Consider.
li <- list(c(1,1,1), c(2,1,1), c(5,1,0))
## [[1]]
## [1] 1 1 1 <-- 0 modulo 3
##
## [[2]]
## [1] 2 1 1
##
## [[3]]
## [1] 5 1 0 <--- 0 modulo 3
I want to keep all items in the list whose sum modulo 3 equals zero (or some other logical expression regarding the item in the list).
Code.
new <- list()
idx <- 1
for (i in seq_along(li) ) {
nxt <- li[[i]]
if ((sum(nxt) %% 3) == 0) {
new[idx] <- list(nxt)
idx <- idx + 1
}
}
## new
## [[1]]
## [1] 1 1 1
##
## [[2]]
## [1] 5 1 0
However. I would like a vectorized solution. I tried.
new <- subset(li, lapply(li, (sum %% 3) == 0 ))
but to no avail.
Edit See also Difference between subset and filter from dplyr for more detailed information.
in Base R:
Filter(\(x)!sum(x)%%3, li)
[[1]]
[1] 1 1 1
[[2]]
[1] 5 1 0
in tidyverse
purrr::keep(li, ~!sum(.x)%%3)
[[1]]
[1] 1 1 1
[[2]]
[1] 5 1 0
purrr::discard(li, ~!!sum(.x)%%3) # !0
[[1]]
[1] 1 1 1
[[2]]
[1] 5 1 0