[R] Regression expression to delete one or more spaces at end of string

[David R Forrest]

Double the [[]] and add a + for one-or-more characters: 

sub("[[:blank:]]+$", "", COLNAMES)



> On Aug 2, 2016, at 12:46 PM, Dennis Fisher <fisher at plessthan.com> wrote:
> 
> R 3.3.1
> OS X
> 
> Colleagues, 
> 
> I have encountered an unexpected regex problem
> 
> I have read an Excel file into R using the readxl package.  Columns names are:
> 
> COLNAMES	<- c("Study ID", "Test and Biological Matrix", "Subject No. ", "Collection Date", 
> "Collection Time", "Scheduled Time Point", "Concentration", "Concentration Units", 
> "LLOQ", "ULOQ", "Comment”)
> 
> As you can see, there is a trailing space in “Subject No. “.  I would like to delete that space.  The following works:
> 	sub(“ $”, “”, COLNAMES)
> However, I would like a more general approach that removes any trailing whitespace.
> 
> I tried variations such as:
> 	sub("[:blank:]$", "", COLNAMES)
> (also, without the $ and ‘space' instead of ‘blank') without success — to my surprise, characters other than the trailing space were deleted but the trailing space remained.
> 
> Guidance on the correct syntax would be appreciated.
> 
> Dennis
> 
> Dennis Fisher MD
> P < (The "P Less Than" Company)
> Phone / Fax: 1-866-PLessThan (1-866-753-7784)
> www.PLessThan.com
> 
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--
Dr. David Forrest
drf at vims.edu
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