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
>
>
> --
> ЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧ
> 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.1988@gmail.com
> ЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧЧ
>
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