On Wed, 28 Jun 2023 20:29:23 +0000
Kevin Zembower via R-help <r-help@r-project.org> wrote:
> I think my algorithm for the labels is:
> 1. keep everything from the last "!!" up to and including the last
> character
> 2. for everything remaining, replace each "!!.*:" group with a single
> space.
If you remove the initial ' !!', the problem becomes a more tractable
"replace each group of non-'!' followed by '!!' with one space":
bg3_race_sum$label |>
(\(.) sub('^ !!', '', .))() |>
(\(.) gsub('[^!]*!!', ' ', .))()
But that solution could have been impossible if the task was slightly
different.
> I can split the label using str_split(label, pattern = "!!") to get a
> vector of strings, but don't know how to work on the last string and
> all the rest of the strings separately.
str_split() would have given you a list of character vectors. You can
use lapply to evaluate a function on each vector inside that list.
Inside the function, use length(x) (if `x` is the argument of the
function) to find out how many spaces to produce and which element of
the vector is the last one. (For code golf points, use rev(x)[1] to get
the last element.)
--
Best regards,
Ivan
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