Permute all unique enumerations of a vector in R

Question

I'm trying to find a function that will permute all the unique permutations of a vector, while not counting juxtapositions within subsets of the same element type. For example:

dat <- c(1,0,3,4,1,0,0,3,0,4)

has

factorial(10)
> 3628800

possible permutations, but only 10!/(2!*2!*4!*2!)

factorial(10)/(factorial(2)*factorial(2)*factorial(2)*factorial(4))
> 18900

unique permutations when ignoring juxtapositions within subsets of the same element type.

I can get this by using unique() and the permn() function from the package combinat

unique( permn(dat) )

but this is computationally very expensive, since it involves enumerating n!, which can be an order of magnitude more permutations than I need. Is there a way to do this without first computing n!?

Answer 1

EDIT: Here's a faster answer; again based on the ideas of Louisa Grey and Bryce Wagner, but with faster R code thanks to better use of matrix indexing. It's quite a bit faster than my original:

> ddd <- c(1,0,3,4,1,0,0,3,0,4)
> system.time(up1 <- uniqueperm(d))
   user  system elapsed 
  0.183   0.000   0.186 
> system.time(up2 <- uniqueperm2(d))
   user  system elapsed 
  0.037   0.000   0.038 

And the code:

uniqueperm2 <- function(d) {
  dat <- factor(d)
  N <- length(dat)
  n <- tabulate(dat)
  ng <- length(n)
  if(ng==1) return(d)
  a <- N-c(0,cumsum(n))[-(ng+1)]
  foo <- lapply(1:ng, function(i) matrix(combn(a[i],n[i]),nrow=n[i]))
  out <- matrix(NA, nrow=N, ncol=prod(sapply(foo, ncol)))
  xxx <- c(0,cumsum(sapply(foo, nrow)))
  xxx <- cbind(xxx[-length(xxx)]+1, xxx[-1])
  miss <- matrix(1:N,ncol=1)
  for(i in seq_len(length(foo)-1)) {
    l1 <- foo[[i]]
    nn <- ncol(miss)
    miss <- matrix(rep(miss, ncol(l1)), nrow=nrow(miss))
    k <- (rep(0:(ncol(miss)-1), each=nrow(l1)))*nrow(miss) + 
               l1[,rep(1:ncol(l1), each=nn)]
    out[xxx[i,1]:xxx[i,2],] <- matrix(miss[k], ncol=ncol(miss))
    miss <- matrix(miss[-k], ncol=ncol(miss))
  }
  k <- length(foo)
  out[xxx[k,1]:xxx[k,2],] <- miss
  out <- out[rank(as.numeric(dat), ties="first"),]
  foo <- cbind(as.vector(out), as.vector(col(out)))
  out[foo] <- d
  t(out)
}

It doesn't return the same order, but after sorting, the results are identical.

up1a <- up1[do.call(order, as.data.frame(up1)),]
up2a <- up2[do.call(order, as.data.frame(up2)),]
identical(up1a, up2a)

For my first attempt, see the edit history.

Answer 2

The following function (which implements the classic formula for repeated permutations just like you did manually in your question) seems quite fast to me:

upermn <- function(x) {
    n <- length(x)
    duplicates <- as.numeric(table(x))
    factorial(n) / prod(factorial(duplicates))
}

It does compute n! but not like permn function which generates all permutations first.

See it in action:

> dat <- c(1,0,3,4,1,0,0,3,0,4)
> upermn(dat)
[1] 18900
> system.time(uperm(dat))
   user  system elapsed 
  0.000   0.000   0.001 

---

UPDATE: I have just realized that the question was about generating all unique permutations not just specifying the number of them - sorry for that!

