R: trim consecutive trailing and leading special characters from set of strings - r

Question

I have a list of character vectors, all equal lengths. Example data:
> a = list('**aaa', 'bb*bb', 'cccc*')
> a = sapply(a, strsplit, '')
> a
[[1]]
[1] "*" "*" "a" "a" "a"
[[2]]
[1] "b" "b" "*" "b" "b"
[[3]]
[1] "c" "c" "c" "c" "*"
I would like to identify the indices of all leading and trailing consecutive occurrences of the character *. Then I would like to remove these indices from all three vectors in the list. By trailing and leading consecutive characters I mean e.g. either only a single occurrence as in the third one (cccc*) or multiple consecutive ones as in the first one (**aaa).
After the removal, all three character vectors should still have the same length.
So the first two and the last character should be removed from all three vectors.
[[1]]
[1] "a" "a"
[[2]]
[1] "*" "b"
[[3]]
[1] "c" "c"
Note that the second vector of the desired result will still have a leading *, which, however became the first character after the operation, so it should be in.
I tried using which to identify the indices (sapply(a, function(x)which(x=='*'))) but this would still require some code to detect the trailing ones.
Any ideas for a simple solution?

Answer 1

I would replace the lead and lag stars with NA:
aa <- lapply(setNames(a,seq_along(a)), function(x) {
star = x=="*"
toNA = cumsum(!star) == 0 | rev(cumsum(rev(!star))) == 0
replace(x, toNA, NA)
})
Store in a data.frame:
DF <- do.call(data.frame, c(aa, list(stringsAsFactors=FALSE)) )
Omit all rows with NA:
res <- na.omit(DF)
# X1 X2 X3
# 3 a * c
# 4 a b c
If you hate data.frames and want your list back: lapply(res,I) or c(unclass(res)), which gives
$X1
[1] "a" "a"
$X2
[1] "*" "b"
$X3
[1] "c" "c"

Answer 2

First of, like Richard Scriven asked in his comment to your question, your output is not the same as the thing you asked for. You ask for removal of leading and trailing characters, but your given ideal output is just the 3rd and 4th element of the character lists.
This would be easily achievable by something like
a <- list('**aaa', 'bb*bb', 'cccc*')
alist = sapply(a, strsplit, '')
lapply(alist, function(x) x[3:4])
Now for an answer as you asked it:
IMHO, sapply() isn't necessary here.
You need a function of the grep family to operate directly on your characters, which all share a help page in R opened by ?grep.
I would propose gsub() and a bit of Regular Expressions for your problem:
a <- list('**aaa', 'bb*bb', 'cccc*')
b <- gsub(pattern = "^(\\*)*", x = a, replacement = "")
c <- gsub(pattern = "(\\*)*$", x = b, replacement = "")
> c
[1] "aaa" "bb*bb" "cccc"
This is doable in one regex, but then you need a backreference for the stuff in between i think, and i didn't get this to work.
If you are familiar with the magrittr package and its excellent pipe operator, you can do this more elegantly:
library(magrittr)
gsub(pattern = "^(\\*)*", x = a, replacement = "") %>%
gsub(pattern = "(\\*)*$", x = ., replacement = "")