rbindfill like merge of list of vectors - r

Question

I have a list of named vectors (see below and at end for dput version) I would like to "merge" together to make a matrix and fill in zeros if a vector doesn't contain a name (character in this case). This doesn't seem that hard but I haven't found a working base solution to the problem. I thought about using match but that seems very costly of time when I'm sure there's a fancy way to use do.call and rbind together.
List of Named Vectors:
$greg
e i k l
1 2 1 1
$sam
! c e i t
1 1 1 2 1
$teacher
? c i k l
1 1 1 1 1
Final Desired Output
! ? c e i k l t
greg 0 0 0 1 2 1 1 0
sam 1 0 1 1 2 0 0 1
teacher 0 1 1 0 1 1 1 0
Likely this is the output people will give and filling NAs with 0 is easy
! ? c e i k l t
greg NA NA NA 1 2 1 1 NA
sam 1 NA 1 1 2 NA NA 1
teacher NA 1 1 NA 1 1 1 NA
Sample Data
L2 <- structure(list(greg = structure(c(1L, 2L, 1L, 1L), .Dim = 4L, .Dimnames = structure(list(
c("e", "i", "k", "l")), .Names = ""), class = "table"), sam = structure(c(1L,
1L, 1L, 2L, 1L), .Dim = 5L, .Dimnames = structure(list(c("!",
"c", "e", "i", "t")), .Names = ""), class = "table"), teacher = structure(c(1L,
1L, 1L, 1L, 1L), .Dim = 5L, .Dimnames = structure(list(c("?",
"c", "i", "k", "l")), .Names = ""), class = "table")), .Names = c("greg",
"sam", "teacher"))

Answer 1

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

Answer 2

I think something like this:
names <- sort(unique(unlist(lapply(L2, names), use.names=FALSE)))
L3 <- t(vapply(L2, function(x) x[names], FUN.VALUE=numeric(length(names))))
colnames(L3) <- names
L3[is.na(L3)] <- 0

Answer 3

reshape2 Solution. This can be readily done with the reshape2 package by melting the list into long form and then using dcast to reshape it back into wide form:
> library(reshape2)
> m <- melt(L2)
> m$Var.1 <- factor(as.character(m$Var.1)) # optional - if columns should be sorted
> dcast(m, L1 ~ Var.1, fill = 0)
L1 ! ? c e i k l t
1 greg 0 0 0 1 2 1 1 0
2 sam 1 0 1 1 2 0 0 1
3 teacher 0 1 1 0 1 1 1 0
Base Solution. And here is a corresponding base solution where the first two lines perform the melt, the next line ensures the columns will be sorted and the last line reshapes from long to wide:
> 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)
Var1
row ! ? c e i k l t
greg 0 0 0 1 2 1 1 0
sam 1 0 1 1 2 0 0 1
teacher 0 1 1 0 1 1 1 0
EDIT: Added a base solution and modified the sort line.

Answer 4

While typing this I thought of this solution but wonder if there's a more efficient one:
chars <- sort(unique(unlist(lapply(L2, names))))
L3 <- lapply(L2, function(x){
nots <- chars[!chars %in% names(x)]
new <- rev(c(x, rep(0, length(nots))))
names(new)[1:length(nots)] <- nots
new[order(names(new))]
})
do.call(rbind, L3)
Yielding:
! ? c e i k l t
greg 0 0 0 1 2 1 1 0
sam 1 0 1 1 2 0 0 1
teacher 0 1 1 0 1 1 1 0