Re: [Neo4j] Cypher: how would you query first- and second-degree friends?
And now it's in code. http://docs.neo4j.org/chunked/snapshot/query-match.html#match-optional-relationship Hope that helps, Andrés On Thu, Aug 25, 2011 at 10:34 PM, Andres Taylor andres.tay...@neotechnology.com wrote: Good and valid requests, Aseem. Consider them heard. We'll see how soon we can get it to you. Andrés On Thu, Aug 25, 2011 at 12:57 PM, Aseem Kishore aseem.kish...@gmail.comwrote: Man, I am loving Cypher. Thanks so much guys for introducing it. I'm a bit stuck on one query, though, and I wanted to ask for help. I think the reasons I'm stuck are related to the two feature requests I made yesterday (optional matches, and returning IDs). I want to fetch first- and second-degree friends in one query, in a way that preserves the first-degree friend(s) for each second-degree friends. Assume an asymmetrical friendship model, like Twitter's following, because that makes things a bit easier. Here's a simple transitive example: - Alice follows Bob and Carol. - Bob follows Carol and Dave. - Carol follows Dave and Alice. I want to fetch Alice's first-degree friends (Bob and Carol) and each of their first-degree friends ((Carol and Dave) for Bob, and (Dave and Alice) for Carol). This example is easy enough: START zero=(Alice) MATCH (zero) -[:FOLLOWS]- (first) -[:FOLLOWS]- (second) RETURN first, second I expect to get back results like: - Bob, Carol - Bob, Dave - Carol, Dave - Carol, Alice This is great because I can get Alice's first-degree friends by unique()'ing the first column, for each person, I know their second-degree friends via the second column. But this doesn't work if any of my first-degree friends don't follow anyone. For that, it would be great if I could specify that the second part of that match was optional, and I'd get back a row like this in that case: - Elizabeth, null If I'm returning nodes, it also duplicates a ton of info: each first-degree friend's info is returned in full for each of their friends. Maybe there's no way around that, but this is also where I would find it convenient to be able to return IDs somehow for all but one of the results. Just thinking out loud here, but I'd greatly appreciate any feedback or ideas for this scenario. Thanks much! Aseem ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
Re: [Neo4j] Cypher: how would you query first- and second-degree friends?
Awesome! Thanks again! Aseem On Wed, Sep 14, 2011 at 12:59 PM, Andres Taylor andres.tay...@neotechnology.com wrote: And now it's in code. http://docs.neo4j.org/chunked/snapshot/query-match.html#match-optional-relationship Hope that helps, Andrés On Thu, Aug 25, 2011 at 10:34 PM, Andres Taylor andres.tay...@neotechnology.com wrote: Good and valid requests, Aseem. Consider them heard. We'll see how soon we can get it to you. Andrés On Thu, Aug 25, 2011 at 12:57 PM, Aseem Kishore aseem.kish...@gmail.com wrote: Man, I am loving Cypher. Thanks so much guys for introducing it. I'm a bit stuck on one query, though, and I wanted to ask for help. I think the reasons I'm stuck are related to the two feature requests I made yesterday (optional matches, and returning IDs). I want to fetch first- and second-degree friends in one query, in a way that preserves the first-degree friend(s) for each second-degree friends. Assume an asymmetrical friendship model, like Twitter's following, because that makes things a bit easier. Here's a simple transitive example: - Alice follows Bob and Carol. - Bob follows Carol and Dave. - Carol follows Dave and Alice. I want to fetch Alice's first-degree friends (Bob and Carol) and each of their first-degree friends ((Carol and Dave) for Bob, and (Dave and Alice) for Carol). This example is easy enough: START zero=(Alice) MATCH (zero) -[:FOLLOWS]- (first) -[:FOLLOWS]- (second) RETURN first, second I expect to get back results like: - Bob, Carol - Bob, Dave - Carol, Dave - Carol, Alice This is great because I can get Alice's first-degree friends by unique()'ing the first column, for each person, I know their second-degree friends via the second column. But this doesn't work if any of my first-degree friends don't follow anyone. For that, it would be great if I could specify that the second part of that match was optional, and I'd get back a row like this in that case: - Elizabeth, null If I'm returning nodes, it also duplicates a ton of info: each first-degree friend's info is returned in full for each of their friends. Maybe there's no way around that, but this is also where I would find it convenient to be able to return IDs somehow for all but one of the results. Just thinking out loud here, but I'd greatly appreciate any feedback or ideas for this scenario. Thanks much! Aseem ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
Re: [Neo4j] Cypher: how would you query first- and second-degree friends?
