luc*_*ano 14 r combinatorics combn
感觉这应该很简单，但我已经经历过堆栈溢出和combn帮助，但看不到解决方案。
下面的玩家需要组成 3 对 2 的队伍。我需要找到所有可能的队伍组合。例如，两个可能的团队是“Ross”、“Bobby”和“Casper”在一个团队中，“Max”和“Jake”在另一团队中。我该如何编码？
players <- c("Ross", "Bobby", "Max", "Casper", "Jake")
Tho*_*ing 18
我认为关键的一步是从5名球员中随机选择3（或2）名产生第一队，剩下的自然分配到第二队。
也许你可以这样尝试
combn(players,
3,
function(x) list(team1 = x, team2 = players[!players %in% x]),
simplify = FALSE
)
或者如果你不介意时间效率，你可以setdiff使用
combn(players,
3,
function(x) list(team1 = x, team2 = setdiff(players, x)),
simplify = FALSE
)
并且其中任何一个都给出
[[1]]
[[1]]$team1
[1] "Ross" "Bobby" "Max"
[[1]]$team2
[1] "Casper" "Jake"
[[2]]
[[2]]$team1
[1] "Ross" "Bobby" "Casper"
[[2]]$team2
[1] "Max" "Jake"
[[3]]
[[3]]$team1
[1] "Ross" "Bobby" "Jake"
[[3]]$team2
[1] "Max" "Casper"
[[4]]
[[4]]$team1
[1] "Ross" "Max" "Casper"
[[4]]$team2
[1] "Bobby" "Jake"
[[5]]
[[5]]$team1
[1] "Ross" "Max" "Jake"
[[5]]$team2
[1] "Bobby" "Casper"
[[6]]
[[6]]$team1
[1] "Ross" "Casper" "Jake"
[[6]]$team2
[1] "Bobby" "Max"
[[7]]
[[7]]$team1
[1] "Bobby" "Max" "Casper"
[[7]]$team2
[1] "Ross" "Jake"
[[8]]
[[8]]$team1
[1] "Bobby" "Max" "Jake"
[[8]]$team2
[1] "Ross" "Casper"
[[9]]
[[9]]$team1
[1] "Bobby" "Casper" "Jake"
[[9]]$team2
[1] "Ross" "Max"
[[10]]
[[10]]$team1
[1] "Max" "Casper" "Jake"
[[10]]$team2
[1] "Ross" "Bobby"
players <- c("Ross", "Bobby", "Max", "Casper", "Jake")
microbenchmark(
f1 = combn(players,
3,
function(x) list(team1 = x, team2 = players[!players %in% x]),
simplify = FALSE
),
f2 = combn(players,
3,
function(x) list(team1 = x, team2 = setdiff(players, x)),
simplify = FALSE
)
)
我们将会看到
Unit: microseconds
expr min lq mean median uq max neval
f1 26.200 28.0510 51.55586 29.4515 32.5015 1935.301 100
f2 97.301 99.8505 119.44610 103.6010 111.1510 1162.501 100
RcppAlgos（我是作者）中有一个函数comboGroups专门为此任务而构建。从版本开始2.8.0，我们现在可以处理不同大小的组。
library(RcppAlgos)\npackageVersion("RcppAlgos")\n#> [1] \'2.8.0\'\n我们使用以下参数指定每个团队的规模grpSizes：
players <- c("Ross", "Bobby", "Max", "Casper", "Jake")\ncomboGroups(players, grpSizes = c(2, 3))\n#> Grp1 Grp1 Grp2 Grp2 Grp2 \n#> [1,] "Ross" "Bobby" "Max" "Casper" "Jake" \n#> [2,] "Ross" "Max" "Bobby" "Casper" "Jake" \n#> [3,] "Ross" "Casper" "Bobby" "Max" "Jake" \n#> [4,] "Ross" "Jake" "Bobby" "Max" "Casper"\n#> [5,] "Bobby" "Max" "Ross" "Casper" "Jake" \n#> [6,] "Bobby" "Casper" "Ross" "Max" "Jake" \n#> [7,] "Bobby" "Jake" "Ross" "Max" "Casper"\n#> [8,] "Max" "Casper" "Ross" "Bobby" "Jake" \n#> [9,] "Max" "Jake" "Ross" "Bobby" "Casper"\n#> [10,] "Casper" "Jake" "Ross" "Bobby" "Max"\n它非常高效且非常灵活。让\xe2\x80\x99s 对更多的玩家进行测试。
