查找列表元素交集的长度答案

【问题标题】：Find length of intersection of list elements查找列表元素交集的长度
【发布时间】：2015-04-02 20:22:02
【问题描述】：

是否有更快（易于扩展）和更简洁的方法从 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)

【问题讨论】：

标签： r list intersection

【解决方案1】：

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

【讨论】：

【解决方案2】：

这更干净，但不是更快：

sapply(x, function(a) {
  sapply(x, function(b) length(intersect(a,b)))
})

时间安排：

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

【讨论】：

【解决方案3】：

我会使用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

【讨论】：