Hi Judith,
This probably isn't the only way to do it, but:
gsub("\\(.*?\\)", "", myvector, perl=TRUE)
seems to do the trick.
The problem is that regular expressions are greedy, so you were matching
everything between the first and last parens, as you noticed. Putting
the question mark there makes it a "minimal" matching operation.
Apparently this is only implemented in perl regex's, or at least in that
syntax. Hence the 'perl=TRUE'.
hth,
allie
On 8/25/2009 3:17 PM, Judith Flores wrote:
> Hello dear R-helpers,
>
> I haven't been able to figure out of find a solution in the R-help
> archives about how to delete all the characters contained in groups of
> parenthesis. I have a vector that looks more or less like this:
>
> myvector<-c("something (80 km/h, sd) & more (6 kg/L,sd)", "somethingelse (48
> m/s, sd) & moretoo (50g/L , sd)")
>
> I want to extract all the strings that are not contained in parenthesis, the
> goal would be to obtain the following new vector:
>
> subvector<-c("something & more", "somethingelse & moretoo")
>
> I tried the following, but this pattern seems to enclose all that is included
> between the first opened parenthesis and the last closed parethesis, which
> makes sense, but it's not what I need:
>
> subvector<-gsub("\\((.*)\\)","",myvector
>
>
> Your help will be very appreciated.
>
> Thank you,
>
> Judith
>
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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>
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