Re: beg for Bag
HaloO, Jonathan Lang wrote: Would (1,2,2,3,4,4) be a Seq or a Bag? Comma constructs a Seq, of course. IMHO, the _only_ way this could work would be if it's a Bag: if it's a Seq, I see no way that one could resolve '(1,2,3) ∪ (3,1,2)'. This is not any different from '3' + '4' resulting in numeric 7. It is the operator that builds Bags from (1,2,3) and (3,1,2) and then calculates the union. Mind you, I'm still not sold on the idea of performing set operations on Seqs - it may be technically feasible to do so, but it strikes me as fundamentally unintuitive. But you have no problem with doing numeric stuff? E.g. (1,2,3) + (3,4) actually means +(1,2,3) + +(3,4) == 3 + 2 == 5. It is the operator that indicates Set/Bag operations. And 'Seq does Bag' comes in handy here. Actually, no set operation is performed on Seq. The Seq is just used to build a Bag. This is like there is no string concatenation for numbers, arrays etc. They are stringified first. One could also write @a (|) @b and the two arrays would be flattened to Seq and then used to build a Bag. The only drawback with Bag being the Seq supertype is that people might want @a (|) @b to mean Set(@a) (|) Set(@b). The Set enforcement could also come after the Bag union: Set(@a (|) @b). Perhaps @a (|) @b delivers the result in an Array instead of a Bag value. So in general the parens ops should not be called set ops but unordered ops because this is their core meaning. They are overloaded for Bag and Set. I'm still bothered by the idea that you have to wrap _every_ ASCII representative of a set operation in parentheses - something which is only necessary when you start applying the full range of set operations to non-Set entities. In particular, I want 'Set - Set' to produce the difference of the two Sets. But that should be the numeric difference of the cardinalities of the Sets. Plain - means numeric. It's the same thing that we don't do string concatenation with + but have a stringish ~ for the task. Hence it works to say 3 ~ 4 and end up with '34'. If set operations also apply to Seq, then (=) is not the same as ===. The former ignores the order of the terms; the latter only does so for Sets and Bags. In a way, this is what started the whole debate. You are correct. This emphasizes the need for (=). You mention a single disjoint union operator: is it supposed to be the disjoint union comparison operator (i.e., it returns true if the sets are disjoint), or the disjoint union composition operator (which returns a Set of Pairs, with each element being keyed according to the Set that it was originally in)? Perhaps we need both. I've chosen (/) for it's visual similarity with (|) like || and //. We could use (/?) for the disjointness test. Or doing visual games again: (%). You see the disjoint sets in the glyph. (:) might work, too. Saying that complement is difficult is an understatement. I suppose you _could_ get it to work by having a 'complemented Set' would keep track of which elements it _doesn't_ have; but this opens a can of worms that I really don't think we want to get into (e.g., A ∪ (!)B). And the notion of a complement with regard to the surrounding set is already handled by the difference operator. Isn't that similar in semantics to the none junction? If you have e.g. a Set of Int (1,2,3) you can express (!)(1,2,3) as Set(*..0,4..*). But I agree that it is difficult in general. Well, (!)$x might actually create a large set of almost all currently known objects in some scope. This might just be written as * (-) $x. IOW, Set(*) means all known objects. As long as a program is not wielding large data sets complement might just work---and the complement of a large set is small, of course. Note that $a (-) $b is equivalent to $a () (!)$b, you have it as the union. Or did you mean $a (|) (!)$b === (!)($b (-) $a)? Hmm, how is Bag complement defined in the first place? Is it just complementing the set and goes for multiplicities of 1? You did highlight some things - for instance, a Set of Pairs is _not_ a Hash: a Hash has a further requirement that every key must be unique, whereas a Set of Pairs allows for duplicate keys (but not duplicate key-value pairs; go with a Bag of Pairs for that). Correct. Regards, TSa. --
Re: beg for Bag
