OK I think I see the source of my confusion
The definition says
-----
Definition: ListEval
ListEval((expr1, ..., exprn), μ) returns a list (e1, ..., en), where ei =
expri(μ) or error.
ListEval retains errors resulting from the evaluation of the list elements.
-----
The "or error", is that supposed to mean that e_i = expr_i(u) or error, or is
it supposed to mean that the entire ListEval() can return error?
I read it the first way, which would allow a list of error entries, but I am
guessing that you are reading the second way?
graham
On Dec 7, 2012, at 3:05 PM, Andy Seaborne wrote:
> On 07/12/12 21:13, Rob Vesse wrote:
>> My understanding has always been that all errors are grouped together into
>> a single group.
>
> Yes.
>
>> Here's a nice simple query that illustrates this:
>>
>> SELECT ?groupkey
>> WHERE
>> {
>> VALUES ( ?a ?b)
>> {
>> ( 1 2 )
>> ( UNDEF 2 )
>> ( 1 UNDEF )
>> ( UNDEF UNDEF )
>> }
>> }
>> GROUP BY (?a + ?b AS ?groupkey)
>>
>> Running this will give two results, one with an unbound value for
>> ?groupkey and one with 3 as the ?groupkey
>>
>> As the editor of the query specification Andy will likely come along and
>> explain the mathematica better when he sees this thread
>
> I could waffle about "one point extension of the value space of the
> expression being grouped"
>
> Or just say that all evaluations yielding 'error' get put into the same group.
>
> If the group key is composite, any error makes the whole key an error ... see
> below ...
>
>>
>> Rob
>>
>>
>>
>> On 12/7/12 12:17 PM, "Graham Matthews" <[email protected]>
>> wrote:
>>
>>> Hi
>>>
>>> This is kind of a JENA question, kind of a generic SPARQL question.
>>>
>>> The SPARQL 1.1 definition says:
>>>
>>> ----
>>> Note that, although the result of a ListEval can be an error, and errors
>>> may be used to group, solutions containing error values are removed at
>>> projection time.
>>>
>>> ListEval((unbound), μ) = (error), as the evaluation of an unbound
>>> expression is an error.
>>>
>>> ----
>>>
>>>
>>> I have looked at this many times and cannot figure out what it means to
>>> say that "errors may be used to group". To make sense of this requires
>>> that errors be comparable for equality (otherwise you can't group …),
>>> which I don't see any definition of anywhere in the SPARQL semantics.
>>>
>>> I don't even understand how it works in a really simple query grouping
>>> solutions from a basic graph pattern, for example:
>>>
>>> select whatever from ..
>>> where
>>> {
>>> ?a ?b ?c
>>> }
>>> group by ?a+1, ?b+2
>>>
>>> Suppose:
>>> -) ?a+1 produces a dynamic error but ?b+2 does not for one choice of ?a
>>> and ?b
>
> ?a+1 is always an error - can't have literals as subjects!
>
>>> -) ?a+1 does not produce a dynamic error but ?b+2 does for a different
>>> choice of ?a and ?b
>
> The group key is the pair (?a+1, ?b+2) and if either are an error, then the
> whole key is an error.
>
>
>>>
>>> Are these two choices place in the same group?
>
> Yes.
>
>>>
>>> And what happens if for two choices of ?a and ?b, ?a+1 and ?b+2 both
>>> produce produce dynamic errors, but different ones (we we have 4
>>> different kinds of errors)? Are they in the same group? And are they in
>>> the same group as the previous example?
>>>
>>> My gut says that what was intended here was SQL style null groupings
>>> (which are horrible, but ..). But the mathematics in the definition isn't
>>> clear.
>
> SQL Null is a bit different - it is a real value [*] you can talk about,
> whereas SPARQL treats the group key as (A,B) and if either A or B are an
> error, the whole key for the group is the "error" key.
>
>
> Andy
>
> [*] aside from the that fact that "null = null" is false
>
>>> Any SPARQL gurus know what is meant to happen for grouping here?
>>>
>>> thanks
>>> graham
>>>
>>
>
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