewS)
        Sales <-cbind(Sales,NewS)

}

此代码产生错误。

LIST(order.rules@rhs)[i]中的错误:下标超出界限

我认为发生这种情况是因为我没有为所有用户提供推荐，但我希望代码继续，并在本例中“没有建议”。

做这件事的最好方法是什么？

Answer 1

我想你想要这样的代码。

阅读数据和地雷规则：

library(arules)

Data <- structure(list(Client = structure(c(1L, 1L, 1L, 2L, 2L, 3L), .Label = c("A", "B",  "C"), class = "factor"), product = structure(c(1L, 4L, 5L, 3L, 4L, 2L), .Label = c("Banana", "Kiwi", "Orange", "Tomato", "Tuna"), class = "factor"), N = c(1L, 1L, 1L, 2L, 3L, 11L), Date = structure(c(1L, 1L, 1L, 1L, 2L, 3L), .Label = c("01/01/2016", "02/01/2016", "08/01/2016"), class = "factor")), .Names = c("Client", "product", "N", "Date"), class = "data.frame", row.names = c(NA, -6L))

trans <- as(split(Data$product, Data$Client), "transactions")

rules <-  apriori(trans, parameter = list(support = 0.001, confidence = 0.5, maxlen = 2))
inspect(rules) 

输出：

lhs         rhs      support   confidence lift
1 {}       => {Tomato} 0.6666667 0.6666667  1.0 
2 {Orange} => {Tomato} 0.3333333 1.0000000  1.5 
3 {Tomato} => {Orange} 0.3333333 0.5000000  1.5 
4 {Tuna}   => {Banana} 0.3333333 1.0000000  3.0 
5 {Banana} => {Tuna}   0.3333333 1.0000000  3.0 
6 {Tuna}   => {Tomato} 0.3333333 1.0000000  1.5 
7 {Tomato} => {Tuna}   0.3333333 0.5000000  1.5 
8 {Banana} => {Tomato} 0.3333333 1.0000000  1.5 
9 {Tomato} => {Banana} 0.3333333 0.5000000  1.5 

提出建议：

reco <- function(rules, newTrans){
     rules.sorted <- sort(rules, by="lift")
     rhs_labels <- unlist(as(rhs(rules.sorted), "list"))

     matches <- is.subset(lhs(rules.sorted), newTrans) &
        !(is.subset(rhs(rules.sorted), newTrans))
     apply(matches, MARGIN = 2, FUN = function(x) unique(rhs_labels[x]))
}

reco(rules, trans)

这三项交易(即客户)的产出：

$`{Banana,Tomato,Tuna}`
[1] "Orange"

$`{Orange,Tomato}`
[1] "Tuna"   "Banana"

$`{Kiwi}`
[1] "Tomato"

几个注意事项：

• 我只挖掘长度为1和2的规则，这样效率更高，不需要寻找多余的规则。
• 我增加了信心。
• 包重新实验室将使用"AR“方法执行这种类型的建议。这是目前工作不正确，但它将很快工作。