Mark -
Since regular expressions in R are just character
strings, it's pretty easy to assemble a regular expression
to delete leading or trailing characters. For example:
delchars = function(str,n,lead=TRUE){
+dots = paste(rep('.',n),collapse='')
+pat = if(lead)paste('^',dots,sep='') else paste(dots,'$',sep='')
+sub(pat,'',str)
+ }
str = this is a test
delchars(str,4)
[1] is a test
delchars(str,4,lead=FALSE)
[1] this is a
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
spec...@stat.berkeley.edu
On Tue, 14 Dec 2010, Mark Na wrote:
Hi R-helpers,
I have a character string, for example:
lm(y ~ X2 + X3 + X4)
from which I would like to strip off the leading and trailing
quotation marks resulting in this:
lm(y ~ X2 + X3 + X4)
I have tried using gsub() but I can't figure out how to specify the
quotation mark using a regular expression.
Alternatively, I would like a function that lets me delete the leading
(or trailing) X characters, and in this case X=1 (but it could be used
more flexibly to delete several leading or trailing characters).
I would appreciate help with either of these potential solutions (gsub
and regex, or delete leading/trailing characters).
Many thanks!
Mark
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