Returns an instance of T calculated by summing all instances in
iter in one pass.
Returns an instance of T calculated by summing all instances in
iter in one pass. Returns None if iter is empty, else
Some[T].
instances of T to be combined
None if iter is empty, else an option value containing the summed T
This is a combinator on semigroups, after you do the plus, you transform B with a fold function This will not be valid for all fold functions. You need to prove that it is still associative.
Clearly only values of (a,b) are valid if fold(a,b) == b, so keep that in mind.
I have not yet found a sufficient condition on (A,B) => B that makes it correct Clearly a (trivial) constant function {(l,r) => r} works. Also, if B is List[T], and (l:A,r:List[T]) = r.sortBy(fn(l)) this works as well (due to the associativity on A, and the fact that the list never loses data).
For approximate lists (like top-K applications) this might work (or be close enough to associative that for approximation algorithms it is fine), and in fact, that is the main motivation of this code: Produce some ordering in A, and use it to do sorted-topK on the list in B.
Seems like an open topic here.... you are obliged to think on your own about this.