[regex] Extract a substring according to a pattern

Suppose I have a list of string:

string = c("G1:E001", "G2:E002", "G3:E003")

Now I hope to get a vector of string that contains only the parts after the colon ":", i.e substring = c(E001,E002,E003).

Is there a convenient way in R to do this? Using substr?

For example using gsub or sub

    gsub('.*:(.*)','\\1',string)
    [1] "E001" "E002" "E003"

If you are using data.table then tstrsplit() is a natural choice:

tstrsplit(string, ":")[[2]]
[1] "E001" "E002" "E003"

The unglue package provides an alternative, no knowledge about regular expressions is required for simple cases, here we'd do :

# install.packages("unglue")
library(unglue)
string = c("G1:E001", "G2:E002", "G3:E003")
unglue_vec(string,"{x}:{y}", var = "y")
#> [1] "E001" "E002" "E003"

^{Created on 2019-11-06 by the reprex package (v0.3.0)}

More info : https://github.com/moodymudskipper/unglue/blob/master/README.md

This should do:

gsub("[A-Z][1-9]:", "", string)

gives

[1] "E001" "E002" "E003"

Late to the party, but for posterity, the stringr package (part of the popular "tidyverse" suite of packages) now provides functions with harmonised signatures for string handling:

string <- c("G1:E001", "G2:E002", "G3:E003")
# match string to keep
stringr::str_extract(string = string, pattern = "E[0-9]+")
# [1] "E001" "E002" "E003"

# replace leading string with ""
stringr::str_remove(string = string, pattern = "^.*:")
# [1] "E001" "E002" "E003"

Another method to extract a substring

library(stringr)
substring <- str_extract(string, regex("(?<=:).*"))
#[1] "E001" "E002" "E003

• (?<=:): look behind the colon (:)

Here is another simple answer

gsub("^.*:","", string)