Converting a list to a data frame
If you require that the 'type' column be a factor with a level for each element of the input list, then you need to do that after calling dplyr::bind_rows(), just as with the base-R solutions.
l <- list( A = data.frame(X=1:2, Y=11:12), B = data.frame(X=integer(),
Y=integer()), C = data.frame(X=3L, Y=13L) )
str(d <- dplyr::bind_rows(l, .id = "Which") )
'data.frame': 3 obs. of 3 variables: $ Which: chr "A" "A" "C" $ X : int 1 2 3 $ Y : int 11 12 13
d$Which <- factor(d$Which, levels=names(l)) str(d)
'data.frame': 3 obs. of 3 variables: $ Which: Factor w/ 3 levels "A","B","C": 1 1 3 $ X : int 1 2 3 $ Y : int 11 12 13 Sometimes you need the names of the of the 0-row data.frames in the output to make plots, etc., comparable across various samples of the data. Bill Dunlap TIBCO Software wdunlap tibco.com
On Thu, May 3, 2018 at 10:28 AM, Hadley Wickham <h.wickham at gmail.com> wrote:
On Wed, May 2, 2018 at 11:53 AM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote:
Another approach:
########
library(tidyr)
L <- list( A = data.frame( x=1:2, y=3:4 )
, B = data.frame( x=5:6, y=7:8 )
)
D <- data.frame( Type = names( L )
, stringsAsFactors = FALSE
)
D$data <- L
unnest(D, data)
#> Type x y
#> 1 A 1 3
#> 2 A 2 4
#> 3 B 5 7
#> 4 B 6 8
########
I think a slightly more idiomatic tidyverse solution is dplyr::bind_rows() l <- list( A = data.frame(x = 1:2, y = 3:4), B = data.frame(x = 5:6, y = 7:8) ) dplyr::bind_rows(l, .id = "type") #> type x y #> 1 A 1 3 #> 2 A 2 4 #> 3 B 5 7 #> 4 B 6 8 This also has the advantage of returning a data frame when the inputs are data frames. Hadley -- http://hadley.nz
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