如何计算数据框中每种组合的比例？

Question

Lets say we have a data.frame as following:

  A B C
1 1 1 1
2 1 0 1
3 1 0 1
4 0 1 0
5 0 0 1

As Output I want something containing this:

ABC = 0.2

AC = 0.4

B = 0.2

C = 0..2

but for a much larger data.frame. Does anyone know an elegant code to do so?. If so please let me know, thank you.

Answer 1

If M is your matrix you can do

table(apply(M, 1, function(v) paste0(names(v[v==1]), collapse = ""))) / nrow(M)

With your example:

> M <- cbind(A = c(1,1,1,0,0), B = c(1,0,0,1,0), C = c(1,1,1,0,1))
> table(apply(M, 1, function(v) paste0(names(v[v==1]), collapse = ""))) / nrow(M)

ABC  AC   B   C 
0.2 0.4 0.2 0.2 

Answer 2

ind = which(d == 1, arr.ind = TRUE)
table(sapply(split(colnames(d)[ind[,2]], ind[,1]), paste, collapse = "-"))/NROW(d)

#A-B-C   A-C     B     C 
#  0.2   0.4   0.2   0.2

DATA

d = structure(list(A = c(1L, 1L, 1L, 0L, 0L),
                   B = c(1L, 0L, 0L, 1L, 0L),
                   C = c(1L, 1L, 1L, 0L, 1L)),
              class = "data.frame",
              row.names = c("1", "2", "3", "4", "5"))

Answer 3

Using imap from purr we can replace the 1s with the column name and the 0s with empty strings. Then the prop.table of the columns pasted together gives the desired output

library(purrr)

df %>% 
  imap(~ifelse(.x, .y, '')) %>% 
  do.call(what = paste0) %>% 
  table %>% 
  prop.table
# .
# ABC  AC   B   C 
# 0.2 0.4 0.2 0.2 

If your data is very large it would be faster to change the names of the table at the end instead of creating three new columns of "A", "B" and "C" instead of ones and zeros first. Same output as above, purrr not needed.

out <- 
  df %>% 
    do.call(what = paste0) %>% 
    table %>% 
    prop.table

names(out) <- sapply(strsplit(names(out), ''), 
                     function(x) paste(LETTERS[which(x == '1')], collapse = ''))
out