Message-ID: <CAF8bMcavAJwinSMMj3WMxp7T3-t8KW_m7G7vqXDkZ0WQh+DS=Q@mail.gmail.com>
Date: 2016-08-02T16:57:12Z
From: William Dunlap
Subject: Regression expression to delete one or more spaces at end of string
In-Reply-To: <DDE9660F-AB7E-4E18-9102-D86CD9B0AA2D@plessthan.com>
First, use [[:blank:]] instead of [:blank:]. that latter matches colon, b,
l,
a, n, and k, the former whitespace.
Second, put + after [[:blank:]] to match one or more of them.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Aug 2, 2016 at 9:46 AM, 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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