Adding a column to every dataframe in a list with the name of the list element

Question

I have a list containing multiple data frames, and each list element has a unique name. The structure is similar to this dummy data

a <- data.frame(z = rnorm(20), y = rnorm(20))
b <- data.frame(z = rnorm(30), y = rnorm(30))
c <- data.frame(z = rnorm(40), y = rnorm(40))
d <- data.frame(z = rnorm(50), y = rnorm(50))

my.list <- list(a,b,c,d)
names(my.list) <- c("a","b","c","d")

I want to create a column in each of the data frames that has the name of it's respective list element. My goal is to merge all the list element into a single data frame, and know which data frame they came from originally. The end result I'm looking for is something like this:

            z           y group
1   0.6169132  0.09803228     a
2   1.1610584  0.50356131     a
3   0.6399438  0.84810547     a
4   1.0878453  1.00472105     b
5  -0.3137200 -1.20707112     b
6   1.1428834  0.87852556     b
7  -1.0651735 -0.18614224     c
8   1.1629891 -0.30184443     c
9  -0.7980089 -0.35578381     c
10  1.4651651 -0.30586852     d
11  1.1936547  1.98858128     d
12  1.6284174 -0.17042835     d

My first thought was to use mutate to assign the list element name to a column in each respective data frame, but it appears that when used within lapply, names() refers to the column names, not the list element names

test <- lapply(my.list, function(x)  mutate(x, group = names(x)))

   Error: Column `group` must be length 20 (the number of rows) or one, not 2

Any suggestions as to how I could approach this problem?

Roman · Accepted Answer · 2020-08-04 19:45:35Z

6

there is no need to mutate just bind using dplyr's bind_rows

library(tidyverse)
my.list %>% 
  bind_rows(.id = "groups")

Obviously requires that the list is named.

answered Aug 4, 2020 at 19:45

Roman

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1 Comment

canderson156 Over a year ago

This answer solution is simple and does exactly what I wanted - thanks! Maybe you can answer this question too....I've never understood the syntax where some arguments have a . in front of them like .id. Do you know why it's called .id and not just id?

TylerH · Accepted Answer · 2025-08-14 15:29:15Z

4

We can use Map from base R:

Map(cbind, my.list, group = names(my.list))

Or with imap from purrr:

library(dplyr)
library(purrr)
imap(my.list, ~ .x %>% mutate(group = .y))

Or if the intention is to create a single data.frame

library(data.table)
rbindlist(my.list. idcol = 'groups')

edited Aug 14 at 15:29

TylerH

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answered Aug 4, 2020 at 19:24

akrun

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