r - 查找列表元素交集的长度

是否有一种更快(易于扩展)和更清洁的方法来从 x 获取 y？

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)

最佳答案

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

关于r - 查找列表元素交集的长度，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/29421774/