TSa wrote: Jonathan Lang wrote: Would (1,2,2,3,4,4) be a Seq or a Bag? Comma constructs a Seq, of course. The context of the question was that you provided the above as the result of unioning two Seqs; as such, I was trying to find out whether you meant that the union of two Seqs should be a Seq, or if that was an unintended artifact of the way you said it. IMHO, the _only_ way this could work would be if it's a Bag: if it's a Seq, I see no way that one could resolve '(1,2,3) ∪ (3,1,2)'. This is not any different from '3' + '4' resulting in numeric 7. It is the operator that builds Bags from (1,2,3) and (3,1,2) and then calculates the union. IOW, you're going with the notion that Seqs should be treated like Bags when you apply set operations to them - that is, the order of their elements becomes irrelevant. Mind you, I'm still not sold on the idea of performing set operations on Seqs - it may be technically feasible to do so, but it strikes me as fundamentally unintuitive. But you have no problem with doing numeric stuff? E.g. (1,2,3) + (3,4) actually means +(1,2,3) + +(3,4) == 3 + 2 == 5. I didn't say that. For the record, I _do_ have a problem with saying that '(1,2,3) + (3,4)' should be equivalent to '+(1,2,3) + +(3,4)'. It's counter-DWIMish, and I'd rather see the compiler complain about the former than to do the latter when that's not what I wanted. As a general rule of thumb, I do _not_ want data types being coerced behind my back. '3' + '4' producing 7 should be an exception to the rule, not the basis for it; and the reason for the exception is that the fact that '3' is supposed to mean the same as 3 is not very hard to spot. Stating that (1,2,3) is supposed to mean the same as 3 is considerably more murky. I'm still bothered by the idea that you have to wrap _every_ ASCII representative of a set operation in parentheses - something which is only necessary when you start applying the full range of set operations to non-Set entities. In particular, I want 'Set - Set' to produce the difference of the two Sets. But that should be the numeric difference of the cardinalities of the Sets. Plain - means numeric. It's the same thing that we don't do string concatenation with + but have a stringish ~ for the task. Hence it works to say 3 ~ 4 and end up with '34'. Plain '-' means numeric when the terms are scalars - and possibly not even then. Perl 6 allows for operator overloading, and it does so because sometimes it makes more sense to have the same operator mean different things depending on what it's operating on. The interplay between Strings and Nums shouldn't be taken as indicative of how everything should work. If set operations also apply to Seq, then (=) is not the same as ===. The former ignores the order of the terms; the latter only does so for Sets and Bags. In a way, this is what started the whole debate. You are correct. This emphasizes the need for (=). ...or for allowing explicit conversion of Seqs to Bags or Sets, and demanding that it be done before set operations are permitted. Note that $a (-) $b is equivalent to $a () (!)$b, you have it as the union. Or did you mean $a (|) (!)$b === (!)($b (-) $a)? No, I goofed and meant A ⋂ !B when I said A ∪ !B. Hmm, how is Bag complement defined in the first place? Is it just complementing the set and goes for multiplicities of 1? Conceptually, it would be going for multiplicities of Inf. Even if you define Set complement, you probably don't want to define Bag complement. -- Jonathan Dataweaver Lang
Re: beg for Bag
HaloO, Jonathan Lang wrote: Note that this would mean that Seq would also have set operations. I count this as an advantage. So one can write (1,2,3) (|) (2,2,3,4,4) to get a result of (1,2,2,3,4,4). As long as the Seq is a Set, that is it has no duplicates, you get Set behavior through the Bag ops: (1,2,3) (|) (2,3,4) === (1,2,3,4); (1,2,3) () (2,3,4) === (2,3). BTW, the set/bag operations are not yet mentioned in S03 as new operators. Here's a list what I think they should be: (|) union () intersection (^) symmetric difference (/) disjoint union? (!) complement, this is difficult because you need the surrounding set (-) difference (+) join, returns a bag (*) cartesian product (**) powerset (in) membership (!in) negated membership () proper subset () proper superset (=) subset (=) superset (=) equality, also with === (!=) inequality, also with !=== Did I forget something? Regards, TSa. --
Re: beg for Bag
TSa wrote: Jonathan Lang wrote: Note that this would mean that Seq would also have set operations. I count this as an advantage. So one can write (1,2,3) (|) (2,2,3,4,4) to get a result of (1,2,2,3,4,4). As long as the Seq is a Set, that is it has no duplicates, you get Set behavior through the Bag ops: (1,2,3) (|) (2,3,4) === (1,2,3,4); (1,2,3) () (2,3,4) === (2,3). Would (1,2,2,3,4,4) be a Seq or a Bag? IMHO, the _only_ way this could work would be if it's a Bag: if it's a Seq, I see no way that one could resolve '(1,2,3) ∪ (3,1,2)'. Mind you, I'm still not sold on the idea of performing set operations on Seqs - it may be technically feasible to do so, but it strikes me as fundamentally unintuitive. BTW, the set/bag operations are not yet mentioned in S03 as new operators. Here's a list what I think they should be: (|) union () intersection (^) symmetric difference (/) disjoint union? (!) complement, this is difficult because you need the surrounding set (-) difference (+) join, returns a bag (*) cartesian product (**) powerset (in) membership (!in) negated membership () proper subset () proper superset (=) subset (=) superset (=) equality, also with === (!=) inequality, also with !