R：删除字符串中分隔符后的所有字母字符

Question

I would like to remove all the letters ([az]) that come after a delimiter (eg "-"), in a string, eg:

s <- "abc-10abc"

So to get:

> s2
[1] "abc-10"

How can I do this? Thank you

Answer 1

gsub("(.*\\d).*", "\\1", s)

The first pattern argument uses () to "capture" a group of characters. Inside the capture we are looking for all wild card character until a digit \\\\d . This "captures" everything up until the last digit here.

Since the pattern argument also includes a multiple-wildcard after the capture group, the entire original string is being targeted for replacement. The replace argument \\\\1 says to use the first (and in this case only) capture expression from the pattern argument.

Let me know if that's not clear, this is my regex gospel for R regex help https://www.cheatography.com/davechild/cheat-sheets/regular-expressions/

Like Rich Scriven pointed out you could substitute the .* with [az]* to target just letters a thru z after the last digit. You may want to add the argument ignore.case = TRUE to gsub() as well incase not everything is lower case:

gsub("(.*\\-\\d*)[a-z]*", "\\1", s, ignore.case = TRUE)

Answer 2

I'm not a regex expert but I believe this follows your pattern.

gsub("(^.*-[^[:alpha:]]*)[[:alpha:]]*", "\\1", s)
#[1] "abc-10"

Explanation:

1. ^ - beginning of string
2. ^.* any character at the beginning of string followed by zero or more repetitions of it.
3. - matches the delimiter in your question
4. [^[:alpha:]]* the circunflex negates the class [:alpha:] , do not match alphabetic characters
5. (all of above) form a pattern group, the first (and only)
6. [[:alpha:]]* match an alphabetic character followed by zero or more repetitions of it

Then in the replacement argument \\\\1 means to replace the pattern with the first group, so the [[:alpha:]]* part is left out.