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

[Duncan Murdoch]

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
>
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