Obtaining the last element of a ProductIterator

General Usage
,
1

Currently there is no way to obtain the last element of an iterator product, for example:

julia> it = Iterators.product(1:3, 2:3)
Base.Iterators.ProductIterator{Tuple{UnitRange{Int64}, UnitRange{Int64}}}((1:3, 2:3))

julia> last(it)
ERROR: MethodError: no method matching lastindex(::Base.Iterators.ProductIterator{Tuple{UnitRange{Int64}, UnitRange{Int64}}})

However should it not be possible to define this by collating the last elements of the component iterators?

julia> Base.last(it::Iterators.ProductIterator) = map(last, it.iterators)

julia> last(it)
(3, 3)
2

In a future version (1.6?) the following works:

julia> it = Iterators.product(1:3, 2:3)
Base.Iterators.ProductIterator{Tuple{UnitRange{Int64}, UnitRange{Int64}}}((1:3, 2:3))

julia> last(it, 1)
1-element Vector{Tuple{Int64, Int64}}:
 (3, 3)

This was added in https://github.com/JuliaLang/julia/pull/34868

This works as long as the iterator implements reverse. There was recently a discussion here about the difficulties of adding a generic last method: Getting the last element of an iterator. I planned to submit a PR but as explained in that thread I couldn’t find an implementation that was performant and backward-compatible.

As a workaround @mschauer suggested foldl((_, y) -> y, itr), which also works fine here:

julia> foldl((_,y)->y, Iterators.product(1:3, 2:3))
(3, 3)
2 Likes
3

I realize that a general O(1) last method is not possible in most cases, and the foldl equivalents would be O(n) if I understand correctly. In this specific case of a ProductIterator, it feels like it should support fetching the last element in O(1) time, unless there are some corner cases that I am missing out on?

The method last(it, 1) seems to rely on actually collecting the elements of the iterator, which may be avoided in many cases?

4

It is O(1) in this case. According to @less Iterators.reverse(it) the implementation is:

last(itr, n::Integer) = reverse!(collect(Iterators.take(Iterators.reverse(itr), n)))

The inner reverse(itr) will call reverse on each iterator in the product, for example reverse(1:3) which returns 3:-1:1 without any collection.

collect operates on the take iterator so it only collects 1 element.