How to create a data-driven list of lists in R

Question

The code below does exactly what I need, but there has to be a better way to do this. I need to create a list of lists with titles so that I can subset like this: position.list[["a.com"]]

There has to be a way to do this more efficiently, as what I have below is pretty horrendous. I saw several similar topics, but none seemed to do exactly what I need... I apologize if this has been answered before.

website.vec <- c("a.com", "b.com", "c.com")
position.list <- vector(len = length(website.vec))
position.list <- setNames(as.list(position.list), website.vec)
for (i in 1:length(position.list)) {
  position.list[[i]] <- list()
}

ah the uppercase "R" threw me off...

zap2008
– zap2008

2013-12-16 00:39:09 +00:00
Commented Dec 16, 2013 at 0:39 — zap2008
– zap2008, Commented Dec 16, 2013 at 0:39

Ricardo Saporta · Accepted Answer · 2013-12-16 00:38:41Z

2

Option 1:

setNames(replicate(length(website.vec), list(), simplify=FALSE), website.vec)

Option 2:

lapply(setNames(nm=website.vec), function(x) list())

answered Dec 16, 2013 at 0:38

Ricardo Saporta

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3 Comments

zap2008 Over a year ago

i tried one just like the lapply() method, but without setNames. Thanks!

zap2008 Over a year ago

just ran a couple tests. looks like the the lapply() method is a bit faster (not that it matters for me, as website.vec will likely be no more than 100 elements long...)

Ricardo Saporta Over a year ago

the names in the results of lapply are the names of lapply's main argument.

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