How to take the union of element in a nested list in R

Question

I have a nested list in say lst(all the elements are of class int). I don't know the length of lst in advance; however I do know that each element of lst is a list of length say k

length(lst[[i]]) # this equals k and is known in advance, 
                 # this is true for i = 1 ... length(lst)

How do I take the union of the 1st element, 2nd element, ..., kth element of all the elements of lst

Specifically, if the length of lst is n, I want (not R code):

# I know that union can only be taken for 2 elements, 
# following is for illustration purposes
listUnion1 <- union(lst[[1, 1]], lst[[2, 1]], ..., lst[[n, 1]])
listUnion2 <- union(lst[[1, 2]], lst[[2, 2]], ..., lst[[n, 2]])
.
.
.
listUnionk <- union(lst[[1, k]], lst[[2, k]], ..., lst[[n, k]])

Any help or pointers are greatly appreciated.

Here is a dataset that can be used, n = 3 and k = 2

list(structure(list(a = 1:5, b = 6:11), .Names = c("a", "b")), 
    structure(list(a = 6:11, b = 1:5), .Names = c("a", "b")), 
    structure(list(a = 12, b = 12), .Names = c("a", "b")))

Answer 1

Here is a general solution, similar in spirit to that of @Ramnath, but avoiding the use of union() which is a binary function. The trick is to note that union() is implemented as:

unique(c(as.vector(x), as.vector(y)))

and the bit inside unique() can be achieved by unlisting the nth component of each list.

The full solution then is:

unionFun <- function(n, obj) {
    unique(unlist(lapply(obj, `[[`, n)))
}
lapply(seq_along(lst[[1]]), FUN = unionFun, obj = lst)

which gives:

[[1]]
 [1]  1  2  3  4  5  6  7  8  9 10 11 12

[[2]]
 [1]  6  7  8  9 10 11  1  2  3  4  5 12

on the data you showed.

A couple of useful features of this are:

• we use `[[` to subset obj in unionFun. This is similar to function(x) x$a in @Ramnath's Answer. However, we don't need an anonymous function (we use `[[` instead). The equivalent to @Ramnath's Answer is: lapply(lst, `[[`, 1)
• to generalise the above, we replace the 1 above with n in unionFun(), and allow our list to be passed in as argument obj.

Now that we have a function that will provide the union of the nth elements of a given list, we can lapply() over the indices k, applying our unionFun() to each sub-element of lst, using the fact that the length of lst[[1]] is the same as length(lst[[k]]) for all k.

If it helps to have the names of the nth elements in the returned object, we can do:

> unions <- lapply(seq_along(lst[[1]]), FUN = unionFun, obj = lst)
> names(unions) <- names(lst[[1]])
> unions
$a
 [1]  1  2  3  4  5  6  7  8  9 10 11 12

$b
 [1]  6  7  8  9 10 11  1  2  3  4  5 12

Answer 2

Here is one solution

# generate dummy data
x1 = sample(letters[1:5], 20, replace = T)
x2 = sample(letters[1:5], 20, replace = T)
df = data.frame(x1, x2, stringsAsFactors = F)

# find unique elements in each column
union_df = apply(df, 2, unique)

Let me know if this works

EDIT: Here is a solution for lists using the data you provided

mylist = list(structure(list(a = 1:5, b = 6:11), .Names = c("a", "b")), 
              structure(list(a = 6:11, b = 1:5), .Names = c("a", "b")), 
              structure(list(a = 12, b = 12), .Names = c("a", "b")))
list_a = lapply(mylist, function(x) x$a)
list_b = lapply(mylist, function(x) x$b)

union_a = Reduce(union, list_a)
union_b = Reduce(union, list_b)

If you have more than 2 elements in your list, we could generalize this code.