How to generate permutations or combinations of object in R?

combinations, multiset, permutation, r, r-faq

Solution

EDIT: I have updated the answer to use a more efficient package `arrangements`

Getting start of using `arrangement`

arrangements contains some efficient generators and iterators for permutations and combinations. It has been demonstrated that `arrangements` outperforms most of the existing packages of similar kind. Some benchmarks could be found here.

Here are the answers to the above questions

# 1) combinations: without replacement: distinct items

combinations(5, 2)

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


# 2) combinations: with replacement: distinct items

combinations(5, 2, replace=TRUE)

      [,1] [,2]
 [1,]    1    1
 [2,]    1    2
 [3,]    1    3
 [4,]    1    4
 [5,]    1    5
 [6,]    2    2
 [7,]    2    3
 [8,]    2    4
 [9,]    2    5
[10,]    3    3
[11,]    3    4
[12,]    3    5
[13,]    4    4
[14,]    4    5
[15,]    5    5



# 3) combinations: without replacement: non distinct items

combinations(x = c("a", "b", "c"), freq = c(2, 1, 1), k = 2)

     [,1] [,2]
[1,] "a"  "a" 
[2,] "a"  "b" 
[3,] "a"  "c" 
[4,] "b"  "c" 



# 4) combinations: with replacement: non distinct items

combinations(x = c("a", "b", "c"), k = 2, replace = TRUE)  # as `freq` does not matter

     [,1] [,2]
[1,] "a"  "a" 
[2,] "a"  "b" 
[3,] "a"  "c" 
[4,] "b"  "b" 
[5,] "b"  "c" 
[6,] "c"  "c" 

# 5) permutations: without replacement: distinct items

permutations(5, 2)

      [,1] [,2]
 [1,]    1    2
 [2,]    1    3
 [3,]    1    4
 [4,]    1    5
 [5,]    2    1
 [6,]    2    3
 [7,]    2    4
 [8,]    2    5
 [9,]    3    1
[10,]    3    2
[11,]    3    4
[12,]    3    5
[13,]    4    1
[14,]    4    2
[15,]    4    3
[16,]    4    5
[17,]    5    1
[18,]    5    2
[19,]    5    3
[20,]    5    4



# 6) permutations: with replacement: distinct items

permutations(5, 2, replace = TRUE)

      [,1] [,2]
 [1,]    1    1
 [2,]    1    2
 [3,]    1    3
 [4,]    1    4
 [5,]    1    5
 [6,]    2    1
 [7,]    2    2
 [8,]    2    3
 [9,]    2    4
[10,]    2    5
[11,]    3    1
[12,]    3    2
[13,]    3    3
[14,]    3    4
[15,]    3    5
[16,]    4    1
[17,]    4    2
[18,]    4    3
[19,]    4    4
[20,]    4    5
[21,]    5    1
[22,]    5    2
[23,]    5    3
[24,]    5    4
[25,]    5    5


# 7) permutations: without replacement: non distinct items

permutations(x = c("a", "b", "c"), freq = c(2, 1, 1), k = 2)

     [,1] [,2]
[1,] "a"  "a" 
[2,] "a"  "b" 
[3,] "a"  "c" 
[4,] "b"  "a" 
[5,] "b"  "c" 
[6,] "c"  "a" 
[7,] "c"  "b" 



# 8) permutations: with replacement: non distinct items

permutations(x = c("a", "b", "c"), k = 2, replace = TRUE)  # as `freq` doesn't matter

      [,1] [,2]
 [1,] "a"  "a" 
 [2,] "a"  "b" 
 [3,] "a"  "c" 
 [4,] "b"  "a" 
 [5,] "b"  "b" 
 [6,] "b"  "c" 
 [7,] "c"  "a" 
 [8,] "c"  "b" 
 [9,] "c"  "c" 

Compare to other packages

There are few advantages of using `arrangements` over the existing packages.

Integral framework: you don't have to use different packages for different methods.

It is very efficient. See https://randy3k.github.io/arrangements/articles/benchmark.html for some benchmarks.

It is memory efficient, it is able to generate all 13! permutation of 1 to 13, existing packages will fail to do so because of the limitation of matrix size. The `getnext()` method of the iterators allow users to get the arrangements one by one.

The generated arrangements are in dictionary order which may be desired for some users.

Problem

How to generate sequences of `r` objects from `n` objects? I'm looking for a way to do either permutations or combinations, with/without replacement, with distinct and non-distinct items (aka multisets). This is related to twelvefold way. The "distinct" solutions could be included in twelvefold way, while the "non-distinct" are not included.

Original source

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