Dear Yao,
You can use a list() as follows (untested):
setwd("~/path/to/your/files")
f <- list.files(pattern = ".txt") # list the files
info <- lapply(f, read.table, header = TRUE)
names(info) <- f
info
To see the information in the first data set just do
info[[1]]
See ?list.files and ?lapply for more information.
HTH,
Jorge
On Thu, Dec 20, 2012 at 3:24 PM, Yao He <> wrote:
> Dear All
>
> I have a lot of files in a directory as follows:
> "02-03.txt" "03-04.txt" "04-05.txt" "05-06.txt" "06-07.txt"
> "07-08.txt" "08-09.txt"
> "09-10.txt" "G0.txt" "G1.txt" "raw_ped.txt"
> ..........................
>
> I want to read them into different objects according to their
> filenames,such as:
> 02-03<-read.table("02-03.txt",header=T)
> 03-04<-read.table("03-04.txt",header=T)
> I don't want to type hundreds of read.table(),so how I read it in one time?
> I think the core problem is that I can't create different objects'
> name in the use of loop or sapply() ,but there may be a better way to
> do what I want.
>
> Thanks a lot
>
> Yao He
>
> Yao He
>
>
> --
>
> Master candidate in 2rd year
> Department of Animal genetics & breeding
> Room 436,College of Animial Science&Technology,
> China Agriculture University,Beijing,100193
> E-mail: yao.h.1...@gmail.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.
>
[[alternative HTML version deleted]]
______________________________________________
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.
