Question

Suppose we would like to store pairs as elements in a list, e.g.,

> x <- list()
> x[[1]] <- list(1,2)
> x[[2]] <- list(3,4)
> x
[[1]]
[[1]][[1]]
[1] 1

[[1]][[2]]
[1] 2


[[2]]
[[2]][[1]]
[1] 3

[[2]][[2]]
[1] 4

If one wants to create the same thing with concatenation, without explicit reference to an index of the outer list, a natural approach (see EDIT below for motivation for this approach) is

> c(list(list(1,2)),list(3,4))
[[1]]
[[1]][[1]]
[1] 1

[[1]][[2]]
[1] 2


[[2]]
[1] 3

[[3]]
[1] 4

Is there a way to avoid the “flattening” of the second argument to c() and thereby use concatenation to produce x?

EDIT: I should have made my intended application clear.

In the end, I would like to add pairs to a list whenever I find one as I go through a loop. Currently I maintain an index for the main list to add new pairs to the end of it when one is found, however using c is perhaps more natural. So using an explicit index the loop may look like this

index <- 1
l <- list()
for (i in 1:10) {
    for (j in (i+1):10) {
        if ( (i+j)%%2 == 0 ) {
            l[[index]] <- list(i,j)
            index <- index + 1
        }
    }
}

And the failed attempt to use c looks like this

l <- list()
for (i in 1:10) {
    for (j in (i+1):10) {
        if ( (i+j)%%2 == 0 )
            l <- c(l,list(i,j))
    }
}

ANSWER: as noted by Tommy below, an extra list() is needed, so the following works:

l <- list()
for (i in 1:10) {
    for (j in (i+1):10) {
        if ( (i+j)%%2 == 0 )
            l <- c(l,list(list(i,j)))
    }
}

Kind of obvious thing I should have figured out. Tommy also notes that this might not be a very good way of doing things.

Answer 1

I guess the simple answer is “no, you need the extra list() wrapper”.

…but in general, it is better (faster/less memory) to add values using indexing to a pre-allocated list than using c()

x <- NULL
for(i in 1:2) x <- c(x, list(list(i,i+1)))
x

x <- vector('list', 2)
for(i in seq_along(x)) x[[i]] <- list(i,i+1)
x

And if lapply can be used, that is usually even more efficient:

lapply(1:2, function(i) list(i, i+1))