Non-redundant version of expand.grid

Solution 1:

How about using outer? But this particular function concatenates them into one character string.

outer( c("aa", "ab", "cc"), c("aa", "ab", "cc") , "paste" )
#     [,1]    [,2]    [,3]   
#[1,] "aa aa" "aa ab" "aa cc"
#[2,] "ab aa" "ab ab" "ab cc"
#[3,] "cc aa" "cc ab" "cc cc"

You can also use combn on the unique elements of the two vectors if you don't want the repeating elements (e.g. aa aa)

vals <- c( c("aa", "ab", "cc"), c("aa", "ab", "cc") )
vals <- unique( vals )
combn( vals , 2 )
#     [,1] [,2] [,3]
#[1,] "aa" "aa" "ab"
#[2,] "ab" "cc" "cc"

Solution 2:

In base R, you can use this:

expand.grid.unique <- function(x, y, include.equals=FALSE)
{
    x <- unique(x)

    y <- unique(y)

    g <- function(i)
    {
        z <- setdiff(y, x[seq_len(i-include.equals)])

        if(length(z)) cbind(x[i], z, deparse.level=0)
    }

    do.call(rbind, lapply(seq_along(x), g))
}

Results:

> x <- c("aa", "ab", "cc")
> y <- c("aa", "ab", "cc")

> expand.grid.unique(x, y)
     [,1] [,2]
[1,] "aa" "ab"
[2,] "aa" "cc"
[3,] "ab" "cc"

> expand.grid.unique(x, y, include.equals=TRUE)
     [,1] [,2]
[1,] "aa" "aa"
[2,] "aa" "ab"
[3,] "aa" "cc"
[4,] "ab" "ab"
[5,] "ab" "cc"
[6,] "cc" "cc"

Solution 3:

If the two vectors are the same, there's the combinations function in the gtools package:

library(gtools)
combinations(n = 3, r = 2, v = c("aa", "ab", "cc"), repeats.allowed = TRUE)

#      [,1] [,2]
# [1,] "aa" "aa"
# [2,] "aa" "ab"
# [3,] "aa" "cc"
# [4,] "ab" "ab"
# [5,] "ab" "cc"
# [6,] "cc" "cc"

And without "aa" "aa", etc.

combinations(n = 3, r = 2, v = c("aa", "ab", "cc"), repeats.allowed = FALSE)