R help - Mar 2007 - Help with grep (and similar functions)

This works:

grep("([A-Za-z]*) ", "Aaaa 3 x 0 Bbbb") # 1

This also works:

grep("([A-Za-z]*) ", "Aaaa 3 x 0 Bbbb", value=T) # Aaaa 3 x
0 Bbbb

However, I want a grep that returns the _matched_ pattern, which, in this 
case, would be Aaaa. How can I do it?

Alberto Monteiro

Try this:

library(gsubfn)
pat <- "([[:upper:]][[:lower:]]*) "
s <- "Aaaa 3 x 0 Bbbb"

strapply(s, pat, backref =-1)[[1]]

or (not quite as general but works in this case):

pat <- "([[:upper:]][[:lower:]]*) "
s <- "Aaaa 3 x 0 Bbbb"

s.idx <- gregexpr(pat, s)[[1]]
substring(s, s.idx, s.idx + attr(s.idx, "match.length") - 2)

On 3/28/07, Alberto Monteiro <albmont at centroin.com.br>
wrote:
> This works:
>
> grep("([A-Za-z]*) ", "Aaaa 3 x 0 Bbbb") # 1
>
> This also works:
>
> grep("([A-Za-z]*) ", "Aaaa 3 x 0 Bbbb", value=T) # Aaaa
3 x 0 Bbbb
>
> However, I want a grep that returns the _matched_ pattern, which, in this
> case, would be Aaaa. How can I do it?
>
> Alberto Monteiro
>
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

Gabor Grothendieck wrote:
>
> Try this:
> 
> library(gsubfn)
> pat <- "([[:upper:]][[:lower:]]*) "
> s <- "Aaaa 3 x 0 Bbbb"
> 
> strapply(s, pat, backref =-1)[[1]]
> 
Ok, I think I got it.

For example, if I want to retrieve Aaaa and Bbbb (which
could be any strings) or the numbers 3 and 0, I would do
this:

library(gsubfn)
pat <- "^([a-zA-Z ]+) ([1-9]*[0-9]) x ([1-9]*[0-9]) ([a-zA-Z ]+)$"
s <- "My team 3 x 0 the team from Outer Space"
x <- strapply(s, pat, c)[[1]]

Then x[1] is s, x[2] is "My team", x[3] is "3", x[4] is
"0"
and x[5] is "the team from Outer Space".

Thanks.

Alberto Monteiro