If they are all formatted as your example, substr() would be simpler:

MONTH <- c("2012-07", "2001-07", "2002-01")
QUARTER <- c("2012-3", "2001-3", "2002-1")
YEAR <- c(2013, 2001, 2002)
Inp <- data.frame(MONTH, QUARTER, YEAR)
Out <- data.frame(MONTH=substr(MONTH, 6, 8),
     QUARTER=substr(QUARTER, 6, 7), YEAR)

This assumes MONTH and QUARTER are character strings and not dates.

----------------------------------------------
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77843-4352



> -----Original Message-----
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Fred G
> Sent: Tuesday, July 24, 2012 12:36 PM
> To: r-help@r-project.org
> Subject: [R] Regular Expression
> 
> Hi--
> 
> I have three columns in an input file:
> MONTH   QUARTER  YEAR
> 2012-07   2012-3        2012
> 2001-07   2001-3        2001
> 2002-01   2002-1        2002
> 
> I want to make output like so:
> MONTH   QUARTER  YEAR
> 07           3                2012
> 07           3                2001
> 01           1                2002
> 
> I was having some trouble getting the regular expression to work.  I
> think
> it should be something like the following:
> tmp <- uncurated$MONTH
> *tmp <- gsub("[^-\\d\\d]","",tmp,perl=TRUE)*
> *tmp[tmp=="-"] <- ""*
> *curated$MONTH <- tmp*
> *
> *
> tmp <- uncurated$QUARTER
> *tmp <- gsub("[^-\\d]","",tmp,perl=TRUE)*
> *tmp[tmp=="-"] <- ""*
> *curated$QUARTER <- tmp*
> *
> *
> *but it's not quite working. I want to be able to isolate any digits
> that
> occur after the hyphen and to delete everything before and including
> the
> hyphen. Would greatly appreciate any clarification anyone can provide.*
> 
>       [[alternative HTML version deleted]]
> 
> ______________________________________________
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> PLEASE do read the posting guide http://www.R-project.org/posting-
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> and provide commented, minimal, self-contained, reproducible code.

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