所有可能的组组合

Question

我有5组：G1，G2，…，G5，每组分别有n1，n2，…，n5个元素。我 select 4 组中的每组 2 个元素和第 5 组中的 1 个元素。如何在 R 中生成所有可能的组合？

Answer 1

(问题中没有具体说明分组是否互斥；所以，假设：
1.组是互斥的
2. 组的子集 (n1, n2, ...) 将使用相同的元素进行填充）
3 只是为了参数 |G1|=|G2|=|G3|=5 （用户可以根据组中不同数量的元素相应地更改以下代码）以下是任何用户都可以概括为任意数量的组的问题的 3 组模拟答案。因此，假设组名是 G1、G2、G3。

library(causfinder)
gctemplate(5,2,2) # Elements are coded as: 1,2,3,4,5; |sub-G1|=2; |sub-G2|=2; |sub-G3|=5-(2+2)=1
# In the following table, each number represents a unique element. (SOLUTION ENDED!)

我的包裹 (causfinder) 不在 CRAN 中。因此，我将在下面给出函数gctemplate的代码。

      [,1] [,2] [,3] [,4] [,5]
 [1,]    1    2    3    4    5   sub-G1={1,2}  sub-G2={3,4} sub-G3={5}
 [2,]    1    2    3    5    4
 [3,]    1    2    4    5    3   sub-G1={1,2}  sub-G2={4,5} sub-G3={3}
 [4,]    1    3    2    4    5
 [5,]    1    3    2    5    4
 [6,]    1    3    4    5    2
 [7,]    1    4    2    3    5
 [8,]    1    4    2    5    3
 [9,]    1    4    3    5    2
[10,]    1    5    2    3    4
[11,]    1    5    2    4    3
[12,]    1    5    3    4    2
[13,]    2    3    1    4    5
[14,]    2    3    1    5    4
[15,]    2    3    4    5    1
[16,]    2    4    1    3    5
[17,]    2    4    1    5    3
[18,]    2    4    3    5    1
[19,]    2    5    1    3    4
[20,]    2    5    1    4    3
[21,]    2    5    3    4    1
[22,]    3    4    1    2    5
[23,]    3    4    1    5    2
[24,]    3    4    2    5    1
[25,]    3    5    1    2    4
[26,]    3    5    1    4    2
[27,]    3    5    2    4    1
[28,]    4    5    1    2    3
[29,]    4    5    1    3    2
[30,]    4    5    2    3    1

gctemplate代码：

gctemplate <- function(nvars, ncausers, ndependents){
independents <- combn(nvars, ncausers)
patinajnumber <-  dim(combn(nvars - ncausers, ndependents))[[2]]
independentspatinajednumber <- dim(combn(nvars, ncausers))[[2]]*patinajnumber 
dependents <- matrix(, nrow = dim(combn(nvars, ncausers))[[2]]*patinajnumber, ncol = ndependents)
for (i in as.integer(1:dim(combn(nvars, ncausers))[[2]])){   
dependents[(patinajnumber*(i-1)+1):(patinajnumber*i),] <- t(combn(setdiff(seq(1:nvars), independents[,i]), ndependents))
}
independentspatinajed <- matrix(, nrow = dim(combn(nvars, ncausers))[[2]]*patinajnumber, ncol = ncausers)
for (i in as.integer(1:dim(combn(nvars, ncausers))[[2]])){   
for (j in as.integer(1:patinajnumber)){   
independentspatinajed[(i-1)*patinajnumber+j,] <- independents[,i]  
}}
independentsdependents <- cbind(independentspatinajed, dependents)
others <- matrix(, nrow = dim(combn(nvars, ncausers))[[2]]*patinajnumber, ncol = nvars - ncausers - ndependents) 
for (i in as.integer(1:((dim(combn(nvars, ncausers))[[2]])*patinajnumber))){  
others[i, ]  <- setdiff(seq(1:nvars), independentsdependents[i,]) 
}
causalitiestemplate <- cbind(independentsdependents, others)
causalitiestemplate
}

现在，G1,G2,G3 的解决方案如上。只需将上述代码推广到具有相同逻辑的 5 变量情况即可！

所有可能的组组合

All possible combinations over groups

r

combinatorics