How can I automate data frame naming in R? - r

Question

Let's say I have the following data frame:
x <- data.frame(let = sample(LETTERS, 100, replace = T),
num = sample(1:10, 100, replace = T))
I want to create several subsets of x where each new data frame is named after the levels of x$let. So far, I've come up with this simple function:
ss <- function(letra){
return(subset(x, let == letra))
}
Which is very rudimentary and doesn't do the naming as I wanted. My question is: how can I automate the following procedure?
a <- ss('A')
b <- ss('B')
c <- ss('C')
...
z <- ss('Z')

Answer 1

To elaborate a bit.
xs <- split(x, x$let)
Now we have a list, xs, of each subset of the original data frame. The names of each list component matches the factor level it was selected on:
xs[['D']]
let num
8 D 8
14 D 1
16 D 9
54 D 5
60 D 6
64 D 8
74 D 8
Most people use either xlsx or XLConnect to write Excel files from R. I happen to use XLConnect, but the solutions would be very similar.
Now we can simply do this:
require(XLConnect)
file_name <- paste0("file",LETTERS,".xlsx")
for (i in seq_len(length(xs))){
wb <- loadWorkbook(file_name[i],create = TRUE)
createSheet(wb,"Sheet1")
writeWorksheet(wb,data = xs[[i]],sheet = 1)
saveWorkbook(wb)
}
I've done this in a for loop so that it's easier to read and understand, but obviously this could all be shoved into an lapply or mapply type solution as well.

Answer 2

Agreed with Joshua that you may want to do something different, but if you're really hooked on your previous idea, you can use:
x <- data.frame(let = sample(LETTERS, 100, replace = T),
num = sample(1:10, 100, replace = T))
ss <- function(letra){
assign(letra, subset(x, let == letra), envir = .GlobalEnv)
# Returning the DF is optional:
# return(subset(x, let == letra))
}
ss('A')
print(A)
Update: taking Joran's suggestion, one can write:
x_split <- split(x,x$let)
for (let in x_split) {
write.csv(let, file = paste0((let$let)[1], ".csv"))
}