Определение относительного размера перекрывающихся групп на основе информации в двух векторах

Эту функцию также можно использовать для выполнения задания. Он возвращает вектор символов, содержащий имена семейств, которые можно удалить.

test_function <- function(dataset){

## split the kidid on the basis of famid
kids_family <- split.default(dataset[['kidid']],f = dataset[['famid']])

family <- names(kids_family)

## This function generates all the possible combinations if we select any two families from family
combn_family <- combn(family,2)

family_removed <- character(0)
apply(combn_family,MARGIN = 2, function(x){

  if (length(setdiff(kids_family[[x[1]]],kids_family[[x[2]]])) == 0)
    family_removed <<- c(family_removed,x[1])
  else if (length(setdiff(kids_family[[x[2]]],kids_family[[x[1]]])) == 0)
    family_removed <<- c(family_removed,x[2])

})

return (family_removed)
}
> df <- data.frame(famid = c("A","A","B","C","C","D","D", "E", "E", "E", "F", "F"),
+                  kidid = c(1, 2, 1, 3, 4, 4, 5, 7, 8, 9, 7, 9))
> test_function(df)
[1] "B" "F"

Я пробовал около setdiff без шансов. Я пришел и опубликовал это трудоемкое решение в надежде, что есть лучший способ.

# dependencies for melting tables and handling data.frames
require(reshape2)
require(dplyr)


# I have added two more cases to your data.frame
# kidid is passed as numeric (with quoted would have been changed to vector by default)
df <- data.frame(famid = c("A","A","B","C","C","D","D", "E", "E", "E", "F", "F"),
                 kidid = c(1, 2, 1, 3, 4, 4, 5, 7, 8, 9, 7, 9))

# let's have a look to it
df
famid kidid
1      A     1
2      A     2
3      B     1
4      C     3
5      C     4
6      D     4
7      D     5
8      E     7
9      E     8
10     E     9
11     F     7
12     F     9

# we build a contingency table
m <- table(df$famid, df$kidid)

# a family A only contains a family B, if A has all the elements of B, 
# and at least one that B doesnt have
m

  1 2 3 4 5 7 8 9
A 1 1 0 0 0 0 0 0
B 1 0 0 0 0 0 0 0
C 0 0 1 1 0 0 0 0
D 0 0 0 1 1 0 0 0
E 0 0 0 0 0 1 1 1
F 0 0 0 0 0 1 0 1

# an helper function to implement that and return a friendly data.frame
family_contained <- function(m){
  res <- list()
  for (i in 1:nrow(m))
    # for each line in m, we calculate the difference to all other lines
    res[[i]] <- t(apply(m[-i, ], 1, function(row) m[i, ] - row))
  # here we test if all values are 0+ (ie if the selected family has all element of the other)
  # and if at least one is >=1 (ie if the selected family has at least one element that the other doesnt have)
  tab <- sapply(res, function(m) apply(m, 1,  function(x) all(x>=0) & any(x>=1)))
  # we format it as a table to have nice names
  tab %>% as.table() %>% 
    # we melt it into a data.frame
    melt()  %>% 
    # only select TRUE and get rid of this column
    filter(value) %>% select(-value) %>% 
    # to make things clear we name columns
    `colnames<-`(c("this_family_is_contained", "this_family_contains"))
}

family_contained(m)
# this_family_is_contained this_family_contains
# 1           B               A
# 2           F               E

# finally you can filter them with
filter(df, !(famid %in% family_contained(m)$this_family_is_contained))