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R查找所有可能的唯一组合

combn combinations r

5

Jian · 技术社区 · 6 年前

我正在努力寻找R中所有可能的独特组合。似乎有很多类似的问题被问到,但我找不到相同的问题。

我的问题是从向量x中找出m个元素的组合,但m可能大于x。例如,从字母[1:2]中选取3个元素,希望可以返回:

combn(letters[1:2],3)
       [,1]  [,2]  [,3]  [,4] 
[1,]   "a"   "a"   "a"    "b"
[2,]   "a"   "a"   "b"    "b"
[3,]   "a"   "b"   "b"    "b"

而是combn函数中的错误n<m、还有类似的功能,包括gtools:置换、扩展。网格

如果之前有人问了同样的问题,请再次道歉,但我没听清楚。谢谢

3 回复 | 直到 6 年前

1

7

Joseph Wood 6 年前

有几个专门为此构建的包。基本前提是我们需要长度重复的组合 m 哪里 m级 可能大于输入向量。我们从经典开始 gtools :

library(gtools)
combinations(2, 3, letters[1:2], repeats.allowed = TRUE)
    [,1] [,2] [,3]
[1,] "a"  "a"  "a" 
[2,] "a"  "a"  "b" 
[3,] "a"  "b"  "b" 
[4,] "b"  "b"  "b"

然后就有了 arrangements 它是 iterpc (上述评论中由@Gregor链接的包):

library(arrangements)
arrangements::combinations(2, 3, letters[1:2], replace = TRUE)
     [,1] [,2] [,3]
[1,] "a"  "a"  "a" 
[2,] "a"  "a"  "b" 
[3,] "a"  "b"  "b" 
[4,] "b"  "b"  "b"

最后还有 RcppAlgos ,作者:

library(RcppAlgos)
comboGeneral(letters[1:2], 3, TRUE)
     [,1] [,2] [,3]
[1,] "a"  "a"  "a" 
[2,] "a"  "a"  "b" 
[3,] "a"  "b"  "b" 
[4,] "b"  "b"  "b"

combn 是一个很棒的功能,作为基本软件包之一随附 R 然而,它的缺点之一是不允许重复(这是这里需要的)。我写了一篇非常全面的概述,内容与下面的问题一模一样: A Walk Through a Slice of Combinatorics in R 。

2

1

Onyambu 6 年前

combn1=function(x,m){
  n=ifelse(length(x)==1,ifelse(is.numeric(x),x,1),length(x))
  if(n>=m) return(combn(x,m))
  a=do.call(expand.grid, rep(list(x),m))
  b=t(unique(t(apply(a,1,sort))))
  `dimnames<-`(b,NULL)
}

combn1(letters[1],3)
     [,1]
[1,] "a" 
[2,] "a" 
[3,] "a" 
> combn1(letters[1:2],3)
     [,1] [,2] [,3] [,4]
[1,] "a"  "a"  "a"  "b" 
[2,] "a"  "a"  "b"  "b" 
[3,] "a"  "b"  "b"  "b" 
> combn1(letters[1:3],3)
     [,1]
[1,] "a" 
[2,] "b" 
[3,] "c"

3

0

patL grad student 2 年前

v1 = letters[1:4]

do.call("c",lapply(seq_along(v1),function(i) utils::combn(v1,i,FUN=list)))