Create combinations of a binary vector

Question

I would like to create all possible combinations of a binary vector made of a fixed number of 0 and 1. For example: dim(v)=5x1; n1=3; n0=2; In this case I'd like to have something like:

  1,1,1,0,0
  1,1,0,1,0
  1,1,0,0,1
  1,0,1,1,0
  1,0,1,0,1
  1,0,0,1,1
  0,1,1,1,0
  0,1,1,0,1
  0,1,0,1,1
  0,0,1,1,1

I found some help reading this post Create all possible combiations of 0,1, or 2 "1"s of a binary vector of length n but i would like to generate only the combinations I need avoiding any waste of space (I think that the problem will increase explonentially with n)

Answer 1

A slightly faster version of Marat's answer:

f.roland <- function(n, m) {
  ind <- combn(seq_len(n), m)
  ind <- t(ind) + (seq_len(ncol(ind)) - 1) * n
  res <- rep(0, nrow(ind) * n)
  res[ind] <- 1
  matrix(res, ncol = n, nrow = nrow(ind), byrow = TRUE)
}

all.equal(f.2(16, 8), f.roland(16, 8))
#[1] TRUE
library(rbenchmark)
benchmark(f(16,8),f.2(16,8),f.roland(16,8))

#             test replications elapsed relative user.self sys.self user.child sys.child
#2      f.2(16, 8)          100   5.693    1.931     5.670    0.020          0         0
#3 f.roland(16, 8)          100   2.948    1.000     2.929    0.017          0         0
#1        f(16, 8)          100   8.287    2.811     8.214    0.066          0         0

Answer 2

You can try this approach:

f <- function(n=5,m=3)
 t(apply(combn(1:n,m=m),2,function(cm) replace(rep(0,n),cm,1)))

f(5,3)
#       [,1] [,2] [,3] [,4] [,5]
#  [1,]    1    1    1    0    0
#  [2,]    1    1    0    1    0
#  [3,]    1    1    0    0    1
#  [4,]    1    0    1    1    0
#  [5,]    1    0    1    0    1
#  [6,]    1    0    0    1    1
#  [7,]    0    1    1    1    0
#  [8,]    0    1    1    0    1
#  [9,]    0    1    0    1    1
# [10,]    0    0    1    1    1

The idea is to generate all combinations of indices for 1, and then to use them to produce the final result.

Another flavor of the same approach:

f.2 <- function(n=5,m=3)
  t(combn(1:n,m,FUN=function(cm) replace(rep(0,n),cm,1)))

The second approach is about twice faster:

library(rbenchmark)
benchmark(f(16,8),f.2(16,8))
#         test replications elapsed relative user.self sys.self user.child sys.child
# 2 f.2(16, 8)          100   5.706    1.000     5.688    0.017          0         0
# 1   f(16, 8)          100  10.802    1.893    10.715    0.082          0         0

Benchmark

f.akrun <- function(n=5,m=3) {

  indx <- combnPrim(1:n,m)

  DT <- setDT(as.data.frame(matrix(0, ncol(indx),n)))
  for(i in seq_len(nrow(DT))){
    set(DT, i=i, j=indx[,i],value=1) 
  }
  DT  
}

benchmark(f(16,8),f.2(16,8),f.akrun(16,8))
#            test replications elapsed relative user.self sys.self user.child sys.child
# 2     f.2(16, 8)          100   5.464    1.097     5.435    0.028          0         0
# 3 f.akrun(16, 8)          100   4.979    1.000     4.938    0.037          0         0
# 1       f(16, 8)          100  10.854    2.180    10.689    0.129          0         0

@akrun's solution (f.akrun) is ~10% faster than f.2.

[EDIT] Another approach, which is even more faster and simple:

f.3 <- function(n=5,m=3) t(combn(n,m,tabulate,nbins=n))

Answer 3

Here is another approach:

func <- function(n, m) t(combn(n, m, function(a) {z=integer(n);z[a]=1;z}))

func(n = 5, m = 2)

     # [,1] [,2] [,3] [,4] [,5]
 # [1,]    1    1    0    0    0
 # [2,]    1    0    1    0    0
 # [3,]    1    0    0    1    0
 # [4,]    1    0    0    0    1
 # [5,]    0    1    1    0    0
 # [6,]    0    1    0    1    0
 # [7,]    0    1    0    0    1
 # [8,]    0    0    1    1    0
 # [9,]    0    0    1    0    1
# [10,]    0    0    0    1    1