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$id <- name
return(x)
}
df.all <- ldply(df.list, fun)
Which returns
a x id
1 1 1.1138062 X[[1L]]
2 2 -0.5742069 X[[1L]]
3 3 0.7546323 X[[1L]]
4 1 1.8358605 X[[2L]]
5 2 0.9107199 X[[2L]]
6 3 0.8313439 X[[2L]]
7 1 0.5827148 X[[3L]]
8 2 -0.9896495 X[[3L]]
9 3 -0.9451503 X[[3L]]
So obviously each element of the list does not contain the name I think it does. Can anyone suggest a way to get what I expected (shown below)?
a x id
1 1 1.1138062 df1
2 2 -0.5742069 df1
3 3 0.7546323 df1
4 1 1.8358605 df2
5 2 0.9107199 df2
6 3 0.8313439 df2
7 1 0.5827148 df3
8 2 -0.9896495 df3
9 3 -0.9451503 df3
Define your list with names and it should give you an .id
column with the data.frame
name
df.list <- list(df1=df1, df2=df2, df3=df3)
df.all <- ldply(df.list, rbind)
Output:
.id a x
1 df1 1 1.84658809
2 df1 2 -0.01177462
3 df1 3 0.58579469
4 df2 1 -0.64748756
5 df2 2 0.24384614
6 df2 3 0.59012676
7 df3 1 -0.63037679
8 df3 2 -1.17416295
9 df3 3 1.09349618
Then you can know the data.frame
name from the column df.all$.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)
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