Use object names within a list in lapply/ldply

Question

In attempting to answer a question earlier, I ran into a problem that seemed like it should be simple, but I couldn\'t figure out.

If I have a list of dataframes:

Answer 1

Define your list with names and it should give you an .id column with the data.frame name

df.list <- list(df1=df1, df2=df2, df3=df3)
df.all <- ldply(df.list, rbind)

Output:

  .id a           x
1 df1 1  1.84658809
2 df1 2 -0.01177462
3 df1 3  0.58579469
4 df2 1 -0.64748756
5 df2 2  0.24384614
6 df2 3  0.59012676
7 df3 1 -0.63037679
8 df3 2 -1.17416295
9 df3 3  1.09349618

Then you can know the data.frame name from the column df.all$.id

Edit: As per @Gary Weissman's comment if you want to generate the names automatically you can do

names(df.list) <- paste0('df',seq_along(df.list)

Answer 2

In your definition, df.list does not have names, however, even then the deparse substitute idiom does not appear to work easilty (as lapply calls .Internal(lapply(X, FUN)) -- you would have to look at the source to see if the object name was available and how to get it

Something like

names(df.list) <- paste('df', 1:3, sep = '')

foo <- function(n, .list){
         .list[[n]]$id <- n
         .list[[n]]
       } 

     a          x id
1 1  0.8204213  a
2 2 -0.8881671  a
3 3  1.2880816  a
4 1 -2.2766111  b
5 2  0.3912521  b
6 3 -1.3963381  b
7 1 -1.8057246  c
8 2  0.5862760  c
9 3  0.5605867  c

Answer 3

if you want to use your function, instead of deparse(substitute(x)) use match.call(), and you want the second argument, making sure to convert it to character

 name <- as.character(match.call()[[2]])

Answer 4

Using base only, one could try something like:

dd <- lapply(seq_along(df.list), function(x) cbind(df_name = paste0('df',x),df.list[[x]]))

do.call(rbind,dd)