Find a word before one of two possible separators

Question

word:12335
anotherword:2323434
totallydifferentword/455
word/32

I need to grab the character string before the : or / usi

Answer 1

Maybe try:

x <- c("word:12335", "anotherword:2323434", "totallydifferentword/455", "word/32")
lapply(strsplit(x, ":|/"), function(z) z[[1]]) #as a list
sapply(strsplit(x, ":|/"), function(z) z[[1]]) #as a string

There are regex solutions with gsub that will work too but in my experiences with similar problems strsplit will be less eloquent but faster.

I supose this regex would work as well:

gsub("([a-z]+)([/|:])([0-9]+)", "\\1", x)

In this case gsub was faster:

Unit: microseconds
        expr    min     lq median     uq     max
1     GSUB() 19.127 21.460 22.392 23.792 106.362
2 STRSPLIT() 46.650 50.849 53.182 54.581 854.162

Answer 2

You could use the package unglue :

library(unglue)
x <- c("word:12335", "anotherword:2323434", "totallydifferentword/455", "word/32")
unglue_vec(x, "{res}{=[:/].*?}")
#> [1] "word"                 "anotherword"          "totallydifferentword"
#> [4] "word"

^{Created on 2019-10-08 by the reprex package (v0.3.0)}

• {res} matches anything and will be returned, it's equivalent to {res=.*?}
• {=[:/].*?} matches anything starting with : or / and won't be returned as we have no lhs to =

Answer 3

This regex seems to work. Can you use that in R?

Answer 4

Something like this would do the trick in Ruby http://rubular.com/r/PzVQVIpKPq

^(\w+)(?:[:\/])

Starting from the front of the string, grab any word characters and capture them, until you reach the non-capturing / or :