The question says it all - I want to take a list object full of data.frames and write each data.frame to a separate .csv file where the name of the .csv file corresponds to
You could use mapply
:
mapply(
write.table,
x=df.daily, file=paste(names(df.daily), "txt", sep="."),
MoreArgs=list(row.names=FALSE, sep=",")
)
There is thread about similar problem on plyr mailing list.
A couple of things:
laply
performs operations on a list. What you're looking for is d_ply
. And you don't have to break it up by day, you can let plyr
do that for you. Also, I would not use names(x) as that returns all of the column names of a data.frame.
d_ply(df, .(theday), function(x) write.csv(x, file=paste(x$theday,".csv",sep=""),row.names=F))
Try this:
sapply(names(df.daily),
function (x) write.table(df.daily[[x]], file=paste(x, "txt", sep=".") ) )
You should see the names ("1", "2", "3") spit out one by one, but the NULLs are the evidence that the side-effect of writing to disk files was done. (Edit: changed [] to [[]].)