Q:

Combine table with different elements

I have items in different lists and I want to count the item in each list and output it to a table.  However, I ran into difficulty when there are different items in the list.   Too illustrate my problem: 
item_1 <- c("A","A","B")
item_2 <- c("A","B","B","B","C")
item_3 <- c("C","A")
item_4 <- c("D","A", "A")
item_5 <- c("B","D")

list_1 <- list(item_1, item_2, item_3)
list_2 <- list(item_4, item_5)

table_1 <- table(unlist(list_1))
table_2 <- table(unlist(list_2))

> table_1

A B C 
4 4 2 
> table_2

A B D 
2 1 2 

What I get from cbind is : 
> cbind(table_1, table_2)

  table_1 table_2
A       4       2
B       4       1
C       2       2

which is clearly wrong.  What I need is: 
  table_1 table_2
A       4       2
B       4       1
C       2       0
D       0       2

Thanks in advance 

A:

It would probably be better to use factors at the start if possible, something like:
L <- list(list_1 = list_1, 
          list_2 = list_2)
RN <- unique(unlist(L))
do.call(cbind, 
        lapply(L, function(x)
          table(factor(unlist(x), RN))))
#   list_1 list_2
# A      4      2
# B      4      1
# C      2      0
# D      0      2

However, going with what you have, a function like the following might be useful.  I've added comments to help explain what's happening in each step. 
myFun <- function(..., fill = 0) {
  ## Get the names of the ...s. These will be our column names
  CN <- sapply(substitute(list(...))[-1], deparse)
  ## Put the ...s into a list
  Lst <- setNames(list(...), CN)
  ## Get the relevant row names
  RN <- unique(unlist(lapply(Lst, names), use.names = FALSE))
  ## Create an empty matrix.  `fill` can be anything--it's set to 0
  M <- matrix(fill, length(RN), length(CN),
              dimnames = list(RN, CN))
  ## Use match to identify the correct row to fill in
  Row <- lapply(Lst, function(x) match(names(x), RN))
  ## use matrix indexing to fill in the unlisted values of Lst
  M[cbind(unlist(Row), 
          rep(seq_along(Lst), vapply(Row, length, 1L)))] <-
    unlist(Lst, use.names = FALSE)
  ## Return your matrix
  M
}

Applied to your two tables, the outcome is like this:
myFun(table_1, table_2)
#   table_1 table_2
# A       4       2
# B       4       1
# C       2       0
# D       0       2

Here's an example with adding another table to the problem.  It also demonstrates use of NA as a fill value. 
set.seed(1) ## So you can get the same results as me
table_3 <- table(sample(LETTERS[3:6], 20, TRUE) )
table_3
# 
# C D E F 
# 2 7 9 2

myFun(table_1, table_2, table_3, fill = NA)
#   table_1 table_2 table_3
# A       4       2      NA
# B       4       1      NA
# C       2      NA       2
# D      NA       2       7
# E      NA      NA       9
# F      NA      NA       2

