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Re: Hyperoperators and the Each role

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Darren Duncan
July 17, 2006 17:23
Re: Hyperoperators and the Each role
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At 6:06 PM -0400 7/17/06, Christopher Jeris wrote:
>I have a couple of questions about what S03 says about hyperoperators
>applying recursively to any object which matches the 'Each' role:
>[from S03]
>Seq(3,8,[2,Seq(9,3)],8) >>-<< (1,1,2,1); # Seq(2,7,[0,Seq(7,1)],7)
>1. What determines the concrete type of the result of a binary
>hyperoperator which is applied to objects of two different sequence types?
>  Does the example mean that binary hyperoperators are not symmetric (in
>type) even when their underlying operator is symmetric?
>2. Will the 'Each' role require objects implementing it to be, not only
>destructurable, but constructible - as the hyperoperator in the example
>constructs a fresh Seq?  Does this mean that hyperoperators won't apply to,
>say, database result sets?

My understanding of binary hyper-operators is that they only work 
with iterating over ordinal types, like the Seq or Array, where their 
elements have an inherent order that can be used when determine how 
to pair up the elements of the 2 arguments.

If you have a non-ordinal collection like a Set or Mapping or Hash, 
you can not use a hyperoperator with it unless you convert it to an 
ordinal type first, and presumably sort it in the process. 
Practically speaking, often you won't be storing data in a 
non-ordinal collection if you intend to use hyper-operators with it, 
but more likely you would mean to use either set operators, which map 
arguments to each other by their values, or reduce operators, as is 
the case.

As for relational database result sets, in the general case these are 
just as unordered as a Set and so similar caveats apply, unless it is 
a sorted result set that is stored in a Seq or Array.

I should also point out that my in-progress Set::Relation module, 
whose operators are a proper super-set of those that the Set type 
has, is designed to make operations with unordered database result 
sets easy that are analagous to what you might use hyper-operators 
for, juch as performing a relational join, which maps corresponding 
result elements and produces derived ones.

-- Darren Duncan

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