Create a dataframe with list elements with dplyr in R

Question

This is my dataframe:

    df<-list(structure(list(Col1 = structure(1:6, .Label = c("A", "B", 
"C", "D", "E", "F"), class = "factor"), Col2 = structure(c(1L, 
2L, 3L, 2L, 4L, 5L), .Label = c("B", "C", "D", "F", "G"), class = "factor")), class = "data.frame", row.names = c(NA, 
-6L)), structure(list(Col1 = structure(c(1L, 4L, 5L, 6L, 2L, 
3L), .Label = c("A", "E", "H", "M", "N", "P"), class = "factor"), 
    Col2 = structure(c(1L, 2L, 3L, 2L, 4L, 5L), .Label = c("B", 
    "C", "D", "F", "G"), class = "factor")), class = "data.frame", row.names = c(NA, 
-6L)), structure(list(Col1 = structure(c(1L, 4L, 6L, 5L, 2L, 
3L), .Label = c("A", "W", "H", "M", "T", "U"), class = "factor"), 
    Col2 = structure(c(1L, 2L, 3L, 2L, 4L, 5L), .Label = c("B", 
    "C", "D", "S", "G"), class = "factor")), class = "data.frame", row.names = c(NA, 
-6L))) 

I want to extract col1=df[[1]][1] as a dataframe. Then col1 of the second position of this list I want to merge to the df[[1]][1], then I will have a dataframe with 2 columns. After this I want to merge the column 1 of the third position of the list to the dataframe with two columns, then I will have a dataframe with 3 columns.

In other words my dataframe should have 3 columns, all the first columns of each entry of my list.

The dplyr package can helpme to do this?

Any help?

Answer 1

You can use lapply to extract the three columns named "Col1 in one go. Then set the names of the result.

col1 <- as.data.frame(lapply(df, '[[', "Col1"))
names(col1) <- letters[seq_along(col1)]

col1
#  a b c
#1 A A A
#2 B M M
#3 C N U
#4 D P T
#5 E E W
#6 F H H

Choose any other column names that you might find better.

A dplyr way could be

df %>% 
  unlist(recursive = FALSE) %>%
  as.data.frame %>%
  select(., starts_with("Col1"))
#  Col1 Col1.1 Col1.2
#1    A      A      A
#2    B      M      M
#3    C      N      U
#4    D      P      T
#5    E      E      W
#6    F      H      H

Answer 2

With map_dfc from purrr:

library(purrr)

map_dfc(df, `[`, 1)

Output:

  Col1 Col11 Col12
1    A     A     A
2    B     M     M
3    C     N     U
4    D     P     T
5    E     E     W
6    F     H     H