Getting all possible two column subsets - r

Question

I am a relative newbie to R and I am now very close to being finished with a rather long script with many thanks to everyone who helped me thus far at various steps. I have another point I am stuck on. I have simplified the issue to this:
Dataset1
ax ay
1 3
2 4
Dataset2
bx by
5 7
6 8
A <- dataset1
B <- dataset2
a <- 2 #number of columns
b <- 1:2
(my datasets will vary in number of columns and so I need to be able to vary this factor)
I want this answer in any order (i.e. all possible combinations of two columns one from each of the two datasets) like this or equivalent.
[[1]]
1 5
2 6
[[2]]
1 7
2 8
[[3]]
3 5
4 6
[[4]]
3 7
4 8
But I am not getting it.
I tried a bunch of things and the closest to what I want was with this:
i <- 1
for( i in 1:a )
{
e <- lapply(B, function(x) as.data.frame(cbind(A, x)))
print(e)
i <- i+1
}
Close, yes. I can take the answer and do some fiddling and subsetting but its not right and there must be an easy way to do this. I have not seen anything like this in my on line searches. Any help much appreciated.

Answer 1

Does something like this work for you?
Dataset1 <- data.frame(ax=1:2,ay=3:4)
Dataset2 <- data.frame(bx=5:6,by=7:8)
apply(
expand.grid(seq_along(Dataset1),seq_along(Dataset2)),
1,
function(x) cbind(Dataset1[x[1]],Dataset2[x[2]])
)
Result:
[[1]]
ax bx
1 1 5
2 2 6
[[2]]
ay bx
1 3 5
2 4 6
[[3]]
ax by
1 1 7
2 2 8
[[4]]
ay by
1 3 7
2 4 8

Answer 2

I think the easiest way to do is very similar to what you tried, use two explicit loops. However, there are still some things I would do differently:
Pre allocate the list space
Use an explicit counter
Use drop=FALSE
Then you can do the following.
A <- read.table(text = "ax ay
1 3
2 4", header = TRUE)
B <- read.table(text = "bx by
5 7
6 8", header = TRUE)
out <- vector("list", length = ncol(A) * ncol(B))
counter <- 1
for (i in 1:ncol(A)) {
for (j in 1:ncol(B)) {
out[[counter]] <- cbind(A[,i, drop = FALSE], B[,j, drop = FALSE])
counter <- counter + 1
}
}
out
## [[1]]
## ax bx
## 1 1 5
## 2 2 6
##
## [[2]]
## ax by
## 1 1 7
## 2 2 8
##
## [[3]]
## ay bx
## 1 3 5
## 2 4 6
##
## [[4]]
## ay by
## 1 3 7
## 2 4 8

Answer 3

If I understand the question, I think you can use combn to select the columns you want. For instance, if wanted all combinations of 8 columns taken 2 at at time, you could do:
combn(1:8, 2)
Which gives (in part for readability):
combn(1:8,2)[,c(1:5, 15:18)]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,] 1 1 1 1 1 3 3 3 3
[2,] 2 3 4 5 6 5 6 7 8
So then columns of this matrix can be used as the indices you want.