r - Use strsplit starting at end of string -

i've been using code split names of individual samples, change part of sample name, , rebind strings together. code works when names same length (ie: names 8 characters long , splits after first 4 characters), when names different lengths, code no longer effective.

essentially, individual names 7 or 8 characters. last 4 characters what's important.
example 8 characters: samp003a
example 7 characters: sam003a

is there way continue using strsplit separate names, start end of string rather beginning, keep last 4 characters (003a)?

current code:

> rowlist <- as.list(rownames(df1))     > rowlistres <- strsplit(as.character(rowlist), "(?<=.{4})", perl = true)     > rowlistres.df <- do.call(rbind, rowlistres)     > rowlistres.df[,1] <- "ly3d"     > dfnames <- apply(rowlistres.df, 1, paste, collapse="")     > rownames(df1) <- dfnames     

it's line 2 i'm trying hard edit, can split according last 4 characters.

any appreciated!

it looks you're bit mixed how use look-around assertions. pattern you're using, "(?<=.{4})", look-behind assertion says "find me inter-character spaces preceded 4 characters of kind", not want.

the pattern want, "(?=.{4}$)", look-ahead assertion finds single inter-character space followed 4 characters of kind followed end of string.

there is, unfortunately, unpleasant twist. reasons discussed in answers this question, strsplit() interacts oddly look-ahead assertions; result, pattern you'll need "(?<=.)(?=.{4}$)". here's looks in action:

x <- c("samp003a", "sam003a") strsplit(x, split="(?<=.)(?=.{4}$)", perl=t) # [[1]] # [1] "samp" "003a" #  # [[2]] # [1] "sam"  "003a" 

if want final 4 characters of each entry, maybe use substr(), this:

x <- c("samp003a", "sam003a") substr(x, start=nchar(x)-3, stop=nchar(x)) # [1] "003a" "003a"