Get column names and dataframe name from a list of dataframes into a single dataframe

Question

I am looking for a way to make column names and dataframe names from a list of dataframes into a single dataframe. They have unequal length of columns. What's the best way to do this?

dlist <- list(mtcars[1:2], mtcars[1:3], mtcars[1:4])
names(dlist) <- c("mtcars1", "mtcars2", "mtcars3")

Tried:
dlist |> map(~colnames(.x))

Expected output:

1 mtcars1 mpg   cyl   NA    NA   
2 mtcars2 mpg   cyl   disp  NA   
3 mtcars3 mpg   cyl   disp  hp  

Laughable

is your expected output a dataframe? It looks like a char vector. Can you please clarify? – Nanine

Sorry, the output is a dataframe but I just didn't know how to make one like the one I wanted. @Nanine – Laughable

Answer 1

You can try:

library(tidyverse)

dlist %>%
  map_df(~names(.x) %>%
           enframe(), .id = "id") %>%
  pivot_wider(names_from = name, id_cols = id)

# A tibble: 3 x 5
  id      `1`   `2`   `3`   `4`  
  <chr>   <chr> <chr> <chr> <chr>
1 mtcars1 mpg   cyl   NA    NA   
2 mtcars2 mpg   cyl   disp  NA   
3 mtcars3 mpg   cyl   disp  hp  

Or same idea in base:

reshape(stack(lapply(dlist, names)), idvar = "ind", timevar = "values", direction = "wide", v.names = "values")

      ind values.mpg values.cyl values.disp values.hp
1 mtcars1        mpg        cyl        <NA>      <NA>
3 mtcars2        mpg        cyl        disp      <NA>
6 mtcars3        mpg        cyl        disp        hp

Prehension

Answer 2

Similar to the accepted solution but more compact thanks to unnest_wider:

library(tidyr)
library(tibble)
library(purrr)

dlist %>% 
  map(colnames) %>% 
  enframe %>% 
  unnest_wider(value, names_sep = "_")

#> # A tibble: 3 x 5
#>   name    value_1 value_2 value_3 value_4
#>   <chr>   <chr>   <chr>   <chr>   <chr>  
#> 1 mtcars1 mpg     cyl     NA      NA     
#> 2 mtcars2 mpg     cyl     disp    NA     
#> 3 mtcars3 mpg     cyl     disp    hp   

Nanine

Answer 3

Maybe there is a better way but this is how I would do it:

library(purrr)

dlist %>%
  map(~ .x %>% 
        names() %>% 
        append(rep(NA_character_, dlist %>%
                     map(~ .x %>% names()) %>%
                     reduce(~ max(length(..1), length(..2))) - length(.x)))) %>%
  exec(rbind, !!!.) %>%
  as.data.frame()

         V1  V2   V3   V4
mtcars1 mpg cyl <NA> <NA>
mtcars2 mpg cyl disp <NA>
mtcars3 mpg cyl disp   hp

Or a bit more concise with map2, similar to may dear friend's elegant solution:

dlist %>% 
  map2_dfr(names(dlist), ~ c(.y, names(.x)) %>% set_names(~ letters[seq_along(.x)])) %>%
  column_to_rownames("a")

          b   c    d    e
mtcars1 mpg cyl <NA> <NA>
mtcars2 mpg cyl disp <NA>
mtcars3 mpg cyl disp   hp

Gutter

Answer 4

A base R solution is:

dlist |> lapply(names) |>
  (\(x){
    res <- t(sapply(x, `length<-`, max(lengths(x))))
    cbind(id = names(x), setNames(data.frame(res), 1:NCOL(res)))
  })()
#R>              id   1   2    3    4
#R> mtcars1 mtcars1 mpg cyl <NA> <NA>
#R> mtcars2 mtcars2 mpg cyl disp <NA>
#R> mtcars3 mtcars3 mpg cyl disp   hp

Since the row names and id are the same, then this might do:

dlist |> lapply(names) |>
  (\(x) sapply(x, `length<-`, max(lengths(x))))() |> 
  t() |> as.data.frame()
#R>          V1  V2   V3   V4
#R> mtcars1 mpg cyl <NA> <NA>
#R> mtcars2 mpg cyl disp <NA>
#R> mtcars3 mpg cyl disp   hp

Hypha

Answer 5

or this one?

library(tidyverse)
dlist <- list(mtcars[1:2], mtcars[1:3], mtcars[1:4])
names(dlist) <- c("mtcars1", "mtcars2", "mtcars3")


map_dfr(dlist, ~names(.x) %>% set_names(paste0('col', seq_along(.x))))

#> # A tibble: 3 x 4
#>   col1  col2  col3  col4 
#>   <chr> <chr> <chr> <chr>
#> 1 mpg   cyl   <NA>  <NA> 
#> 2 mpg   cyl   disp  <NA> 
#> 3 mpg   cyl   disp  hp

If you also want rownames back

imap_dfr(dlist, ~c(.y, names(.x)) %>% set_names('cc',paste0('col', seq_along(.x)))) %>%
  column_to_rownames('cc')

#>         col1 col2 col3 col4
#> mtcars1  mpg  cyl <NA> <NA>
#> mtcars2  mpg  cyl disp <NA>
#> mtcars3  mpg  cyl disp   hp

^{Created on 2021-07-02 by the reprex package (v2.0.0)}

Stupe

Answer 6

Another way would be:

do.call('rbind',
        lapply(seq_len(length(dlist)), 
               function(i) c(names(dlist)[i], 
                             names(dlist[[i]]),
                             rep(NA, max(sapply(dlist, length)) - ncol(dlist[[i]]))
               )
        )
)

Labialize