remove part of string after pattern

Question

I'm working with ; separated strings like

a="My name is Laura;My name is Martin;My name is Carl;"

I want to remove the part after name, and have this as result:

a="My name;My name;My name;"

Answer 1

Using gsub:

gsub("(?<=name)[^;]*", "", a, perl = TRUE)
# [1] "My name;My name;My name;"

Alternatively, we can use gregexpr and `regmatches<-` for this:

gre <- gregexpr("(?<=name)[^;]*", a, perl = TRUE)
regmatches(a, gre)
# [[1]]
# [1] " is Laura"  " is Martin" " is Carl"  
regmatches(a, gre) <- ""
a
# [1] "My name;My name;My name;"

Walk-through:

• (?<=name) is a lookbehind regex that does not "consume" the name string, just uses it as a starting-point; this is a Perl extension, ergo perl=TRUE;
• (?<=name)[^;]* find all non-; text after name
• regmatches(a, gre) returns the values; we don't need to run this line of code for anything other than validation of the text that will be removed;
• `regmatches<-` is a form of regmatches for the LHS of an assignment operator (not all functions have this form); we use this to replace the matched portions.

Answer 2

It worked like this

gsub("My name[^;]*","My name",a)

Answer 3

Solution:

With gsub and perl=TRUE:

> gsub("(?<=name).*", "", unlist(strsplit(a, ";")), perl=TRUE)
[1] "My name" "My name" "My name"
> 

Explanation:

This detects name in a lookbehind and removes everything (.*) after it. First of all it uses strsplit to split it into a list by ;, then I use unlist to make it a vector.

Use:

> paste(gsub("(?<=name).*", "", unlist(strsplit(a, ";")), perl=TRUE), collapse=";")
[1] "My name;My name;My name"
>