Save all data frames in list to separate .csv files

Question

I have a list of data frames that I want to save to independent .csv files.

Currently I have a new line for each data frame:

write.csv(lst$df1, "C:/Users/.../df1")
write.csv(lst$df2, "C:/Users/.../df2")
ad nauseam

Obviously this isn't ideal: a change in the names would mean going through every case and updating it. I considered using something like

lapply(lst, f(x) write.csv(x, "C:/Users/.../x") 

but that clearly won't work. How do I save each data frame in the list as a separate .csv file?

Larynx

Answer 1

You can do this:

N <- names(lst)
for (i in seq_along(N)) write.csv(lst[[i]], file=paste0("C:/Users/.../"), N[i], ".csv")

Following the comment of Heroka a shorter version:

for (df in names(lst)) write.csv(lst[[df]], file=paste0("C:/Users/.../"), df, ".csv")

or

lapply(names(lst), function(df) write.csv(lst[[df]], file=paste0("C:/Users/.../"), df, ".csv") )

Laclair

As you can also access list items by name, you could even iterate over the names themselves. – Potage

Answer 2

The mapply function and its Map wrapper are the multi-argument versions of lapply. You have the list of data.frames; you need to build a vector of file names. Like this:

filenames<-paste0("C:/Users/.../",names(lst), ".csv")
Map(write.csv,lst,filenames)

What Map does? It calls the function provided as first argument multiple times and in each iteration its arguments are taken from the elements of the other arguments provided. Something on the line of:

list(write.csv(lst[[1]],filenames[[1]]),write.csv(lst[[2]],filenames[[2]]),...)

Hugohugon