CODEWITHSUNDEEP
performancerlistmemory
Alex
Alex

Reputation: 19803

preallocate list in R

It is inefficient in R to expand a data structure in a loop. How do I preallocate a list of a certain size? matrix makes this easy via the ncol and nrow arguments. How does one do this in lists? For example:

x <- list()
for (i in 1:10) {
    x[[i]] <- i
}

I presume this is inefficient. What is a better way to do this?

Upvotes: 59

Views: 29211

Answers (4)

stevec
stevec

Reputation: 52238

All 3 existing answers are great.

The reason the vector() function can create a list is explained in JennyBC's purrr tutorial:

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.

To preallocate a list

list <- vector(mode = "list", length = 10)

To preallocate a vector

vec <- rep(NA, 10)

Upvotes: 1

Justin
Justin

Reputation: 43255

To expand on what @Jilber said, lapply is specially built for this type of operation.

instead of the for loop, you could use:

x <- lapply(1:10, function(i) i)

You can extend this to more complicated examples. Often, what is in the body of the for loop can be directly translated to a function which accepts a single row that looks like a row from each iteration of the loop.

Upvotes: 25

Jilber Urbina
Jilber Urbina

Reputation: 61154

Something like this:

   x <- vector('list', 10)

But using lapply is the best choice

Upvotes: 6

Luciano Selzer
Luciano Selzer

Reputation: 10016

vector can create empty vector of the desired mode and length.

x <- vector(mode = "list", length = 10)

Upvotes: 98

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