No, and yes, depending what you mean.
No, because you have to supply the file name to open it... you cannot directly
use wildcards to open files.
Yes, because the list.files function can be used to match all file names
fitting a regex pattern, and you can use those filenames to open the files.
E.g.
fns <- list.files( pattern="dat(\\.[^.]+)$" )
dtaL <- lapply( fns, function(fn){ read.csv( fn, stringsAsFactors=FALSE ) } )
If you only expect one file to be in any given directory, you can skip the
lapply and just read the file, or you can extract the data frame from the list
using dtaL[[ 1 ]].
?list.files
?regex for help on patterns
--
Sent from my phone. Please excuse my brevity.
On November 28, 2016 2:23:23 PM PST, Ashta <sewa...@gmail.com> wrote:
>Hi all,
>
>I have a script that reads a file (dat.csv) from several folders.
>However, in some folders the file name is (dat) with out csv and in
>other folders it is dat.csv. The format of data is the same(only the
>file name differs with and without "csv".
>
>Is it possible to read these files depending on their name in one?
>like read.csv("dat.csv"). How can I read both type of file names?
>
>Thank you in advance
>
>______________________________________________
>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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>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 -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
