On 5/22/2009 10:45 AM, am...@xs4all.nl wrote:
I am choosing a file like this:
#Bring up file selection box
fn<-file.choose()
fp<-file.path(fn,fsep='\\')
file.path() constructs a path from component parts, it doesn't extract
the path from a filename.
You want dirname(fn). You may also want to look at normalizePath, to
convert short names to long ones, or shortPathName, for the reverse.
Duncan Murdoch
Unfortunately, the file path contains the short file name and extension as
well. I had hoped to get only the path so I could make my own long
filenames (for output graphs) by concatenation with this file path.
Of course I can split the string and assemble the components from the
returned list:
fp<-strsplit(fp,'\\',fixed=TRUE)
But there must be a better way?
Thanks in advance,
Alex van der Spek
______________________________________________
R-help@r-project.org mailing list
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.
______________________________________________
R-help@r-project.org mailing list
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.
