Basic question: In R, how can I make a list and later populate it with vector elements?
l <- list() l[1] <- c(1,2,3)
This gives the error "number of items to replace is not a multiple of replacement length", so R is trying to unpack the vector. The only way I have found to work so far is to add the vectors when making the list.
l <- list(c(1,2,3), c(4,5,6))
list() function in R creates a list of the specified arguments. The vectors specified as arguments in this function may have different lengths. Syntax: list(arg1, arg2, ..)
How to Create Lists in R? We can use the list() function to create a list. Another way to create a list is to use the c() function. The c() function coerces elements into the same type, so, if there is a list amongst the elements, then all elements are turned into components of a list.
A list is actually still a vector in R, but it's not an atomic vector. We construct a list explicitly with list() but, like atomic vectors, most lists are created some other way in real life.
According to ?"["
(under the section "recursive (list-like) objects"):
Indexing by ‘[’ is similar to atomic vectors and selects a list of the specified element(s). Both ‘[[’ and ‘$’ select a single element of the list. The main difference is that ‘$’ does not allow computed indices, whereas ‘[[’ does. ‘x$name’ is equivalent to ‘x[["name", exact = FALSE]]’. Also, the partial matching behavior of ‘[[’ can be controlled using the ‘exact’ argument.
Basically, for lists, [
selects more than one element, so the replacement must be a list (not a vector as in your example). Here's an example of how to use [
on lists:
l <- list(c(1,2,3), c(4,5,6)) l[1] <- list(1:2) l[1:2] <- list(1:3,4:5)
If you only want to replace one element, use [[
instead.
l[[1]] <- 1:3
Use [[1]]
as in
l[[1]] <- c(1,2,3) l[[2]] <- 1:4
and so. Also recall that preallocation is much more efficient, so if you know how long your list is going to be, use something like
l <- vector(mode="list", length=N)
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