Identify which objects of list are contained (subset of) in another list in R

问题

Thank you for your kind reply to my previous questions. I have two lists: list1 and list2. I would like to know if each object of list1 is contained in each object of list2. For example:

> list1
[[1]]
[1] 1

[[2]]
[1] 2

[[3]]
[1] 3

> list2
[[1]]
[1] 1 2 3

[[2]]
[1] 2 3

[[3]]
[1] 2 3

Here are my questions: 1.) How do you I ask R to check if an object is a subset of another object in a list? For instance I would like to check if list2[[3]]={2,3} is contained in (subset of) list1[[2]]={2}. When I do list2[[3]] %in% list1[[2]], I get [1] TRUE FALSE. However, this is not what I desire to do?! I just want to check if list2[[3]] is a subset of list1[[2]], i.e. is {2,3} \subset of {3} as in the set theoretic notion? I do not want to perform elementwise check as R seems to be doing with the %in% command. Any suggestions?

2.) Is there some sort of way to efficiently make all pairwise subset comparisons (i.e. list1[[i]] subset of list2[[j]], for all i,j combinations? Would something like outer(list1,list2, func.subset) work once question number 1 is answered? Thank you for your feedback!

回答1:

setdiff compares unique values

length(setdiff(5, 1:5)) == 0

Alternatively, all(x %in% y) will work nicely.

To do all comparisons, something like this would work:

dt <- expand.grid(list1,list2)
dt$subset <- apply(dt,1, function(.v) all(.v[[1]] %in% .v[[2]]) )


  Var1    Var2 subset
1    1 1, 2, 3   TRUE
2    2 1, 2, 3   TRUE
3    3 1, 2, 3   TRUE
4    1    2, 3  FALSE
5    2    2, 3   TRUE
6    3    2, 3   TRUE
7    1    2, 3  FALSE
8    2    2, 3   TRUE
9    3    2, 3   TRUE

Note that the expand.grid isn't the fastest way to do this when dealing with a lot of data (dwin's solution is better in that regard) but it allows you to quickly check visually whether this is doing what you want.

回答2:

You can use the sets package as follows:

library(sets)
is.subset <- function(x, y) as.set(x) <= as.set(y)

outer(list1, list2, Vectorize(is.subset))
#      [,1]  [,2]  [,3]
# [1,] TRUE FALSE FALSE
# [2,] TRUE  TRUE  TRUE
# [3,] TRUE  TRUE  TRUE

@Michael or @DWin's base version of is.subset will work just as well, but for part two of your question, I'd maintain that outer is the way to go.

回答3:

is.subset <- function(x,y) {length(setdiff(x,y)) == 0}

First the combos of list1 elements that are subsets of list2 items:

> sapply(1:length(list1), function(i1) sapply(1:length(list2), 
                 function(i2) is.subset(list1[[i1]], list2[[i2]]) ) )
      [,1] [,2] [,3]
[1,]  TRUE TRUE TRUE
[2,] FALSE TRUE TRUE
[3,] FALSE TRUE TRUE

Then the unsurprising lack of any of the list2 items (all of length > 1) that are subsets of list one items (all of length 1):

> sapply(1:length(list1), function(i1) sapply(1:length(list2), 
                 function(i2) is.subset(list2[[i2]], list1[[i1]]) ) )
      [,1]  [,2]  [,3]
[1,] FALSE FALSE FALSE
[2,] FALSE FALSE FALSE
[3,] FALSE FALSE FALSE

回答4:

adding to @Michael's, here's a neat way to avoid the messiness of expand.grid using the AsIs function:

list2 <- list(1:3,2:3,2:3)
a <- data.frame(list1 = 1:3, I(list2))
a$subset <- apply(a, 1, function(.v) all(.v[[1]] %in% .v[[2]]) )

  list1   list2 subset
1     1 1, 2, 3   TRUE
2     2    2, 3   TRUE
3     3    2, 3   TRUE

来源：https://stackoverflow.com/questions/14410050/identify-which-objects-of-list-are-contained-subset-of-in-another-list-in-r

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