list.files() will give you all the file names in your working directory (you can also give it a pattern argument) then can loop over those with something like:
lapply(list.files(), read.table) which will put all your files in a list object. This is generally considered much more convenient than trying to create a whole bunch of objects with different names programmatically. Michael On Fri, Apr 27, 2012 at 11:23 AM, jiangxijixzy <jiangxiji...@163.com> wrote: > I want to import data from about 2000 text files, and hope to create a data > frame to make it easy to quote the data. > For example, the files like this > Oil_20030801.txt, Oil_20030804.txt, Oil_20030805.txt … Oil_20120427.txt > The dates aren’t continuous. I want to create the data frame called “Oil”, > like that > Oil20030801<-read.table(“E:/Oil/ Oil_20030801.txt”) > Oil20030804<-read.table(“E:/Oil/ Oil_20030804.txt”) > Oil20030805<-read.table(“E:/Oil/ Oil_20030805.txt”) > … > Oil20120427<-read.table(“E:/Oil/ Oil_20120427.txt”) > It is a time consuming way. How can I perform a convenient way? Thank you! > > > -- > View this message in context: > http://r.789695.n4.nabble.com/Batch-importing-data-tp4592997p4592997.html > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > 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.
