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@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.
