Permute all unique enumerations of a vector in R
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
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.
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!
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.