Reputation: 63
So I have a list, say:
L1 <- list(1:10, 5:14, 10:19)
Now I am trying to get the output of the list as dataframe such that my output looks:
1. 1 2 3 4 5 6 7 8 9 10
2. 5 6 7 8 9 10 11 12 13 14
3. 10 11 12 13 14 15 16 17 18 19
I am using
as.data.frame(L1, row.names = TRUE)
and
list_vect2df(L1)
But none of them are giving the required output
Upvotes: 6
Views: 26101
Reputation: 1
I have found a tool on rbloggers that is really helpful.Please refer to details on the link below:
https://www.r-bloggers.com/2013/01/converting-a-list-to-a-data-frame/
Basically, you get the tool by the code below,
require(devtools)
source_gist(4676064)
Then, use "as.data.frame" to convert list to dataframe. df<-as.data.frame(list)
Another way is to use the code below and then transpose it. This code could apply to list with different lengths. "data.frame(lapply(list, "length<-", max(lengths(list))))"
Upvotes: -1
Reputation: 17648
another purrr
solution
purrr:::reduce(L1, rbind.data.frame)
Using rbind
instead gives a matrix.
Upvotes: 1
Reputation: 39154
Here is an idea using t
, as.data.frame
, and map_dfr
from the purrr package.
L1 <- list(1:10, 5:14, 10:19)
library(purrr)
map_dfr(L1, ~as.data.frame(t(.x)))
# V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
# 1 1 2 3 4 5 6 7 8 9 10
# 2 5 6 7 8 9 10 11 12 13 14
# 3 10 11 12 13 14 15 16 17 18 19
The same idea but completely in base R.
do.call(rbind, lapply(L1, function(x) as.data.frame(t(x))))
# V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
# 1 1 2 3 4 5 6 7 8 9 10
# 2 5 6 7 8 9 10 11 12 13 14
# 3 10 11 12 13 14 15 16 17 18 19
Another base R idea that uses as.data.frame
twice. We can change the row names later.
as.data.frame(t(as.data.frame(L1)))
# V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
# X1.10 1 2 3 4 5 6 7 8 9 10
# X5.14 5 6 7 8 9 10 11 12 13 14
# X10.19 10 11 12 13 14 15 16 17 18 19
Finally, the same idea but use transpose
function from the data.table.
data.table::transpose(as.data.frame(L1))
# V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
# 1 1 2 3 4 5 6 7 8 9 10
# 2 5 6 7 8 9 10 11 12 13 14
# 3 10 11 12 13 14 15 16 17 18 19
Upvotes: 5
Reputation: 47300
You can unlist
and use matrix
, then converting to data.frame
. It seems to be faster for this case.
as.data.frame(matrix(unlist(L1),nrow=length(L1),byrow=TRUE)
microbenchmark::microbenchmark(
a= map_dfr(L1, ~as.data.frame(t(.x))),
b= do.call(rbind, lapply(L1, function(x) as.data.frame(t(x)))),
c= as.data.frame(t(as.data.frame(L1))),
d= data.table::transpose(as.data.frame(L1)),
e= as.data.frame(matrix(unlist(L1),nrow=length(L1),byrow=TRUE)),
times = 100,unit = "relative")
# Unit: relative
# expr min lq mean median uq max neval
# a 9.146545 8.548656 8.859087 8.859051 9.449237 7.265274 100
# b 13.879833 11.523000 11.433790 10.924726 10.797251 24.012107 100
# c 12.719835 10.635809 10.442108 10.229913 10.259789 7.020377 100
# d 10.439881 9.143530 9.205734 8.859026 9.176125 6.624454 100
# e 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 100
Upvotes: 5