rxk011 - 5 months ago 28

R Question

Suppose

`v=c(-1,-1,-1,0,0,0,1,1,1)`

`9!/(3!3!3!)=1680`

Currently, I'm using the following line to get that, but it is not very efficient.

`unique(replicate(100000, sample(v), simplify=FALSE))`

This is one possible case. I want to generalize this. Suppose

`n`

`k`

`a1,a2,...,ak`

`n1,n2,...,nk (n1+n2+...+nk=n)`

`n!/(n1!n2!...nk!)`

How do I do this efficiently?

Answer

One approach is to use a permutations function to generate all possibilities then filter. If you need a guarantee of all possibilities this has advantages over the sample method in your question however it can 'crash' if you have a long vector.

```
v=c(-1,-1,-1,0,0,0,1,1,1)
unique(combinat::permn(v))
```

This method is much less quick than @sirallen's approach. A generalization of that approach might be:

```
u <- unique(v)
Pv <- expand.grid(rep(list(u),length(v)))
for(i in seq_along(u)) {
Pv <- Pv[rowSums(Pv==u[i])==sum(u[i]==v),]
}
dim(Pv)
```