I am choosing a file like this: #Bring up file selection box fn<-file.choose() fp<-file.path(fn,fsep='\\') 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
On 5/22/2009 10:45 AM, amvds at 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 at 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.
Are you looking for choose.dir() ? Or basename() and dirname()? Uwe Ligges amvds at xs4all.nl wrote:> I am choosing a file like this: > > #Bring up file selection box > fn<-file.choose() > fp<-file.path(fn,fsep='\\') > > 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 at 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.