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
On Mon, Nov 28, 2016 at 4:23 PM, Ashta <sewashm at 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 > >?I'd do something like this:> files=c('dat.csv','dat') > file2read=files[file.exists(files)][1] > file2read[1] "dat.csv" You put the possible file names into the variable in the order of preference. E.g. I prefer "dat.csv" over "dat" if by chance both exist.> files=c('not.csv','not') > file2read=files[file.exists(files)][1] > file2read[1] NA ?The above shows the result should none of the files exist. So if "file2read" has an NA, then you go on to the next directory.? -- Heisenberg may have been here. Unicode: http://xkcd.com/1726/ Maranatha! <>< John McKown? [[alternative HTML version deleted]]
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(\\.[^.]+)$" )
dtaL <- lapply( fns, function(fn){ read.csv( fn, stringsAsFactors=FALSE ) } )
If you only expect one file to be in any given directory, you can skip the
lapply and just read the file, or you can extract the data frame from the list
using dtaL[[ 1 ]].
?list.files
?regex for help on patterns
--
Sent from my phone. Please excuse my brevity.
On November 28, 2016 2:23:23 PM PST, Ashta <sewashm at gmail.com>
wrote:>Hi all,
>
>I have a script that reads a file (dat.csv) from several folders.
>However, in some folders the file name is (dat) with out csv and in
>other folders it is dat.csv. The format of data is the same(only the
>file name differs with and without "csv".
>
>Is it possible to read these files depending on their name in one?
>like read.csv("dat.csv"). How can I read both type of file names?
>
>Thank you in advance
>
>______________________________________________
>R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>https://stat.ethz.ch/mailman/listinfo/r-help
>PLEASE do read the posting guide
>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.
Hi Jeff and John,
Thank you for your response.
In each folder, I am expecting a single file name (either dat or
dat.csv).v so will this work?
Is the following correct?
fns <- list.files(mydir)
if (is.element(pattern="dat(\\.[^.]+)$",fns ))
Thank you again.
On Mon, Nov 28, 2016 at 7:20 PM, Jeff Newmiller
<jdnewmil at dcn.davis.ca.us> wrote:> 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 <sewashm at 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 at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>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 at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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.