CODEWITHSUNDEEP
ralgorithmvectorpermutation
holic
holic

Reputation: 13

unique permutations of zero-one-vector in R

My problem is as follows:

Imagine we have a vector (1,1,1,...,0,0) of length n with k ones in the beginning. Think of this vector as of vector with realizations of some variables L1 till Ln. What i need to calculate is

sum over all unique permutations of (1,1,1,...,0,0) of Function(L1,...,Ln)

I have searched for solutions of my problem and yes, there are some, which work as long as n isn't too big.

As long as n is under 30 my PC doesn't die and following idea works:

1) creating a data.frame of all unique permutations with a help of following code (found it here)

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)
}

2) sum over components of this data.frame

Sadly in my problems n is 100 and above. Good news for me are that i actually do not need whole data.frame in my RAM. An algorithm which would remember last permutation, use it to evaluate Funktion(L1,...,Ln)and compute next permutation and so on in a loop would be enough. Any help is appreciated.

EDIT Hack-R asked for an example, here what i get

    > d <- c()
    > d[1:25]=0
    > d[25:50]=1
    > d
     [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
> uniqueperm2(d)
Error: cannot allocate vector of size 905608.1 Gb
In addition: Warning messages:
1: In vector("list", count) :
  Reached total allocation of 8109Mb: see help(memory.size)
2: In vector("list", count) :
  Reached total allocation of 8109Mb: see help(memory.size)
3: In vector("list", count) :
  Reached total allocation of 8109Mb: see help(memory.size)
4: In vector("list", count) :
  Reached total allocation of 8109Mb: see help(memory.size)

Upvotes: 1

Views: 412

Answers (2)

Randy Lai
Randy Lai

Reputation: 3174

iterpc is another solution

k <- 5
n <- 10
library(iterpc)
it <- iterpc(c(k, n-k), ordered=TRUE)
while (!is.null(x <- getnext(it))){
  print(x)
}

PS: The default labels are 1 and 2 instead of 0 and 1.

Simple benchmark shows that iterpc is at least 2x faster than next_x when n=10, k=5

Unit: milliseconds
   expr       min        lq      mean    median        uq       max neval
 next_x 11.663353 12.599623 13.665913 13.532414 14.411556 17.619208   100
 iterpc  4.987268  5.325663  5.939558  5.613265  6.572008  8.685916   100

Upvotes: 0

MrFlick
MrFlick

Reputation: 206197

Here's one way to walk the permutations. I still think there is a better way but haven't figured it out yet.

This function looks at an array of 1's an 0's and tries to move the right most 1 to the left if possible. (Basically thinking of the vector as a binary number and trying to find the next largest number with exactly n bits)

next_x <- function(x) {
    i <- tail(which(diff(x)==1),1)
    if (length(i)>0) {
        x[c(i, i+1)]<-c(1,0)
        x[(i+1):length(x)] <- sort(x[(i+1):length(x)])
    } else {
        stop("no more moves")
    }
    x
}

You start out with x all to the right and you can iterate with

x <- c(0,0,0,0,1,1,1)
while(!all(x==c(1,1,1,0,0,0,0))) {
    x <- next_x(x)
    print(x)
}

Upvotes: 1

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