R help - Apr 2011 - regular expression in gsub() for strings with leading backslash

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.

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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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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>
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