Question

I regulary have the problem that I need to access the actual id variable when using d*ply or l*ply. A simple (yet nonsense) example would be:

df1 <- data.frame( p = c("a", "a", "b", "b"), q = 1:4 )
df2 <- data.frame( m = c("a", "b" ), n = 1:2 )

d_ply( df1, "p", function(x){
 actualId <- unique( x$p )
 print( mean(x$q)^df2[ df2$m == actualId, "n" ] )
})

So in case of d*ply functions I can help myself with unique( x$p ). But when it comes to l*ply, I have no idea how to access the name of the according list element.

l_ply( list(a = 1, b = 2, c = 3), function(x){
  print( <missing code> )
})
# desired output
[1] "a"
[1] "b"
[1] "c"

Any suggestions? Anything I am ignoring?

Answer 1

One way I’ve gotten around this is to loop over the index (names) and do the subsetting within the function.

l <- list(a = 1, b = 2, c = 3)
l_ply(names(l), function(x){
  print(x)
  myl <- l[[x]]
  print(myl)
})

myl will then be the same as

l_ply(l, function(myl) {
  print(myl)
})