Reputation: 3021
I have the following named list output from a analysis. The reproducible code is as follows:
list(structure(c(-213.555409754509, -212.033637890131, -212.029474755074,
-211.320398316741, -211.158815833294, -210.470525157849), .Names = c("wasn",
"chappal", "mummyji", "kmph", "flung", "movie")), structure(c(-220.119433774144,
-219.186901747536, -218.743319709963, -218.088361753899, -217.338920075687,
-217.186050877079), .Names = c("crazy", "wired", "skanndtyagi",
"andr", "unveiled", "contraption")))
I want to convert this to a data frame. I have tried unlist to data frame options using reshape2, dplyr and other solutions given for converting a list to a data frame but without much success. The output that I am looking for is something like this:
Col1 Val1 Col2 Val2
1 wasn -213.55 crazy -220.11
2 chappal -212.03 wired -219.18
3 mummyji -212.02 skanndtyagi -218.74
so on and so forth. The actual out put has multiple columns with paired values and runs into many rows. I have tried the following codes already:
do.call(rbind, lapply(df, data.frame, stringsAsFactors = TRUE))
works partially provides all the character values in a column and numeric values in the second.
data.frame(Reduce(rbind, df))
didn't work - provides the names in the first list and numbers from both the lists as tow different rows
colNames <- unique(unlist(lapply(df, names)))
M <- matrix(0, nrow = length(df), ncol = length(colNames),
dimnames = list(names(df), colNames))
matches <- lapply(df, function(x) match(names(x), colNames))
M[cbind(rep(sequence(nrow(M)), sapply(matches, length)),
unlist(matches))] <- unlist(df)
M
didn't work correctly.
Can someone help?
Upvotes: 3
Views: 970
Reputation: 66819
Here's another way:
as.data.frame( c(col = lapply(x, names), val = lapply(x,unname)) )
How it works. lapply
returns a list; two lists combined with c
make another list; and a list is easily coerced to a data.frame, since the latter is just a list of vectors having the same length.
Better than coercing to a data.frame is just modifying its class, effectively telling the list "you're a data.frame now":
L = c(col = lapply(x, names), val = lapply(x,unname))
library(data.table)
setDF(L)
The result doesn't need to be assigned anywhere with =
or <-
because L
is modified "in place."
Upvotes: 2
Reputation: 193687
Since the list elements are all of the same length, you should be able to stack
them and then combine them by columns.
Try:
do.call(cbind, lapply(myList, stack))
Upvotes: 3