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
Returning only a file path on Windows
3 messages · amvds at xs4all.nl, Duncan Murdoch, Uwe Ligges
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.