=== Initial thought: overkill. Several of these operations (e.g., the Cartesian product) are obscure and only of interest to mathematicians. This isn't a reason to exclude them; but a non-mathematician should not be made to feel like he needs to get a math degree in order to use sets. (On a tangent, he also shouldn't be made to feel like he has to learn Type Theory in order to use Perl6's type system.) I'm still bothered by the idea that you have to wrap _every_ ASCII representative of a set operation in parentheses - something which is only necessary when you start applying the full range of set operations to non-Set entities. In particular, I want 'Set - Set' to produce the difference of the two Sets. Setting aside the issue of the notation to be used, there are several concerns that I have with this: If set operations also apply to Seq, then (=) is not the same as ===. The former ignores the order of the terms; the latter only does so for Sets and Bags. In a way, this is what started the whole debate. You mention a single disjoint union operator: is it supposed to be the disjoint union comparison operator (i.e., it returns true if the sets are disjoint), or the disjoint union composition operator (which returns a Set of Pairs, with each element being keyed according to the Set that it was originally in)? Saying that complement is difficult is an understatement. I suppose you _could_ get it to work by having a 'complemented Set' would keep track of which elements it _doesn't_ have; but this opens a can of worms that I really don't think we want to get into (e.g., A ∪ (!)B). And the notion of a complement with regard to the surrounding set is already handled by the difference operator. A Cartesian product would return a Set (or Bag, depending on the left term) of Pairs, keyed by the elements of the left term. A powerset strikes me as something that you'd want to do as a 0-ary method, rather than as an operator. Did I forget something? You did highlight some things - for instance, a Set of Pairs is _not_ a Hash: a Hash has a further requirement that every key must be unique, whereas a Set of Pairs allows for duplicate keys (but not duplicate key-value pairs; go with a Bag of Pairs for that). -- Jonathan Dataweaver Lang
beg for Bag
HaloO, as a spin-off of the 'Set-returning .keys (was Re: Smart Matching clarification)' thread I want to propose the addition of a Bag type that completes the set of immutable types. It shall have the following properties. 1) It is a multiset generalization of Set 2) It is a supertype of Set and Seq (a Set can of course be build from a Seq). That is 'Set does Bag' and 'Seq does Bag'. Note that a Seq is a ready-made Bag and if it happens to have no duplicates it behaves like a Set. 3) It has set operations as generalizations of the Set operations 4) It provides some Bag specific ops like (+) that return a Bag even when called with Sets 5) It provides the iteration interface (which in turn is applicable to the subtypes Set and Seq, of course) 6) %hash.values returns a Bag 7) %hash.keys returns a Set The wording in section 'Immutable Types' in S06 concerning Set as Unordered Seqs that allow no duplicates is a bit misleading because it hints as Set being a subtype of Seq. The Mapping could be explained as Set of Pair. The Bag type could be implemented in a module but then we need a way to do supertyping. The added Bag type would need to add a multiplicity accessor that returns 1 to the Set implementation and add lazy multiplicity counting to Seq. And I don't know how Hash::values would be augmented to return a Bag. This Bag return type would guarantee %h1.values === %h2.values to yield true when the two hashes happen to return their values in different orders. Comments? --
Re: beg for Bag
TSa writes: I want to propose the addition of a Bag type Different from the CBag that's already mentioned in Synopsis 3? Smylers
Re: beg for Bag
At 7:08 PM + 11/28/06, Smylers wrote: TSa writes: I want to propose the addition of a Bag type Different from the CBag that's already mentioned in Synopsis 3? Smylers TSa wasn't the first person to ask for an explicit Bag type. I did too, a few weeks ago. And one reason for that was exactly what you mention. Various other parts of the Synopsis documents mention Bag in examples and such, but the list of built-in types in Synopsis 6 does not include it. In my mind, unless Bag appears in the Synopsis 6 list, its references elsewhere count as nothing more than an example stand-in for some arbitrary user-defined type. -- Darren Duncan
Re: beg for Bag
TSa wrote: 1) It is a multiset generalization of Set 2) It is a supertype of Set and Seq (a Set can of course be build from a Seq). That is 'Set does Bag' and 'Seq does Bag'. Note that a Seq is a ready-made Bag and if it happens to have no duplicates it behaves like a Set. 3) It has set operations as generalizations of the Set operations Note that this would mean that Seq would also have set operations. 4) It provides some Bag specific ops like (+) that return a Bag even when called with Sets or Seqs. -- Jonathan Dataweaver Lang
