On 11/17/09 5:02 AM, Peng Yu wrote:
This may not easy to do, when the filename are not hard coded strings. For example, the variable 'filename' is a vector of strings.for (i in 1:length(filename)){ do something... save(....,file=filename[i]) }
That's right. I don't think there is a feasible general solution. You might have more success with a convention-based approach for your scripts that would allow a simple parser to identify output files by name convention, for example.
+ seth ______________________________________________ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
