Extract substring in R using grepl

Question

I have a table with a string column formatted like this

abcdWorkstart.csv
abcdWorkcomplete.csv

And I would like to extract the last word in that filename. So I think the beginning pattern would be the word "Work" and ending pattern would be ".csv". I wrote something using grepl but not working.

grepl("Work{*}.csv", data$filename)

Basically I want to extract whatever between Work and .csv

desired outcome:

start
complete

Answer 1

I think you need sub or gsub (substitute/extract) instead of grepl (find if match exists). Note that when not found, it will return the entire string unmodified:

fn <- c('abcdWorkstart.csv', 'abcdWorkcomplete.csv', 'abcdNothing.csv')
out <- sub(".*Work(.*)\\.csv$", "\\1", fn)
out
# [1] "start"           "complete"        "abcdNothing.csv"

You can work around this by filtering out the unchanged ones:

out[ out != fn ]
# [1] "start"    "complete"

Or marking them invalid with NA (or something else):

out[ out == fn ] <- NA
out
# [1] "start"    "complete" NA        

Answer 2

Here is an option using regmatches/regexpr from base R. Using a regex lookaround to match all characters that are not a . after the string 'Work', extract with regmatches

regmatches(v1, regexpr("(?<=Work)[^.]+(?=[.]csv)", v1, perl = TRUE))
#[1] "start"    "complete"

data

v1 <- c('abcdWorkstart.csv', 'abcdWorkcomplete.csv', 'abcdNothing.csv')