On Fri, 29 Apr 2011, Duncan Murdoch wrote:
On 29/04/2011 7:41 PM, Miao wrote:
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.
If we were dealing with a leading backslash, I guess this would do it:
gsub("^\\\\.*", "", txt)
R would display a double backslash, but I believe that represents a single
backslash. So if the string were saved using write.table, say, only a
single backslash would be stored.
a <- "\\This is a string."
a
[1] "\\This is a string."
gsub("^\\\\", "", a)
[1] "This is a string."
a
[1] "\\This is a string."
gsub("^\\\\.*", "", a)
[1] ""
gsub("^\\\\.*", "", c(a,"Another string","\\more"))
[1] "" "Another string" ""
write.table(a, file="a.txt", quote=F, row.names=F, col.names=F)
$ cat a.txt
\This is a string.
Apparently this is not what the OP really wanted. The OP probably wanted
to remove characters that were not from the regular ASCII set.
Mike
--
Michael B. Miller, Ph.D.
Minnesota Center for Twin and Family Research
Department of Psychology
University of Minnesota
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