再帰的ロジックに従うことで、すべての組み合わせを繰り返しなしで計算でき、最初にすべてを計算する必要もありません。choose(nx-1,ning-1) が整数を返す限り、かなりうまく機能します。そうでない場合、可能性を計算するのは少しばかげています。
これは再帰的なプロセスであるため、時間がかかる可能性があり、ベクトルが特定の制限を超えるとメモリの問題が発生します。しかし、繰り返しますが、14 要素のセットを 7 つのグループに分割すると、すでに 135135 のユニークな可能性が得られます。この種のことでは、物事はすぐに手に負えなくなります。
擬似的なロジック (擬似コードとは呼びません)
nb = number of groups
ning = number of elements in every group
if(nb == 2)
1. take first element, and add it to every possible
combination of ning-1 elements of x[-1]
2. make the difference for each group defined in step 1 and x
to get the related second group
3. combine the groups from step 2 with the related groups from step 1
if(nb > 2)
1. take first element, and add it to every possible
combination of ning-1 elements of x[-1]
2. to define the other groups belonging to the first groups obtained like this,
apply the algorithm on the other elements of x, but for nb-1 groups
3. combine all possible other groups from step 2
with the related first groups from step 1
これを R に翻訳すると、次のようになります。
perm.groups <- function(x,n){
nx <- length(x)
ning <- nx/n
group1 <-
rbind(
matrix(rep(x[1],choose(nx-1,ning-1)),nrow=1),
combn(x[-1],ning-1)
)
ng <- ncol(group1)
if(n > 2){
out <- vector('list',ng)
for(i in seq_len(ng)){
other <- perm.groups(setdiff(x,group1[,i]),n=n-1)
out[[i]] <- lapply(seq_along(other),
function(j) cbind(group1[,i],other[[j]])
)
}
out <- unlist(out,recursive=FALSE)
} else {
other <- lapply(seq_len(ng),function(i)
matrix(setdiff(x,group1[,i]),ncol=1)
)
out <- lapply(seq_len(ng),
function(i) cbind(group1[,i],other[[i]])
)
}
out
}
それが機能することを示すには:
> perm.groups(1:6,3)
[[1]]
[,1] [,2] [,3]
[1,] 1 3 5
[2,] 2 4 6
[[2]]
[,1] [,2] [,3]
[1,] 1 3 4
[2,] 2 5 6
[[3]]
[,1] [,2] [,3]
[1,] 1 3 4
[2,] 2 6 5
[[4]]
[,1] [,2] [,3]
[1,] 1 2 5
[2,] 3 4 6
[[5]]
[,1] [,2] [,3]
[1,] 1 2 4
[2,] 3 5 6
[[6]]
[,1] [,2] [,3]
[1,] 1 2 4
[2,] 3 6 5
[[7]]
[,1] [,2] [,3]
[1,] 1 2 5
[2,] 4 3 6
[[8]]
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 4 5 6
[[9]]
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 4 6 5
[[10]]
[,1] [,2] [,3]
[1,] 1 2 4
[2,] 5 3 6
[[11]]
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 5 4 6
[[12]]
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 5 6 4
[[13]]
[,1] [,2] [,3]
[1,] 1 2 4
[2,] 6 3 5
[[14]]
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 6 4 5
[[15]]
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 6 5 4