CODEWITHSUNDEEP
rlistapply
Will T-E
Will T-E

Reputation: 637

How do I apply a list of argument values to a function in R?

Say you have a number of objects in your R environment, such as:

a <- 4
b <- 3
c <- 2
aa <- 2
bb <- 6
cc <- 9

Now say you would like to remove the objects in your environment that are named with the letters 'a' or 'b'. This can be achieved with 

rm(list = ls(pattern = "a"))
rm(list = ls(pattern = "b"))

However, imagine trying to solve this problem on a much bigger scale where you would like to remove all objects whose values appear in a list such as:

custom <- list("a", "b")

How do I apply this list as a 'looped' argument to the ls() function?

I have experimented with:

rm(lapply(custom, function(x) ls(pattern = x)))

But this does not seem to do anything.

This feels like quite a common problem, so I fear there is an answer to this issue elsewhere on stackoverflow. Unfortunately I could not find it.

Upvotes: 1

Views: 58

Answers (3)

Rich Scriven
Rich Scriven

Reputation: 99331

Option 1: This is a vectorized approach. Paste the custom list together using an "or" regex separator (|) and pass it to pattern.

rm(list = ls(pattern = paste(custom, collapse = "|")))

Option 2: If you still want to use lapply(), you would have to include the envir argument in both ls() and rm() since lapply() itself forms a local environment. And you would also want to put rm() inside the lapply() call.

lapply(custom, function(x) {
    e <- .GlobalEnv
    rm(list = ls(pattern = x, envir = e), envir = e)
})

Upvotes: 2

dww
dww

Reputation: 31452

can be done with a loop easily enough

test.a <- 4
test.b <- 3
test.c <- 2
test.aa <- 2
test.bb <- 6
test.cc <- 9

custom <- c("test.a", "test.b")
for (x in custom) rm(list = ls(pattern = x))

NB I added test. to the start of the object names, to avoid messing with peoples environment if they run this code. We don't wnat to inadvertently delete peoples actual objects named a or b etc.

Upvotes: 1

lmo
lmo

Reputation: 38500

Perhaps grep will solve your problem:

rm(list=grep("^[ab]$", ls(), value=T))

will remove objects a and b, but nothing else.

rm(list=grep("^[ab]", ls(), value=T))

will remove objects a, aa, b, and bb, but leave c and cc in the environment.

Upvotes: 0

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