How to extract a substring in R according to a pattern?

Question

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`? Thanks!

Answer 1

There are a couple ways to do what you'd like:

1. With `substr()``, it can be `substr(string, 4, nchar(string))`; with `substring()`, it can be shortened to `substring(string, 4)`.

2. If you want to pattern-match more explicitly, I would suggest `gsub()` (I also recommend reading the documentation for this function: https://stat.ethz.ch/R-manual/R-devel/library/base/html/grep.html). See below for how to use this function for your case.

```

gsub('.*:', '', string)

# . = any character

# * = any number of times

# Result:

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

```

What the above code block does is that it looks for any character (denoted by the period) that occurs any number of times (denoted by the asterisk) followed by a colon. Then it substitutes it with a blank character (''). Result should be the vector `c(E001, E002, E003)`.

Hope this helps!