rbindfill like merge of list of vectors

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