[SOLVED] Extract a substring according to a pattern

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?

Solution

Here are a few ways:

1) sub

sub(".*:", "", string)
## [1] "E001" "E002" "E003"

2) strsplit

sapply(strsplit(string, ":"), "[", 2)
## [1] "E001" "E002" "E003"

3) read.table

read.table(text = string, sep = ":", as.is = TRUE)$V2
## [1] "E001" "E002" "E003"

4) substring

This assumes second portion always starts at 4th character (which is the case in the example in the question):

substring(string, 4)
## [1] "E001" "E002" "E003"

4a) substring/regex

If the colon were not always in a known position we could modify (4) by searching for it:

substring(string, regexpr(":", string) + 1)

5) strapplyc

strapplyc returns the parenthesized portion:

library(gsubfn)
strapplyc(string, ":(.*)", simplify = TRUE)
## [1] "E001" "E002" "E003"

6) read.dcf

This one only works if the substrings prior to the colon are unique (which they are in the example in the question). Also it requires that the separator be colon (which it is in the question). If a different separator were used then we could use sub to replace it with a colon first. For example, if the separator were _ then string <- sub("_", ":", string)

c(read.dcf(textConnection(string)))
## [1] "E001" "E002" "E003"

7) separate

7a) Using tidyr::separate we create a data frame with two columns, one for the part before the colon and one for after, and then extract the latter.

library(dplyr)
library(tidyr)
library(purrr)

DF <- data.frame(string)
DF %>% 
  separate(string, into = c("pre", "post")) %>% 
  pull("post")
## [1] "E001" "E002" "E003"

7b) Alternately separate can be used to just create the post column and then unlist and unname the resulting data frame:

library(dplyr)
library(tidyr)

DF %>% 
  separate(string, into = c(NA, "post")) %>% 
  unlist %>%
  unname
## [1] "E001" "E002" "E003"

8) trimws We can use trimws to trim word characters off the left and then use it again to trim the colon.

trimws(trimws(string, "left", "\\w"), "left", ":")
## [1] "E001" "E002" "E003"

Note

The input string is assumed to be:

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