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Use object names within a list in lapply/ldply

Tags:

r

plyr

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:

df1 <- data.frame(a=1:3, x=rnorm(3))
df2 <- data.frame(a=1:3, x=rnorm(3))
df3 <- data.frame(a=1:3, x=rnorm(3))

df.list <- list(df1, df2, df3)

That I want to rbind together, I can do the following:

df.all <- ldply(df.list, rbind)

However, I want another column that identifies which data.frame each row came from. I expected to be able to use the deparse(substitute(x)) method (here and elsewhere) to get the name of the relevant data.frame and add a column. This is how I approached it:

fun <- function(x) {
  name <- deparse(substitute(x))
  x$id <- name
  return(x)
}
df.all <- ldply(df.list, fun)

Which returns

  a          x      id
1 1  1.1138062 X[[1L]]
2 2 -0.5742069 X[[1L]]
3 3  0.7546323 X[[1L]]
4 1  1.8358605 X[[2L]]
5 2  0.9107199 X[[2L]]
6 3  0.8313439 X[[2L]]
7 1  0.5827148 X[[3L]]
8 2 -0.9896495 X[[3L]]
9 3 -0.9451503 X[[3L]]

So obviously each element of the list does not contain the name I think it does. Can anyone suggest a way to get what I expected (shown below)?

  a          x  id
1 1  1.1138062 df1
2 2 -0.5742069 df1
3 3  0.7546323 df1
4 1  1.8358605 df2
5 2  0.9107199 df2
6 3  0.8313439 df2
7 1  0.5827148 df3
8 2 -0.9896495 df3
9 3 -0.9451503 df3
like image 927
alexwhan Avatar asked Mar 05 '13 01:03

alexwhan


1 Answers

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)
like image 50
iTech Avatar answered Nov 22 '22 05:11

iTech