\nlibrary(microbenchmark)\nmore_players <- c(players, "Kai", "Eliana", "Jayden", "Luca",\n "Rowan", "Nova", "Amara", "Finn", "Zion", "Mia")\n\nmicrobenchmark(\n f1 = combn(more_players,\n 7,\n function(x) list(team1 = x, team2 = more_players[!more_players %in% x]),\n simplify = FALSE\n ),\n f2 = combn(more_players,\n 7,\n function(x) list(team1 = x, team2 = setdiff(more_players, x)),\n simplify = FALSE\n ),\n f3_rcpp = comboGeneral(\n v = more_players,\n m = 7,\n repetition = FALSE,\n FUN = function(x) list(team1 = x, team2 = more_players[!more_players %in% x])\n ),\n f4 = comboGroups(more_players, grpSizes = c(7, 8)),\n unit = "relative"\n)\n#> Unit: relative\n#> expr min lq mean median uq max neval cld\n#> f1 16.22184 19.03265 23.68141 20.78979 26.59770 92.08256 100 a \n#> f2 41.34947 45.55672 57.30847 54.70320 59.88822 123.09864 100 b \n#> f3_rcpp 11.83424 14.48575 17.65936 16.24856 21.43574 36.27697 100 c \n#> f4 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 100 d\n更奇特的群体又如何呢？对于大多数其他方法，效率和代码维护将是一个问题。
\n举例来说，给定more_players（总共 15 名玩家），如果我们想要找到包含 2 支规模为 3 的球队、一支规模为 4 的球队和一支规模为 5 的球队的所有可能分组，该怎么办？
有了comboGroups，就没有问题了：
system.time(t <- comboGroups(more_players, grpSizes = c(3, 3, 4, 5)))\n#> user system elapsed \n#> 0.508 0.040 0.548 \n\nhead(t, n = 2)\n#> Grp1 Grp1 Grp1 Grp2 Grp2 Grp2 Grp3 Grp3 Grp3 Grp3 \n#> [1,] "Ross" "Bobby" "Max" "Casper" "Jake" "Kai" "Eliana" "Jayden" "Luca" "Rowan"\n#> [2,] "Ross" "Bobby" "Max" "Casper" "Jake" "Kai" "Eliana" "Jayden" "Luca" "Nova" \n#> Grp4 Grp4 Grp4 Grp4 Grp4 \n#> [1,] "Nova" "Amara" "Finn" "Zion" "Mia"\n#> [2,] "Rowan" "Amara" "Finn" "Zion" "Mia"\n\ntail(t, n = 2)\n#> Grp1 Grp1 Grp1 Grp2 Grp2 Grp2 Grp3 Grp3 Grp3 \n#> [6306299,] "Rowan" "Zion" "Mia" "Nova" "Amara" "Finn" "Jake" "Eliana" "Jayden"\n#> [6306300,] "Rowan" "Zion" "Mia" "Nova" "Amara" "Finn" "Kai" "Eliana" "Jayden"\n#> Grp3 Grp4 Grp4 Grp4 Grp4 Grp4 \n#> [6306299,] "Luca" "Ross" "Bobby" "Max" "Casper" "Kai" \n#> [6306300,] "Luca" "Ross" "Bobby" "Max" "Casper" "Jake"\n如果您只需要可能团队的样本，请尝试comboGroupsSample：
comboGroupsSample(\n more_players, grpSizes = c(3, 3, 4, 5), n = 2,\n seed = 42, namedSample = TRUE\n)\n#> Grp1 Grp1 Grp1 Grp2 Grp2 Grp2 Grp3 Grp3 Grp3 \n#> 3207141 "Bobby" "Jake" "Mia" "Luca" "Amara" "Zion" "Max" "Casper" "Eliana"\n#> 4248729 "Max" "Casper" "Nova" "Rowan" "Amara" "Finn" "Ross" "Bobby" "Kai" \n#> Grp3 Grp4 Grp4 Grp4 Grp4 Grp4 \n#> 3207141 "Rowan" "Ross" "Kai" "Jayden" "Nova" "Finn"\n#> 4248729 "Luca" "Jake" "Eliana" "Jayden" "Zion" "Mia"\n只是为了构建@ThomasIsCoding 很好的答案。通过使用出色的 RcppAlgos 库和函数可以获得进一步的速度comboGeneral：
players <- c("Ross", "Bobby", "Max", "Casper", "Jake")
microbenchmark(
f1 = combn(players,
3,
function(x) list(team1 = x, team2 = players[!players %in% x]),
simplify = FALSE
),
f2 = combn(players,
3,
function(x) list(team1 = x, team2 = setdiff(players, x)),
simplify = FALSE
),
f3_rcpp = RcppAlgos::comboGeneral(
v = players,
m = 3,
repetition = FALSE,
FUN = function(x) list(team1 = x, team2 = players[!players %in% x])
)
)
基准
Unit: microseconds
expr min lq mean median uq max neval
f1 34.3 37.95 63.693 39.00 41.45 2016.7 100
f2 143.5 152.40 184.351 155.30 158.95 1961.7 100
f3_rcpp 29.3 32.40 61.820 33.95 42.40 2205.0 100