grepl with regex

Question

I have a difficulty using grepl with regex.

Here is a small example:

I have a character vector:

text <- c(

  "D_Purpose__Repairs" ,
  "Age" ,
  "F_Job"  
)

And I want to select the words that start with D_ or F_. So I write:

grepl("\\>D_.+ | \\>F_.+", text)

grepl("\\D_.+ | \\F_.+", text)

grepl("\\^D_.+ | \\^F_.+", text)

However this returns:

[1] FALSE FALSE FALSE

Could you help me understand what I am doing wrong and how should I correct my code?

Your advice will be appreciated.

Answer 1

You don't need to (and must not) escape the caret character with backslashes, and you can't put extra whitespace in your regex around the |. This works as you intend:

> grepl("^D_.+|^F_.+", text)
[1]  TRUE FALSE  TRUE

Answer 2

Some comments on your patterns:

• \>D_.+ | \>F_.+ - here, \> matches the end of word position while the actual position here is a start of a word (so, you might want to try with \<'). Also, the spaces around|are meaningful, you should not add them unless you use aperl=TRUEwith a(?x)` modifier.

• \D_.+ | \F_.+ is a malformed patter since \F is an unknown regex escape. \D matches any char but a digit, and is clearly something you did not expect.

• \^D_.+ | \^F_.+ is the closest, but there are redundant spaces again, and the escaped ^ match literal caret symbols. If you do not escape carets they match the start of string positions.

Now, the most efficient pattern here is

grepl("^[DF]_.+", text)

Meaning:

• ^ - start of string anchor
• [DF] - either D or F letters
• _ - a literal underscore
• .+ - any 1+ chars as many as possible up to the end of the string.