Re: [R] regular expression in gsub() for strings with leading backslash

From: Duncan Murdoch <murdoch.duncan_at_gmail.com>
Date: Fri, 29 Apr 2011 21:37:48 -0400

On 29/04/2011 9:34 PM, Miao wrote:
> Thanks Duncan for clarifying this. I'm pretty a newbie to such type of
> characters and special characters. In R's gsub() what regular
> expressions shall I use to handle all these situations?

I don't know. This might work:

gsub("[\x01-\x1f\x7f-\xff]", "", x)

(i.e. the range from character 1 to character 31, and 127 to 255) but I don't know if our regular expression matcher will accept those characters.

Duncan Murdoch

>
>
> On Fri, Apr 29, 2011 at 6:07 PM, Duncan Murdoch
> <murdoch.duncan_at_gmail.com <mailto:murdoch.duncan_at_gmail.com>> wrote:
>
> On 29/04/2011 7:41 PM, Miao wrote:
>
> Hello,
>
> Can anyone help on gsub() in R? I have a string like something
> below, and
> wanted to delete all the strings with leading backslash,
> including "\xa0On",
> "\023, "\xab", and many others. How should I write a regular
> expression
> pattern in gsub()? I don't care how many characters following
> backslash.
>
>
>
> If those are R strings, none of them contain a backslash. In R, a
> backslash would always be printed as \\.
>
> \x is the introduction to a hexadecimal encoding for a character;
> the next two characters show the hex digits. So your first string
> contains a single character \xa0, the third one contains \xab, and
> so on.
>
> The \023 is an octal encoding for a single character.
>
> Duncan Murdoch
>
>
>
> txt<- "Is This Thing\xa0On? http://bit.ly/jAbKem wait \023 for
> people \xab
> and be patient :"
>
> Thanks in advance,
> Miao
>
> [[alternative HTML version deleted]]
>
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>
>
>
>
> --
> proceed everyday



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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. Received on Sat 30 Apr 2011 - 01:45:36 GMT

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