CODEWITHSUNDEEP
rlapply
Leif Andersen
Leif Andersen

Reputation: 22332

Filter and map over a list in R

I can do a map over a list in R with lapply:

lapply(1:10,function(y) { if (y %% 2 == 0) {y+42}})

(Which gives me a list between 1 and 10 where every other object is NULL).

Furthermore, I can then remove each of these NULL objects with Filter:

Filter(function(x) !is.null(x),
       lapply(1:10,function(y) { if (y %% 2 == 0) {y+42}}))

(Which gives me a list of even numbers between 1 and 10).

Another example, let's say we want to take a list of strings, and filter out all of the strings that start with "a", and then append "Foo" to the remaining ones. The similar technique could be used:

Filter(function(x) !is.null(x),
       lapply(c("foo","abar","baz"),
              function(y) { if (grepl("^a",y)) {paste("foo",y)}}))

Is there a better way to do a fold over a list in R.

Upvotes: 2

Views: 1752

Answers (1)

akrun
akrun

Reputation: 887038

You could try

y <- 1:10
as.list(y[y%%2==0]+42)

grep can also work in a vector

grepl(paste(c(2,4,6), collapse="|"), y)
#[1] FALSE  TRUE FALSE  TRUE FALSE  TRUE FALSE FALSE FALSE FALSE

y1 <- letters[1:10]
paste('foo', y1[grepl('^a', y1)])
#[1] "foo a"

Upvotes: 1

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