Joint Occurrence of variables in R - r

Question

I want to count individual and combine occurrence of variables (1 represents presence and 0 represents absence). This can be obtained by multiple uses of table function (See MWE below). Is it possible to use a more efficient approach to get the required output given below?
set.seed(12345)
A <- rbinom(n = 100, size = 1, prob = 0.5)
B <- rbinom(n = 100, size = 1, prob = 0.6)
C <- rbinom(n = 100, size = 1, prob = 0.7)
df <- data.frame(A, B, C)
table(A)
A
0 1
48 52
table(B)
B
0 1
53 47
table(C)
C
0 1
34 66
table(A, B)
B
A 0 1
0 25 23
1 28 24
table(A, C)
C
A 0 1
0 12 36
1 22 30
table(B, C)
C
B 0 1
0 21 32
1 13 34
table(A, B, C)
, , C = 0
B
A 0 1
0 8 4
1 13 9
, , C = 1
B
A 0 1
0 17 19
1 15 15
Required Output
I am requiring something like the following:
A = 52
B = 45
C = 66
A + B = 24
A + C = 30
B + C = 34
A + B + C = 15

Answer 1

Expanding on Sumedh's answer, you can also do this dynamically without having to specify the filter every time. This will be useful if you have more than only 3 columns to combine.
You can do something like this:
lapply(seq_len(ncol(df)), function(i){
# Generate all the combinations of i element on all columns
tmp_i = utils::combn(names(df), i)
# In the columns of tmp_i we have the elements in the combination
apply(tmp_i, 2, function(x){
dynamic_formula = as.formula(paste("~", paste(x, "== 1", collapse = " & ")))
df %>%
filter_(.dots = dynamic_formula) %>%
summarize(Count = n()) %>%
mutate(type = paste0(sort(x), collapse = ""))
}) %>%
bind_rows()
}) %>%
bind_rows()
This will:
1) generate all the combinations of the columns of df. First the combinations with one element (A, B, C) then the ones with two elements (AB, AC, BC), etc.
This is the external lapply
2) then for every combination will create a dynamic formula. For AB for instance the formula will be A==1 & B==1, exactly as Sumedh suggested. This is the dynamic_formula bit.
3) Will filter the dataframe with the dynamically generated formula and count the number of rows
4) Bind all together (the two bind_rows)
The output will be
Count type
1 52 A
2 47 B
3 66 C
4 24 AB
5 30 AC
6 34 BC
7 15 ABC

Answer 2

EDITED TO ADD: I see now that you don't want to get the exclusive counts (i.e. A and AB should both include all As).
I got more than a little nerd-sniped by this today, particularly as I wanted to solve it using base R with no packages. The below should do that.
There is a very easy (in principle) solution that simply uses xtabs(), which I've illustrated below. However, to generalize it for any potential number of dimensions, and then to apply it to a variety of combinations, actually was harder. I strove to avoid using the dreaded eval(parse()).
set.seed(12345)
A <- rbinom(n = 100, size = 1, prob = 0.5)
B <- rbinom(n = 100, size = 1, prob = 0.6)
C <- rbinom(n = 100, size = 1, prob = 0.7)
df <- data.frame(A, B, C)
# Turn strings off
options(stringsAsFactors = FALSE)
# Obtain the n-way frequency table
# This table can be directly subset using []
# It is a little tricky to pass the arguments
# I'm trying to avoid eval(parse())
# But still give a solution that isn't bound to a specific size
xtab_freq <- xtabs(formula = formula(x = paste("~",paste(names(df),collapse = " + "))),
data = df)
# Demonstrating what I mean
# All A
sum(xtab_freq["1",,])
# [1] 52
# AC
sum(xtab_freq["1",,"1"])
# [1] 30
# Using lapply(), we pass names(df) to combn() with m values of 1, 2, and 3
# The output of combn() goes through list(), then is unlisted with recursive FALSE
# This gives us a list of vectors
# Each one being a combination in which we are interested
lst_combs <- unlist(lapply(X = 1:3,FUN = combn,x = names(df),list),recursive = FALSE)
# For nice output naming, I just paste the values together
names(lst_combs) <- sapply(X = lst_combs,FUN = paste,collapse = "")
# This is a function I put together
# Generalizes process of extracting values from a crosstab
# It does it in this fashion to avoid eval(parse())
uFunc_GetMargins <- function(crosstab,varvector,success) {
# Obtain the dimname-names (the names within each dimension)
# From that, get the regular dimnames
xtab_dnn <- dimnames(crosstab)
xtab_dn <- names(xtab_dnn)
# Use match() to get a numeric vector for the margins
# This can be used in margin.table()
tgt_margins <- match(x = varvector,table = xtab_dn)
# Obtain a margin table
marginal <- margin.table(x = crosstab,margin = tgt_margins)
# To extract the value, figure out which marginal cell contains
# all variables of interest set to success
# sapply() goes over all the elements of the dimname names
# Finds numeric index in that dimension where the name == success
# We subset the resulting vector by tgt_margins
# (to only get the cells in our marginal table)
# Then, use prod() to multiply them together and get the location
tgt_cell <- prod(sapply(X = xtab_dnn,
FUN = match,
x = success)[tgt_margins])
# Return as named list for ease of stacking
return(list(count = marginal[tgt_cell]))
}
# Doing a call of mapply() lets us get the results
do.call(what = rbind.data.frame,
args = mapply(FUN = uFunc_GetMargins,
varvector = lst_combs,
MoreArgs = list(crosstab = xtab_freq,
success = "1"),
SIMPLIFY = FALSE,
USE.NAMES = TRUE))
# count
# A 52
# B 47
# C 66
# AB 24
# AC 30
# BC 34
# ABC 15
I ditched the prior solution that used aggregate.

Answer 3

Using dplyr,
Occurrence of only A:
library(dplyr)
df %>% filter(A == 1) %>% summarise(Total = nrow(.))
Occurrence of A and B:
df %>% filter(A == 1, B == 1) %>% summarise(Total = nrow(.))
Occurence of A, B, and C
df %>% filter(A == 1, B == 1, C == 1) %>% summarise(Total = nrow(.))