You could improve the unique(perm(...)) part with specifying unique permutations for one less element and later adding the uniqe elements in front of them. Well, my explanation may fail, so let the source speak:

uperm <- function(x) {
u <- unique(x)                    # unique values of the vector
result <- x                       # let's start the result matrix with the vector
for (i in 1:length(u)) {
    v <- x[-which(x==u[i])[1]]    # leave the first occurance of duplicated values
    result <- rbind(result, cbind(u[i], do.call(rbind, unique(permn(v)))))
}
return(result)
}

This way you could gain some speed. I was lazy to run the code on the vector you provided (took so much time), here is a small comparison on a smaller vector:

> dat <- c(1,0,3,4,1,0,0)
> system.time(unique(permn(dat)))
   user  system elapsed 
  0.264   0.000   0.268 
> system.time(uperm(dat))
   user  system elapsed 
  0.147   0.000   0.150 

I think you could gain a lot more by rewriting this function to be recursive!

---

UPDATE (again): I have tried to make up a recursive function with my limited knowledge:

uperm <- function(x) {
    u <- sort(unique(x))
    l <- length(u)
    if (l == length(x)) {
        return(do.call(rbind,permn(x)))
    }
    if (l == 1) return(x)
    result <- matrix(NA, upermn(x), length(x))
    index <- 1
    for (i in 1:l) {
        v <- x[-which(x==u[i])[1]]
        newindex <- upermn(v)
        if (table(x)[i] == 1) {
            result[index:(index+newindex-1),] <- cbind(u[i], do.call(rbind, unique(permn(v))))
            } else {
                result[index:(index+newindex-1),] <- cbind(u[i], uperm(v))
            }
        index <- index+newindex
    }
    return(result)
}

Which has a great gain:

> system.time(unique(permn(c(1,0,3,4,1,0,0,3,0))))
   user  system elapsed 
 22.808   0.103  23.241 

> system.time(uperm(c(1,0,3,4,1,0,0,3,0)))
   user  system elapsed 
  4.613   0.003   4.645 

Please report back if this would work for you!

Answer 3

One option that hasn't been mentioned here is the allPerm function from the multicool package. It can be used pretty easily to get all the unique permutations:

library(multicool)
perms <- allPerm(initMC(dat))
dim(perms)
# [1] 18900    10
head(perms)
#      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
# [1,]    4    4    3    3    1    1    0    0    0     0
# [2,]    0    4    4    3    3    1    1    0    0     0
# [3,]    4    0    4    3    3    1    1    0    0     0
# [4,]    4    4    0    3    3    1    1    0    0     0
# [5,]    3    4    4    0    3    1    1    0    0     0
# [6,]    4    3    4    0    3    1    1    0    0     0

In benchmarking I found it to be faster on dat than the solutions from the OP and daroczig but slower than the solution from Aaron.

Answer 4

I don't actually know R, but here's how I'd approach the problem:

Find how many of each element type, i.e.

4 X 0
2 X 1
2 X 3
2 X 4

Sort by frequency (which the above already is).

Start with the most frequent value, which takes up 4 of the 10 spots. Determine the unique combinations of 4 values within the 10 available spots. (0,1,2,3),(0,1,2,4),(0,1,2,5),(0,1,2,6) ... (0,1,2,9),(0,1,3,4),(0,1,3,5) ... (6,7,8,9)

Go to the second most frequent value, it takes up 2 of 6 available spots, and determine it's unique combinations of 2 of 6. (0,1),(0,2),(0,3),(0,4),(0,5),(1,2),(1,3) ... (4,6),(5,6)

Then 2 of 4: (0,1),(0,2),(0,3),(1,2),(1,3),(2,3)

And the remaining values, 2 of 2: (0,1)

Then you need to combine them into each possible combination. Here's some pseudocode (I'm convinced there's a more efficient algorithm for this, but this shouldn't be too bad):

lookup = (0,1,3,4)
For each of the above sets of combinations, example: input = ((0,2,4,6),(0,2),(2,3),(0,1))
newPermutation = (-1,-1,-1,-1,-1,-1,-1,-1,-1,-1)
for i = 0 to 3
  index = 0
  for j = 0 to 9
    if newPermutation(j) = -1
      if index = input(i)(j)
        newPermutation(j) = lookup(i)
        break
      else
        index = index + 1