R help - May 2011 - Remove duplicate elements in lists via recursive indexing

Dear list,

I'm trying to solve something pretty basic here, but I can't really come
up with a good solution. Basically, I would just like to remove 
duplicated named elements in lists via a their respective recursive 
indexes (given that I have a routine that identifies these recursive 
indexes). Here's a little example:

# VECTORS
# Here, it's pretty simple to remove duplicated entries
y <- c(1,2,3,1,1)
idx.dupl <- which(duplicated(y))
y <- y[-idx.dupl]
# /

# LISTS
x <- list(a=list(a.1.1=1, a.1.1=2, a.1.1=3))

x[[c(1,1)]]
x[[c(1,2)]] # Should be removed.
x[[c(1,3)]] # Should be removed.

# Let's say a 'checkDuplicates' routine would give me:
idx.dupl <- list(c(1,2), c(1,3))

# Remove first duplicate:
x[[idx.dupl[[1]]]] <- NULL
x
# Problem:
# Once I remove the first duplicate, my duplicate index would have to be
# updated as well as there is not third element anymore.
x[[idx.dupl[[2]]]] <- NULL

# So something like this would not work:
sapply(idx.dupl, function(x.idx){
     x[[x.idx]] <<- NULL
})
# /

Sorry if I'm missing something totally obvious here, but do you have any 
idea how to solve this?

Thanks a lot,
Janko

Dear Janko,
I think requires a for loop. The approach I took here was mark the dups, then
dump them all in one hit:

testData = expand.grid(letters[1:4],c(1:3))
testData$keep=F
uniqueIDS = unique(testData$Var1)
for(thisID in uniqueIDS) {
	firstCaseOnly = match(thisID,testData$Var1)
	testData[firstCaseOnly,"keep"]=T
}

(testData = testData[testData$keep==T,])


On 23 May 2011, at 11:59 AM, Janko Thyson wrote:
> Dear list,
> 
> I'm trying to solve something pretty basic here, but I can't really
come up with a good solution. Basically, I would just like to remove duplicated
named elements in lists via a their respective recursive indexes (given that I
have a routine that identifies these recursive indexes). Here's a little
example:
> 
> # VECTORS
> # Here, it's pretty simple to remove duplicated entries
> y <- c(1,2,3,1,1)
> idx.dupl <- which(duplicated(y))
> y <- y[-idx.dupl]
> # /
> 
> # LISTS
> x <- list(a=list(a.1.1=1, a.1.1=2, a.1.1=3))
> 
> x[[c(1,1)]]
> x[[c(1,2)]] # Should be removed.
> x[[c(1,3)]] # Should be removed.
> 
> # Let's say a 'checkDuplicates' routine would give me:
> idx.dupl <- list(c(1,2), c(1,3))
> 
> # Remove first duplicate:
> x[[idx.dupl[[1]]]] <- NULL
> x
> # Problem:
> # Once I remove the first duplicate, my duplicate index would have to be
> # updated as well as there is not third element anymore.
> x[[idx.dupl[[2]]]] <- NULL
> 
> # So something like this would not work:
> sapply(idx.dupl, function(x.idx){
>    x[[x.idx]] <<- NULL
> })
> # /
> 
> Sorry if I'm missing something totally obvious here, but do you have
any idea how to solve this?
> 
> Thanks a lot,
> Janko
> 
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