R List of lists to dataframe with list name as extra column
This might work
library(purrr)
ans <- map_df(past_earnings_lists, ~as.data.frame(.x), .id="id")
It uses map_df
, which will map over lists and convert the result to data frames (if possible). Use the .id
argument to add names
to each data frame as a column.
as @dshkol commented, the easiest is to use dplyr::bind_rows
:
d = data.frame(letter = LETTERS, number = 1:26)
d.list = list(d1 = d, d2 = d)
d.all = dplyr::bind_rows(d.list, .id = "variable")
You can also do this in base R with rbind
and do.call
:
d.all = do.call(rbind, d.list)
However, this will not give you a column containing the list names. You could parse it from the row.names
though:
d.all["variable"] = unlist(lapply(
strsplit(row.names(d.all), ".", fixed = TRUE), function(x) x[[1]])
)
Alternatively, loop through your data frames frames and add the label manually prior to binding:
for (n in names(d.list))
d.list[[n]]['name'] = n
d.all = do.call(rbind, d.list)
However, it looks like your data frames don't have column names. I think you'll need to fix that for either solution to work.
@mikeck was in right track. Splitting string using .
is tricky as .
regex matches any character. So we need escape character \
before the .
. For anyone who wants to accomplish this with base R, you may try this:
df <- do.call(rbind, list)
df$listname <- lapply(strsplit(row.names(df), "\\."), '[[', 1)