On 12/05/2009, at 8:41 AM, Andreas Christoffersen wrote:
I really dont know hot to use it - but have a look at ?source
No; ``source'' is irrelevant here.
On Mon, May 11, 2009 at 10:37 PM, Mark Na <mtb...@gmail.com> wrote:
Hi R-helpers,
I would like to read into R all the .csv files that are in my working
directory, without having to use a read.csv statement for each file.
Each .csv would be read into a separate dataframe which would acquire
the filename of the .csv.
As an example:
Mark<-read.csv("Mark.csv")
...but the code (or command) would do this automatically for every
.csv in the working directory without my specifying each file.
I'd appreciate any help or ideas you might have.
datalist <- list()
files <- list.files(pattern="\\.csv$")
for(file in files) {
stem <- gsub("\\.csv$","",file)
datalist[[stem]] <- read.csv(file)
}
The foregoing does things the *right* way, putting the resulting data
sets together
into a list whereby they can be accessed in an organized fashion. If
you wrong-headedly
*insist* on putting the data sets separately into your workspace then
do:
files <- list.files(pattern="\\.csv$")
for(file in files) {
stem <- gsub("\\.csv$","",file)
assign(stem,read.csv(file))
}
cheers,
Rolf Turner
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