Combine table with different elements - r

Question

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

Answer 1

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

Answer 2

To fix your existing problem, you can put the two tables into a list and add the missing values an names back in. Here, nm is a vector of the table names unique to each table, tbs is a list of the tables, and we can use sapply to append and reorder the missing values.
> nm <- unique(unlist(mget(paste("item", 1:5, sep = "_"))))
> tbs <- list(t1 = table_1, t2 = table_2)
> sapply(tbs, function(x) {
x[4] <- 0L
names(x)[4] <- nm[!nm %in% names(x)]
x[nm]
})
t1 t2
A 4 2
B 4 1
C 2 0
D 0 2
A general solution, for when you have unknowns, and so that you can keep NA values, is
> sapply(tbs, function(x) {
length(x) <- length(nm)
x <- x[match(nm, names(x))]
setNames(x, nm)
})
t1 t2
A 4 2
B 4 1
C 2 NA
D NA 2
But you could have avoided this entirely by going straight from items to table. You put the items into a list and then unlisted them in the very next step. There is a useNA argument in table that will keep the factor levels even when they're zero.
> t1 <- table(c(item_1, item_2, item_3), useNA = "always")
> t2 <- table(c(item_4, item_5), useNA = "always")
> table(c(item_4, item_5), useNA = "always")
A B D <NA>
2 1 2 0

Answer 3

A quick fix to your problem is to make the tables into data frames and then merge them:
d1 <- data.frame(value=names(table_1), table_1=as.numeric(table_1))
d2 <- data.frame(value=names(table_2), table_2=as.numeric(table_2))
merge(d1,d2, all=TRUE)
This will create NA's where you might want 0's. That can be fixed with
M <- merge(d1,d2, all=TRUE)
M[is.na(M)] <- 0