Question 2a)
I am also working with arules package and I have the following problem
let suppose the matrix b like:
b<-matrix(c(1,1,1,1,1,1,0,0,1,1,1,1,0,0,1,1,0,1,1,1,1,1,1,1),nrow=6)
rownames(b)=c("T1", "T2", "T3", "T4", "T5", "T6")
colnames(b)=c("It1", "It2", "It3", "It4")
bt<-as(b, "transactions")
rules<-apriori(bt, parameter = list(maxlen=2))
k<-inspect(rules)
k
##we obtain
lhs rhs support confidence lift
1 {} => {It4} 1.0000000 1 1.0
2 {} => {It1} 1.0000000 1 1.0
3 {It3} => {It2} 0.5000000 1 1.5
4 {It3} => {It4} 0.5000000 1 1.0
5 {It3} => {It1} 0.5000000 1 1.0
6 {It2} => {It4} 0.6666667 1 1.0
7 {It2} => {It1} 0.6666667 1 1.0
8 {It4} => {It1} 1.0000000 1 1.0
9 {It1} => {It4} 1.0000000 1 1.0
WRITE(rules)
##we obtain
"rules" "support" "confidence" "lift"
"1" "{} => {It4}" 1 1 1
"2" "{} => {It1}" 1 1 1
"3" "{It3} => {It2}" 0.5 1 1.5
"4" "{It3} => {It4}" 0.5 1 1
"5" "{It3} => {It1}" 0.5 1 1
"6" "{It2} => {It4}" 0.666666666666667 1 1
"7" "{It2} => {It1}" 0.666666666666667 1 1
"8" "{It4} => {It1}" 1 1 1
"9" "{It1} => {It4}" 1 1 1
I want to convert this in a matrix or data frame and the result should be
something like this
##we obtain
rules support confidence lift
1 {}=>{It4} 1.0000000 1 1.0
2 {}=>{It1} 1.0000000 1 1.0
3 {It3}=>{It2} 0.5000000 1 1.5
4 {It3}=>{It4} 0.5000000 1 1.0
5 {It3}=>{It1} 0.5000000 1 1.0
6 {It2}=>{It4} 0.6666667 1 1.0
7 {It2}=>{It1} 0.6666667 1 1.0
8 {It4}=>{It1} 1.0000000 1 1.0
9 {It1}=>{It4} 1.0000000 1 1.0
R> data.frame(rules = labels(rules), quality(rules))
rules support confidence lift
1 {} => {It4} 1.0000000 1 1.0
2 {} => {It1} 1.0000000 1 1.0
3 {It3} => {It2} 0.5000000 1 1.5
4 {It3} => {It4} 0.5000000 1 1.0
5 {It3} => {It1} 0.5000000 1 1.0
6 {It2} => {It4} 0.6666667 1 1.0
7 {It2} => {It1} 0.6666667 1 1.0
8 {It4} => {It1} 1.0000000 1 1.0
9 {It1} => {It4} 1.0000000 1 1.0
In another hand is it possible to obtain all the rules where lHS<=>rhs?. In
our last example that means to obtain
lhs rhs support confidence lift
8 {It4} => {It1} 1.0000000 1 1.0
9 {It1} => {It4} 1.0000000 1 1.0
This is a little more tricky.
We can create a new set of rules with rhs and lhs reversed and then see
if these reversed rules match some of the original rules.
> rulesRev <- new("rules", rhs=lhs(rules), lhs=rhs(rules))
> m <- match(rules, rulesRev, nomatch=0)
> inspect(rules[m,])
lhs rhs support confidence lift
1 {It1} => {It4} 1 1 1
2 {It4} => {It1} 1 1 1
Maybe there is an easier way. I have to think about it.
-Michael
Thanks again Alberto
--
Michael Hahsler
email: mich...@hahsler.net
web: http://michael.hahsler.net
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