是否有更快的（容易擴展的）和更簡潔的方式從x中獲取y？查找列表元素交集的長度
x <- list(c("a", "b", "c", "d"),
c("a", "b", "e"),
c("x", "y"),
c("z", "x"))
y <- vector(mode = "list", length = length(x))
for(i in 1:length(x)){
for(j in 1:length(x)){
y[[i]] <- append(y[[i]], length(intersect(x[[i]], x[[j]])))}}
y <- do.call(rbind, y)
un <- unique(unlist(x))
crossprod(sapply(x,function(y)un%in%y))
# [,1] [,2] [,3] [,4]
# [1,] 4 2 0 0
# [2,] 2 3 0 0
# [3,] 0 0 2 1
# [4,] 0 0 1 2
microbenchmark::microbenchmark(user1389960(), times = 1000)
# Unit: microseconds
# expr min lq mean median uq max neval
# user1389960() 172.631 181.5195 243.918 187.1015 198.716 45083.95 1000
microbenchmark::microbenchmark(eipi10(), times = 1000)
# Unit: microseconds
# expr min lq mean median uq max neval
# eipi10() 218.625 225.9635 246.9797 234.469 245.4545 1175.439 1000
microbenchmark::microbenchmark(Julius(), times = 1000)
# Unit: microseconds
# expr min lq mean median uq max neval
# Julius() 30.322 32.511 37.61541 34.0175 37.957 1026.268 1000
microbenchmark::microbenchmark(ColonelBeauvel(), times = 1000)
# Unit: microseconds
# expr min lq mean median uq max neval
# ColonelBeauvel() 162.103 169.548 183.9076 175.683 183.677 1052.435 1000
這是清潔的，但不是更快：
sapply(x, function(a) {
sapply(x, function(b) length(intersect(a,b)))
})
時序：
microbenchmark::microbenchmark(sapply(x, function(a) {
sapply(x, function(b) length(intersect(a,b)))}))
min lq mean median uq max neval
377.513 392.2505 406.1243 404.318 416.22 511.877 100
microbenchmark::microbenchmark(for(i in 1:length(x)){
for(j in 1:length(x)){
y[[i]] <- append(y[[i]], length(intersect(x[[i]], x[[j]])))}})
min lq mean median uq max neval
350.471 375.7305 422.0248 388.695 414.41 2386.736 100
我會用mapply：
n = length(x)
matrix(mapply(function(u,v) length(intersect(u,v)), rep(x, n), rep(x, each=n)), ncol=n)
# [,1] [,2] [,3] [,4]
#[1,] 4 2 0 0
#[2,] 2 3 0 0
#[3,] 0 0 2 1
#[4,] 0 0 1 2