I have a list of names that look like this, with some ending in an "*" and others not: Steve Young* Joe Montana* Tom Brady Daunte Culpepper ... I want to create a variable that = 1 if the name ends in "*" and = 0 otherwise. Please help! -- View this message in context: http://r.789695.n4.nabble.com/Extract-character-from-the-end-of-a-string-tp3621732p3621732.html Sent from the R help mailing list archive at Nabble.com.
Hi,
This answer is predicated on the assumption that you are using "list"
in its English sense.
## character vector with the names
x <- c("Steve Young*", "Joe Montana*", "Tom
Brady", "Daunte Culpepper")
## using grepl() and taking advantage of how logical data is stored to
get 0/1 data
as.numeric(grepl("\\*$", x))
See ?regexp for more details on regular expressions, ?grepl and ?TRUE
for documentation on logical values
Hope this helps,
Josh
On Thu, Jun 23, 2011 at 11:06 PM, jwehr <jglandwehr at gmail.com>
wrote:> I have a list of names that look like this, with some ending in an
"*" and
> others not:
>
> Steve Young*
> Joe Montana*
> Tom Brady
> Daunte Culpepper
> ...
>
> I want to create a variable that = 1 if the name ends in "*" and
= 0
> otherwise. Please help!
>
> --
> View this message in context:
http://r.789695.n4.nabble.com/Extract-character-from-the-end-of-a-string-tp3621732p3621732.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
Thank you, Joshua!
I got what I was looking for using this command:
as.numeric(grepl("\\*$", qb$Player))
Where the variable name is "Player" in dataset "qb".
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
1. Please always quote the original message when sending messages to this *mailing list*. 2. Please send your answers and comments also to the original poster who may not be aware you posted to the list. Uwe Ligges On 25.06.2011 06:55, jwehr wrote:> Thank you, Joshua! > > I got what I was looking for using this command: > > as.numeric(grepl("\\*$", qb$Player)) > > Where the variable name is "Player" in dataset "qb". > > -- > View this message in context: http://r.789695.n4.nabble.com/Extract-character-from-the-end-of-a-string-tp3621732p3624124.html > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > R-help at r-project.org 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.