Extracting part of string using regular expressions

Question

I’m struggling to get a bit of regular expressions code to work. I have a long list of strings that I need to partially extract. I need only strings that starting with “WER” and I only need the last part of the string commencing (including) on the letter.

test <- c("abc00012Z345678","WER0004H987654","WER12400G789456","WERF12","0-0Y123")

Here is the line of code which is working but only for one letter. However in my list of strings it can have any letter.

ifelse(substr(test,1,3)=="WER",gsub("^.*H.*?","H",test),"")

What I’m hoping to achieve is the following:

H987654
G789456
F12

Wiktor Stribiżew · Accepted Answer · 2016-03-16 11:16:13Z

5

You can use the following pattern with gsub:

> gsub("^(?:WER.*([a-zA-Z]\\d*)|.*)$", "\\1", test)
[1] ""        "H987654" "G789456" "F12"     ""

See the regex demo

This pattern matches:

^ - start of a string
(?: - start of an alternation group with 2 alternatives:
- WER.*([a-zA-Z]\\d*) - WER char sequence followed with 0+ any characters (.*) as many as possible up to the last letter ([a-zA-Z]) followed by 0+ digits (\\d*) (replace with \\d+ to match 1+ digits, to require at least 1 digit)
- | - or
- `.* - any 0+ characters
)$ - closing the alternation group and match the end of string with $.

With str_match from stringr, it is even tidier:

> library(stringr)
> res <- str_match(test, "^WER.*([a-zA-Z]\\d*)$")
> res[,2]
[1] NA        "H987654" "G789456" "F12"     NA       
>

See another regex demo

If there are newlines in the input, add (?s) at the beginning of the pattern: res <- str_match(test, "(?s)^WER.*([a-zA-Z]\\d*)$").

edited Mar 16, 2016 at 11:16

answered Mar 16, 2016 at 11:07

Wiktor Stribiżew

631k41 gold badges502 silver badges632 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Wiktor Stribiżew Over a year ago

Just FYI: "^.*H.*?" matches the whole chunk of a string from its beginning till the last H, and stops right there as .*? at the end does not consume/return any characters (because it is a lazy subpattern and can match an empty string, so it matches the empty location after H and calls it a day). This matched chunk is replaced with "H" by gsub.

talat · Accepted Answer · 2016-03-16 11:12:45Z

3

If you don't want empty strings or NA for strings that don't start with "WER", you could try the following approach:

sub(".*([A-Z].*)$", "\\1", test[grepl("^WER", test)])
#[1] "H987654" "G789456" "F12"

answered Mar 16, 2016 at 11:12

talat

70.5k22 gold badges130 silver badges158 bronze badges

Collectives™ on Stack Overflow

Extracting part of string using regular expressions

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related