Intersect all possible combinations of list elements - r

Question

I have a list of vectors:
> l <- list(A=c("one", "two", "three", "four"), B=c("one", "two"), C=c("two", "four", "five", "six"), D=c("six", "seven"))
> l
$A
[1] "one" "two" "three" "four"
$B
[1] "one" "two"
$C
[1] "two" "four" "five" "six"
$D
[1] "six" "seven"
I would like to calculate the length of the overlap between all possible pairwise combinations of the list elements, i.e. (the format of the result doesn't matter):
AintB 2
AintC 2
AintD 0
BintC 1
BintD 0
CintD 1
I know combn(x, 2) can be used to get a matrix of all possible pairwise combinations in a vector and that length(intersect(a, b)) would give me the length of the overlap of two vectors, but I can't think of a way to put the two things together.
Any help is much appreciated! Thanks.

Answer 1

If I understand correctly, you can look at crossprod and stack:
crossprod(table(stack(l)))
# ind
# ind A B C D
# A 4 2 2 0
# B 2 2 1 0
# C 2 1 4 1
# D 0 0 1 2
You can extend the idea if you want a data.frame of just the relevant values as follows:
Write a spiffy function
listIntersect <- function(inList) {
X <- crossprod(table(stack(inList)))
X[lower.tri(X)] <- NA
diag(X) <- NA
out <- na.omit(data.frame(as.table(X)))
out[order(out$ind), ]
}
Apply it
listIntersect(l)
# ind ind.1 Freq
# 5 A B 2
# 9 A C 2
# 13 A D 0
# 10 B C 1
# 14 B D 0
# 15 C D 1
Performance seems pretty decent.
Expand the list:
L <- unlist(replicate(100, l, FALSE), recursive=FALSE)
names(L) <- make.unique(names(L))
Set up some functions to test:
fun1 <- function(l) listIntersect(l)
fun2 <- function(l) apply( combn( l , 2 ) , 2 , function(x) length( intersect( unlist( x[1]) , unlist(x[2]) ) ) )
fun3 <- function(l) {
m1 <- combn(names(l),2)
val <- sapply(split(m1, col(m1)),function(x) {x1 <- l[[x[1]]]; x2 <- l[[x[2]]]; length(intersect(x1, x2))})
Ind <- apply(m1,2,paste,collapse="int")
data.frame(Ind, val, stringsAsFactors=F)
}
Check out the timings:
system.time(F1 <- fun1(L))
# user system elapsed
# 0.33 0.00 0.33
system.time(F2 <- fun2(L))
# user system elapsed
# 4.32 0.00 4.31
system.time(F3 <- fun3(L))
# user system elapsed
# 6.33 0.00 6.33
Everyone seems to be sorting the result differently, but the numbers match:
table(F1$Freq)
#
# 0 1 2 4
# 20000 20000 29900 9900
table(F2)
# F2
# 0 1 2 4
# 20000 20000 29900 9900
table(F3$val)
#
# 0 1 2 4
# 20000 20000 29900 9900

Answer 2

combn works with list structures as well, you just need a little unlist'ing of the result to use intersect...
# Get the combinations of names of list elements
nms <- combn( names(l) , 2 , FUN = paste0 , collapse = "" , simplify = FALSE )
# Make the combinations of list elements
ll <- combn( l , 2 , simplify = FALSE )
# Intersect the list elements
out <- lapply( ll , function(x) length( intersect( x[[1]] , x[[2]] ) ) )
# Output with names
setNames( out , nms )
#$AB
#[1] 2
#$AC
#[1] 2
#$AD
#[1] 0
#$BC
#[1] 1
#$BD
#[1] 0
#$CD
#[1] 1

Answer 3

Try:
m1 <- combn(names(l),2)
val <- sapply(split(m1, col(m1)),function(x) {x1 <- l[[x[1]]]; x2 <- l[[x[2]]]; length(intersect(x1, x2))})
Ind <- apply(m1,2,paste,collapse="int")
data.frame(Ind, val, stringsAsFactors=F)
# Ind val
# 1 AntB 2
# 2 AntC 2
# 3 AntD 0
# 4 BntC 1
# 5 BntD 0
# 6 CntD 1