Comparison under more conditions - r

Question

according to my last question i have an new belonging question. After Editing my post and ask there and wait abot a week i want to try it here again.
This time with a better example:
Equip<- c(1,1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,6,6,6)
Notif <-c(1,1,3,4,2,2,2,5,6,7,9,9,15,10,11,12,13,14,16,17,18,19)
rank <- c(1,1,2,3,1,1,1,1,2,3,1,1,2,1,2,3,1,2,3,4,5,6)
Component <- c("Ventil","Motor","Ventil","Ventil","Vergaser","Vergaser","Bremse",
"Lichtmaschine","Bremse","Lichtmaschine","Bremse","Motor","Lichtmaschine",
"Bremse","Bremse","Motor","Vergaser","Motor","Vergaser","Motor",
"Vergaser","Motor")
df <- data.frame(Equip,Notif,rank,Component)
Equip is my subject and rank the actual visit number. Component is the subject what have to be looked for.
I want to have an output like this:
If an Equip(subject) was visited 2 times( rank 1 and 2) look by all Equips with rank 1&2 , if there is any Component which was regarded the first and the second time.
If an Equip(subject) was visited 3 times (rank 1 ,2 and 3) for this look by all Equips, if there is any Component list up 3 times like Equip 1, rank 1, Component Motor, Equip 1, rank 2, Component Motor, Equip 1, rank 3, Component Motor
The output should have the name of the Component, like True "Motor"
I have a code but with this, i can just compare the 1 and the 2 visit, the 2 and the 3 together and so on( i cannot split up again with the ranks, like Equips with 2 ranks, Equips with 3 ranks and so on)
the code is this:
a <- lapply(split(df,df$Equip),function(x){
ll <- split(x,x$rank)
if(length(ll)>1 )
ii <- intersect(ll[[1]]$Component,ll[[2]]$Component ) ## test intersection
else
ii <- NA
c(length(ii)> 0 && !is.na(ii),ii)
})
b <- unlist(a)
c <- table(b,b)
rowSums(c)
Hopefully you can help me. Please ask if there are any questions.
according to your question about the output, and to your way of solution,
Equip Component V1 idx
1: 1 Ventil TRUE 3
2: 2 NA False 1
3: 3 NA False 3
4: 4 NA FALSE 2
5: 5 NA FALSE 3
6: 6 NA FALSE 6
Something like that, but if its easier, Equip and idx is not neccessarilly needed
for Equip with 2 ranks:
TRUE FALSE
0 1
for Equip with 3 ranks:
TRUE FALSE
1 2
for Equip with 6 ranks:
TRUE FALSE
0 1

Answer 1

Here's the output I think would be of interest to you. Its using data.table.
First, we create a data.table from your data.frame df with keys = Equip, Component as follows.
require(data.table) # load package
# then create the data.table with keys as specified above
# Check that both these columns are already sorted out for you!
dt <- data.table(df, key=c("Equip", "Component"))
Second, we create a function that'll give the desired output for a given rank query (2, 3 etc..)
this.check <- function(idx) {
chk <- seq(1, idx)
o <- subset(dt[, all(chk %in% rank), by=c("Equip", "Component")], V1 == TRUE)
if (nrow(o) > 0) o[, idx:=idx]
}
What does this do? Let's run this for rank=1,2. We run this by:
> this.check(2)
# output
Equip Component V1 idx
1: 1 Ventil TRUE 2
2: 5 Bremse TRUE 2
This tells you that for Equip = 1 and 5, there are Components = Ventil and Bremse with rank = 1 and 2, respectively (indicated with idx=2). You also get the column V1 = TRUE, even though I, as #Carl pointed out already, don't understand the need for this. If you require, you can change the column names of this output by using setnames
Third, we use this function to query ranks=1,2, then ranks=1,2,3 .. and so on. This can be accomplished with a simple lapply as follows:
# Let's run the function for idx = 2 to 6.
# This will check from rank = 1,2 until rank=1,2,3,4,5,6
o <- lapply(2:6, function(idx) {
this.check(idx)
})
> o
[[1]]
Equip Component V1 idx
1: 1 Ventil TRUE 2
2: 5 Bremse TRUE 2
[[2]]
Equip Component V1 idx
1: 1 Ventil TRUE 3
[[3]]
NULL
[[4]]
NULL
[[5]]
NULL
It shows that for rank=1,2 and rank=1,2,3 you have some Component. For others there's nothing = NULL.
Finally, we can bind all of these together using rbind to get one single data.table as follows:
o <- do.call(rbind, o)
> o
Equip Component V1 idx
1: 1 Ventil TRUE 2
2: 5 Bremse TRUE 2
3: 1 Ventil TRUE 3
Here, idx=2 are the Component that satisfies rank=1,2 and idx=3 are the ones that satisfy rank=1,2,3.
Putting it all together:
this.check <- function(idx) {
chk <- seq(1, idx)
o <- subset(dt[, all(chk %in% rank), by=c("Equip", "Component")], V1 == TRUE)
if (nrow(o) > 0) o[, idx:=idx]
}
o <- do.call(rbind, lapply(2:6, function(idx) {
this.check(idx)
}))
I hope this helps.
Edit: (After series of exchanges in comments, this is the new solution I propose. I hope this is what you are after.)
require(data.table)
dt <- data.table(df, key=c("Equip", "Component"))
dt[, `:=`(e.max=max(rank)), by=Equip]
dt[, `:=`(ec.max=max(rank)), by=c("Equip", "Component")]
setkey(dt, "e.max", "ec.max")
this.check <- function(idx) {
t1 <- dt[J(idx,idx)]
t2 <- t1[, identical(as.numeric(seq_len(idx)), as.numeric(rank)),
by=c("Equip", "Component")]
o <- table(t2$V1)
if (length(o) == 1)
o <- c(o, "TRUE"=0)
o <- c("idx"=idx, o)
}
o <- do.call(rbind, lapply(2:6, function(idx) this.check(idx)))
> o
# idx FALSE TRUE
# [1,] 2 1 0
# [2,] 3 2 1
# [3,] 4 1 0
# [4,] 5 1 0
# [5,] 6 1 0

Answer 2

If I make an array of your data, columnwise, as
foo<-cbind(Equip,Notif, rank, Component)
eqp<-1 # later, loop over all values
foo[c( which( foo[,1]==eqp & (foo[,3]==1 | foo[,3]==2) ) ),4]
[1] "Ventil" "Motor" "Ventil"
Feed those results to table and extract items with count ==2
Clearly any item which shows up twice is what you want.
This is not an answer I'd recommend using, since tools like ddply and aggregate will do this much more cleanly, but I want to be sure that this is the answer you're after, assuming a loop over eqp values in the original Equip .