Binding lists of data frames while preserving index - r

Question

There are many posts on converting list of data frames to single data frame (ex. here). However, I need to do that while preserving information from which list
resulting rows are, i.e. I need to preserve an index of original list. If the list is unnamed, I need just an index number, while if the list is named I need to save the name of original list element. How can I do that?
With the data like below:
foo <- list(data.frame(x=c('a', 'b', 'c'),y = c(1,2,3)),
data.frame(x=c('d', 'e', 'f'),y = c(4,5,6)))
I need an output like this:
index x y
1 1 a 1
2 1 b 2
3 1 c 3
4 2 d 4
5 2 e 5
6 2 f 6
while with named elements of a list:
foo <- list(df1 = data.frame(x=c('a', 'b', 'c'),y = c(1,2,3)),
df2 = data.frame(x=c('d', 'e', 'f'),y = c(4,5,6)))
the output would be:
index x y
1 df1 a 1
2 df1 b 2
3 df1 c 3
4 df2 d 4
5 df2 e 5
6 df2 f 6

Answer 1

We can use imap to get the index
library(purrr)
imap_dfr(foo, ~ .x %>%
mutate(index = .y))
Or with map
map_dfr(foo, .f = cbind, .id = 'index')
# index x y
#1 1 a 1
#2 1 b 2
#3 1 c 3
#4 2 d 4
#5 2 e 5
#6 2 f 6
Or use Map from base R, where we loop through the elements of 'foo' and a corresponding sequence of 'foo', cbind to create a new column and then rbind the list elements
do.call(rbind, Map(cbind, index = seq_along(foo), foo))
# index x y
#1 1 a 1
#2 1 b 2
#3 1 c 3
#4 2 d 4
#5 2 e 5
#6 2 f 6

Answer 2

You could name the list if it is unnamed and then loop over them using lapply to create list of dataframes and rbind them together as one dataframe.
names(foo) <- if (is.null(names(foo))) seq_len(length(foo)) else names(foo)
do.call("rbind", lapply(names(foo), function(x) cbind(index = x, foo[[x]])))
# index x y
#1 1 a 1
#2 1 b 2
#3 1 c 3
#4 2 d 4
#5 2 e 5
#6 2 f 6
For named list
do.call("rbind", lapply(names(foo), function(x) cbind(index = x, foo[[x]])))
# index x y
#1 df1 a 1
#2 df1 b 2
#3 df1 c 3
#4 df2 d 4
#5 df2 e 5
#6 df2 f 6