19

I have a list of lists that have names.

I want to add them all together into a dataframe but keep all the columns

past_earnings_lists[1]

successfully returns one list from the list of lists

names(past_earnings_lists)[1]

successfully returns the name of the list

past_earnings <- melt(past_earnings_lists)

puts all the data in one data frame but doesn't keep structure

past_earnings <- as.data.frame.matrix(past_earnings_lists$ADBE)

Successfully takes one list and keeps the structure but doesn't add the name of the list to the dataframe.

For example, adbe has 7 columns and 30 rows; I want it to add an 8th column with the name, adbe, and append it to a dataframe with all the other lists doing the same.

structure

I want a dataframe with the results being:
  sym  v1 v2 v3 v4 v5 v6 v7
1 adbe  1  2  3  4  5  6  7
2 adbe  1  2  3  4  5  6  7
3 air   1  2  3  4  5  6  7
4 air   1  2  3  4  5  6  7
5 alog  1  2  3  4  5  6  7
and so on
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5
  • 7
    this isn't going to fly without a reproducible example Commented Sep 19, 2017 at 21:55
  • 1
    Agreed that its unlikely you will find help without providing data structure to make it a reproducible example, but you may find the bind_rows function from dplyr helpful using the .id parameter to concatenate your lists while generating a new variable identifying the name of the list. dplyr.tidyverse.org/reference/bind.html Commented Sep 19, 2017 at 22:38
  • there, i've added a screenshot Commented Sep 19, 2017 at 22:49
  • It's still not clear. Do you have a list, or a list of lists? Can you include the output of str or dput of past_earnings_list? Commented Sep 20, 2017 at 5:38
  • Also note that doing mylist[1] returns a sublist, whereas mylist[[1]] returns the element itself (see the double brackets). Commented Sep 20, 2017 at 5:40

3 Answers 3

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42

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.

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2 Comments

That's great:)! How can I revert this one though?
What do you mean?
6

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.

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1 Comment

it doesn't seem to be working, added more information on what output i would like
2

@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)
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