The function you're after is order (how I arrived at this conclusion -- my first thought was "well, sorting, what about sort?". Tried sort(arr) which looks like it sorts arr as a vector instead of row-wise. Looking at ?sort, I see in the "See Also: order for sorting on or reordering multiple variables.").
Looking at ?order, I see that order(x,y,z, ...) will order by x, breaking ties by y, breaking further ties by z, and so on. Great - all I have to do is pass in each column of arr to order to do this. (There is even an example for this in the examples section of ?order):
order( arr[,1], arr[,2], arr[,3] )
# gives 3 2 1: row 3 first, then row 2, then row 1.
# Hence:
arr[ order( arr[,1], arr[,2], arr[,3] ), ]
# [,1] [,2] [,3]
#[1,] 1 1 2
#[2,] 1 2 3
#[3,] 2 1 3
Great!
But it is a bit annoying that I have to write out arr[,i] for each column in arr - what if I don't know how many columns it has in advance?
Well, the examples show how you can do this too: using do.call. Basically, you do:
do.call( order, args )
where args is a list of arguments into order. So if you can make a list out of each column of arr then you can use this as args.
One way to do this is is to convert arr into a data frame and then into a list -- this will automagically put one column per element of the list:
arr[ do.call( order, as.list(as.data.frame(arr)) ), ]
The as.list(as.data.frame is a bit kludgy - there are certainly other ways to create a list such that list[[i]] is the ith column of arr, but this is just one.
