CODEWITHSUNDEEP
rcombinations
Michael
Michael

Reputation: 7397

Combinations of multiple vectors in R

I'm not sure if permutations is the correct word for this. I want to given a set of n vectors (i.e. [1,2],[3,4] and [2,3]) permute them all and get an output of

[1,3,2],[1,3,3],[1,4,2],[1,4,3],[2,3,2] etc.

Is there an operation in R that will do this?

Upvotes: 9

Views: 16385

Answers (3)

SEAnalyst
SEAnalyst

Reputation: 1211

As an alternative to expand.grid() you could use rep() to produce the desired combination. Consider the following simplified example using the original data from this question:

a <- c(1,2)
b <- c(3,4)
c <- c(2,3)

To get the expand.grid()-like effect, use rep() with a times= argument equal to the product of the length of the other vectors (or 4). The middle vector would use a nested rep() with products of vector length to either side (or 2 and 2). The end vector is like the first but with each= argument in order to pattern correctly. This is trivial to calculate when each vector is length of 2. Example:

#tibble of all combinations of a, b and c
tibble::tibble(
var1 = rep(a, times = 4),
var2 = rep(rep(b, each= 2), times = 2), #nested rep()
var3 = rep(c, each= 4)
)

For an unknown number of input vectors (or unknown vector lengths), we can get all combinations with rep() in a function like this:

#Produces a tibble of all combinations of input vectors
expand_tibble <- function(...){ 
  x <- list(...) #all input vectors stored here
  l <- lapply(x,length)|> unlist() #vector showing length of each input vector
  t <- length(l) #total input vector count
  r <-list() #empty list
  for(i in 1:t){
    if(i==1){ #first input vector
      first <-l[2:length(l)] |> prod()
       r[[i]]<-rep(x[[i]], each = first)
    }else{   #last input vector
    if(i==t){
    last  <- l[1:t-1] |> prod()
    r[[i]]<-rep(x[[i]], last)
    }else{   #all middle input vectors
      m1 <- l[1:(i-1)] |> prod()
      m2 <- l[(i+1):t] |> prod()  
    r[[i]] <- rep(rep(x[[i]], each=m1),m2)
        }
    }
    names(r)[i]<-paste0("var",i)
    }
  tibble::as_tibble(r)
}

output:

expand_tibble(a,b,c)
 var1  var2  var3
  <dbl> <dbl> <dbl>
1     1     3     2
2     1     3     3
3     1     4     2
4     1     4     3
5     2     3     2
6     2     3     3
7     2     4     2
8     2     4     3

Upvotes: 0

Gavin Simpson
Gavin Simpson

Reputation: 174938

This is a useful case for storing the vectors in a list and using do.call() to arrange for an appropriate function call for you. expand.grid() is the standard function you want. But so you don't have to type out or name individual vectors, try:

> l <- list(a = 1:2, b = 3:4, c = 2:3)
> do.call(expand.grid, l)
  a b c
1 1 3 2
2 2 3 2
3 1 4 2
4 2 4 2
5 1 3 3
6 2 3 3
7 1 4 3
8 2 4 3

However, for all my cleverness, it turns out that expand.grid() accepts a list:

> expand.grid(l)
  a b c
1 1 3 2
2 2 3 2
3 1 4 2
4 2 4 2
5 1 3 3
6 2 3 3
7 1 4 3
8 2 4 3

Upvotes: 25

Andrie
Andrie

Reputation: 179558

This is what expand.grid does.

Quoting from the help page: Create a data frame from all combinations of the supplied vectors or factors. The result is a data.frame with a row for each combination.

expand.grid(
    c(1, 2),
    c(3, 4),
    c(2, 3)
)

  Var1 Var2 Var3
1    1    3    2
2    2    3    2
3    1    4    2
4    2    4    2
5    1    3    3
6    2    3    3
7    1    4    3
8    2    4    3

Upvotes: 14

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