- From: Andy Seaborne <andy.seaborne@epimorphics.com>
- Date: Mon, 21 Nov 2011 14:28:56 +0000
- To: birte.glimm@uni-ulm.de
- CC: Steve Harris <steve.harris@garlik.com>, SPARQL Working Group <public-rdf-dawg@w3.org>
Steve, Could you walk through the translation of: SELECT ?x {?x :p ?v } GROUP BY ?x This is the case where there is no AS because the variable is part of the group key. Andy On 17/11/11 09:00, Birte Glimm wrote: > On 16 November 2011 19:08, Steve Harris<steve.harris@garlik.com> wrote: >> On 2011-11-16, at 16:58, Birte Glimm wrote: >> >>> [snip] >>>>>> As syntax, (SAMPLE(?x) AS ?x) isn't legal because AS has to introduce a new >>>>>> variable. This happens in SELECT expression processing a few subsections. >>>>> >>>>> Yes, that occured to me as well. Unless it is made legal for >>>>> intermediate queries, which I don't like, there seems no way around >>>>> creating solutions in the aggregate join that also contain the grouped >>>>> variables. >>>> >>>> Yes, that why I ended up with the messy agg_i thing, to avoid conflating aggregate results with variable names. >>>> >>>> Reading through this again, I think that the text as written is correct: >>>> >>>> For each variable V appearing outside of an aggregate >>>> Replace V with Sample(V) in Q >>>> End >>>> >>>> ensures that there's only aggregates being projected, then >>>> >>>> For each aggregate X(args ; scalarvals) now in E >>>> # note scalarvals may be omitted, then it's equivalent to the empty set >>>> Ai := Aggregation(args, X, scalarvals, G) >>>> Replace X(...) with aggi in Q >>>> i := i + 1 >>>> End >>>> >>>> Defines A_i/agg_i for the Sample(V) above. I could well have spec blindness though. >>> >>> That still does not solve the problem that you loose the original >>> variable name, so results will contain agg_i and even worse, if you >>> have a having clause, the variable used there might no longer exist >>> since it was replaced by agg_i. >> >> I believe that's taken care of by AggregateJoin: >> >> Write A = (A1, A2, ...) where Ai = Aggregation(exprListi, funci, scalarvarsi, P) >> >> eval(D(G), AggregateJoin(A)) = { (agg1, v1), ..., (aggn, vn) | vi such that ( k, vi ) in eval(D(G), Ai) >> for some k and each 1<= i<= n } >> >> vi is your var below, I think. > > I don't think. vi is the aggregated value, so you ( k, vi ) with k the > key that was used to group the values and vi the aggregated value. > I'll do one complete example to make clear where I think we have a > problem: > > Assume data: > ex:Birte ex:mark 4, 5 . > ex:Steve ex:mark 3, 5 . > and query: > SELECT ?name (MAX(?mark) AS ?max) (AVG(?mark) AS ?avg) > WHERE { ?name ex:mark ?mark } GROUP BY ?name > > Algebra translation up to group gives: > A=Group((?name), Bgp(...)) > Replace ?name with SAMPLE(?name) > SELECT SAMPLE(?name) (MAX(?mark) AS ?max) (AVG(?mark) AS ?avg) > WHERE { ?name ex:mark ?mark } GROUP BY ?name > continue transformation: > A= AggregateJoin( > Aggregation((?name), SAMPLE, {}, A), > Aggregation((?mark), MAX, {}, A), > Aggregation((?mark), AVG, {}, A) > ) > plus we rewrite the aggregates with aggi: > SELECT ?agg1 (?agg2 AS ?max) (?agg3 AS ?avg) > WHERE { ?name ex:mark ?mark } GROUP BY ?name > We then translate the expressions: > A=Extend(A, ?max, ?agg2) > A=Extend(A, ?avg, ?agg3) > Finally projection: > Projcet(A, {?agg1, ?max, ?avg}) > > Evaluation the BGP gives 4 solutions: > mu1 : ?name -> ex:Birte, ?mark -> 4 > mu2 : ?name -> ex:Birte, ?mark -> 5 > mu3 : ?name -> ex:Steve, ?mark -> 3 > mu4 : ?name -> ex:Steve ?mark -> 5 > After grouping we get: > { ex:Birte -> {mu1, mu2}, ex:Steve -> {mu3, mu4} } > The three aggregates give > SAMPLE: { (ex:Birte)->ex:Birte, (ex:Steve)->ex:Steve } > MAX: { (ex:Birte)->5, (ex:Steve)->5 } > AVG: { (ex:Birte)->4.5, (ex:Steve)->4 } > Joining the aggregates gives 2 new solutions: > mu_a: ?agg1->ex:Birte, ?agg2->5, ?agg3->4.5 > mu_b : ?agg1->ex:Steve, ?agg2->5, ?agg3->4 > We extend the solutions: > mu_a' = mu_a union ?max->5, ?avg->4.5 > mu_b' = mu_a union ?max->5, ?avg->4 > Project then throws out ?agg2 and ?agg3 and we are done. > > We have, however, ?agg1 instead of ?name and if we had used HAVING > with some condition on ?name we would have a problem. > > Do I do something wrong here or do we indeed have ?agg1 in the result > and ?name got lost? > > Birte > > > > > >> >> The way the document is structured moves these apart in an unfortunate way. >> >> - Steve >> >>> >>> How about having two loops >>> For each aggregate (X(args ; scalarvals) AS var) now in E >>> # note scalarvals may be omitted, then it's equivalent to the empty set >>> Ai := Aggregation(args, X, scalarvals, G) >>> Replace X(...) with aggi in Q >>> i := i + 1 >>> End >>> For each aggregate X(args ; scalarvals) now in E >>> # note scalarvals may be omitted, then it's equivalent to the empty set >>> Ai := Aggregation(args, X, scalarvals, G) >>> Replace X(var; scalarvals) with (aggi AS var) in Q >>> i := i + 1 >>> End >>> >>> This way, we never have an illegal syntax form, we guarantee that all >>> variables are still available after the aggregation and since AS is >>> only processed later all seems to be fine. One could of course think >>> about handling both cases in one loop although for the spec having two >>> loops seems fine to me. >>> >>> Birte >>> >>>> - Steve >>>> >>>>>> There is no definition of "Aggregation". It's mentioned in 11.2 but the >>>>>> link goes to "Definition: Evaluation of Aggregation". There should a >>>>>> definition (just after group?) in 18.4. >>>>> >>>>> Yes, I also wondered about that. It is somehow clear how to evaluate, >>>>> but it would be much more consistent if there were a definition. >>>>> >>>>>> I looked because I wondered if we could just have an "?x" as the >>>>>> "aggregate". >>>>> >>>>> Not sure I understand this. >>>>> >>>>>> But I think, as Birte shows, as because it's done by syntactic >>>>>> rewriting, just leaving it as "?x" would work. >>>>> >>>>> As I don't understand the sentence above. I just want to make my point >>>>> again that we need a binding for ?x if ?x is grouped but not in an >>>>> aggregate as it can be used in the HAVING clause. If, at the point of >>>>> evaluating HAVING, we only have agg_1, we can't filter on ?x. >>>>> >>>>>>> I wanted to convert the plain ?x projection to an aggregate so it was >>>>>>> consistent with the rest of the projections, but expressing it explicitly >>>>>>> would be equivalent I think. >>>>>>> >>>>>>> I will have a run through the aggregation text and see if I can make that >>>>>>> change with a relatively small change to the document. >>>>>>> >>>>>>> Cheers, >>>>>>> Steve >>>>>> >>>>>> I also noticed; >>>>>> >>>>>> [[ >>>>>> Definition: Evaluation of AggregateJoin >>>>>> ... >>>>>> Note that if eval(D(G), Ai) is an error, it is ignored. >>>>>> ]] >>>>>> >>>>>> An error causes an error doesn't it? (AS causes it to be unbound) >>>>> >>>>> AS is transformed into Extend(), which is evaluated: >>>>> Extend(μ, var, expr) = μ ∪ { (var,value) | var not in dom(μ) and value >>>>> = eval(expr) } >>>>> Extend(μ, var, expr) = μ if var not in dom(μ) and eval(expr) is an error >>>>> >>>>> The latter makes the solution just not contain a mapping for the >>>>> variable as I understand it. >>>>> >>>>> But while we are at it, there is a lowercase extend in the Definition of Extend: >>>>> Extend(Ω , var, term) = { extend(μ, var, term) | μ in Ω } >>>>> >>>>> It is also lowercase in the evaluation semantics: >>>>> Definition: Evaluation of Extend >>>>> eval(D(G), extend(var, expr, P)) = extend(var, expr , eval(D(G), P)) >>>>> Furthermore, here we swap the order. It should be >>>>> eval(D(G), Extend(P, var, expr)) = Extend(eval(D(G), P), var, expr) >>>>> or the algorithm for translating queries into the algrebra is wrong >>>>> and has to be changed. >>>>> >>>>> Birte >>>>>> Andy >>>>>> >>>>> >>>>> >>>>> >>>>> -- >>>>> Jun. Prof. Dr. Birte Glimm Tel.: +49 731 50 24125 >>>>> Inst. of Artificial Intelligence Secr: +49 731 50 24258 >>>>> University of Ulm Fax: +49 731 50 24188 >>>>> D-89069 Ulm birte.glimm@uni-ulm.de >>>>> Germany >>>>> >>>> >>>> -- >>>> Steve Harris, CTO, Garlik Limited >>>> 1-3 Halford Road, Richmond, TW10 6AW, UK >>>> +44 20 8439 8203 http://www.garlik.com/ >>>> Registered in England and Wales 535 7233 VAT # 849 0517 11 >>>> Registered office: Thames House, Portsmouth Road, Esher, Surrey, KT10 9AD >>>> >>>> >>> >>> >>> >>> -- >>> Jun. Prof. Dr. Birte Glimm Tel.: +49 731 50 24125 >>> Inst. of Artificial Intelligence Secr: +49 731 50 24258 >>> University of Ulm Fax: +49 731 50 24188 >>> D-89069 Ulm birte.glimm@uni-ulm.de >>> Germany >>> >> >> -- >> Steve Harris, CTO, Garlik Limited >> 1-3 Halford Road, Richmond, TW10 6AW, UK >> +44 20 8439 8203 http://www.garlik.com/ >> Registered in England and Wales 535 7233 VAT # 849 0517 11 >> Registered office: Thames House, Portsmouth Road, Esher, Surrey, KT10 9AD >> >> > > >
Received on Monday, 21 November 2011 14:29:40 UTC