Here's a fairly straight forward base solution:
# first determine all possible column names
cols <- sort(unique(unlist(lapply(L2,names), use.names=FALSE)))
# initialize the output
out <- matrix(0, length(L2), length(cols), dimnames=list(names(L2),cols))
# loop over list and fill in the matrix
for(i in seq_along(L2)) {
out[names(L2)[i], names(L2[[i]])] <- L2[[i]]
}
UPDATE with benchmarks:
f1 <- function(L2) {
cols <- sort(unique(unlist(lapply(L2,names), use.names=FALSE)))
out <- matrix(0, length(L2), length(cols), dimnames=list(names(L2),cols))
for(i in seq_along(L2)) out[names(L2)[i], names(L2[[i]])] <- L2[[i]]
out
}
f2 <- function(L2) {
L.names <- sort(unique(unlist(sapply(L2, names))))
L3 <- t(sapply(L2, function(x) x[L.names]))
colnames(L3) <- L.names
L3[is.na(L3)] <- 0
L3
}
f3 <- function(L2) {
m <- do.call(rbind, lapply(L2, as.data.frame))
m$row <- sub("[.].*", "", rownames(m))
m$Var1 <- factor(as.character(m$Var1))
xtabs(Freq ~ row + Var1, m)
}
library(rbenchmark)
benchmark(f1(L2), f2(L2), f3(L2), order="relative")[,1:5]
# test replications elapsed relative user.self
# 1 f1(L2) 100 0.022 1.000 0.020
# 2 f2(L2) 100 0.051 2.318 0.052
# 3 f3(L2) 100 0.788 35.818 0.760
set.seed(21)
L <- replicate(676, {n=sample(10,1); l=sample(26,n);
setNames(sample(6,n,TRUE), letters[l])}, simplify=FALSE)
names(L) <- levels(interaction(letters,LETTERS))
benchmark(f1(L), f2(L), order="relative")[,1:5]
# test replications elapsed relative user.self
# 1 f1(L) 100 1.84 1.000 1.828
# 2 f2(L) 100 4.24 2.304 4.220