I've run into another scenario which I think is related to this feature request. Take the IMDB example: the graph has movie nodes, and each movie has one or more genres/categories. You could express this as a property on each node, or you could have formal genre nodes that movies are connected to. Either way, the scenario is this: we want to show an overview page that shows the most popular movies for each category. E.g. it shows the top 5 comedy movies, the top 5 action movies, drama, horror, and so on. We can't figure out how to do this in one Cypher query. We can ORDER BY and LIMIT, but that's for the whole set, where we want to do this per category. The closest we've come up with is to return all movie+category pairs -- both as just UUIDs to be efficient (see other thread where I've asked for first-class ID support) -- along with the stats for each movie, and then we sort on the client, then we load the movie data for the top 5 per category. This isn't a bad result, but it feels a bit hacky/manual. Would love if Cypher supported this kind of thing directly. It's almost like grouping results and being able to order/limit within groups. But it's also just another first-degree / second-degree scenario. Food for thought. =) Thanks for the consideration! Aseem On Thu, Aug 25, 2011 at 2:00 PM, Aseem Kishore aseem.kish...@gmail.comwrote: Thanks Andrés! Couple of random extensions to the thoughts above: - To be consistent, perhaps for optional matches, the results could always include a row where that optional match was ignored. That way, Elizabeth, null wouldn't be an inconsistency; *every* first-degree friend would have an [friend], null row. - And perhaps *that* row could be the one that has the full node data, and every subsequent row for the second optional match would just reference the URL/ID of the first-degree node. After chewing on this a lot more, though, I've realized that what I really want in this scenario is not tabular data, but a real subgraph. Hence my email just now. =) Aseem On Thu, Aug 25, 2011 at 1:34 PM, Andres Taylor andres.tay...@neotechnology.com wrote: Good and valid requests, Aseem. Consider them heard. We'll see how soon we can get it to you. Andrés On Thu, Aug 25, 2011 at 12:57 PM, Aseem Kishore aseem.kish...@gmail.com wrote: Man, I am loving Cypher. Thanks so much guys for introducing it. I'm a bit stuck on one query, though, and I wanted to ask for help. I think the reasons I'm stuck are related to the two feature requests I made yesterday (optional matches, and returning IDs). I want to fetch first- and second-degree friends in one query, in a way that preserves the first-degree friend(s) for each second-degree friends. Assume an asymmetrical friendship model, like Twitter's following, because that makes things a bit easier. Here's a simple transitive example: - Alice follows Bob and Carol. - Bob follows Carol and Dave. - Carol follows Dave and Alice. I want to fetch Alice's first-degree friends (Bob and Carol) and each of their first-degree friends ((Carol and Dave) for Bob, and (Dave and Alice) for Carol). This example is easy enough: START zero=(Alice) MATCH (zero) -[:FOLLOWS]- (first) -[:FOLLOWS]- (second) RETURN first, second I expect to get back results like: - Bob, Carol - Bob, Dave - Carol, Dave - Carol, Alice This is great because I can get Alice's first-degree friends by unique()'ing the first column, for each person, I know their second-degree friends via the second column. But this doesn't work if any of my first-degree friends don't follow anyone. For that, it would be great if I could specify that the second part of that match was optional, and I'd get back a row like this in that case: - Elizabeth, null If I'm returning nodes, it also duplicates a ton of info: each first-degree friend's info is returned in full for each of their friends. Maybe there's no way around that, but this is also where I would find it convenient to be able to return IDs somehow for all but one of the results. Just thinking out loud here, but I'd greatly appreciate any feedback or ideas for this scenario. Thanks much! Aseem ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
[Neo4j] Cypher: how would you query first- and second-degree friends?
Man, I am loving Cypher. Thanks so much guys for introducing it. I'm a bit stuck on one query, though, and I wanted to ask for help. I think the reasons I'm stuck are related to the two feature requests I made yesterday (optional matches, and returning IDs). I want to fetch first- and second-degree friends in one query, in a way that preserves the first-degree friend(s) for each second-degree friends. Assume an asymmetrical friendship model, like Twitter's following, because that makes things a bit easier. Here's a simple transitive example: - Alice follows Bob and Carol. - Bob follows Carol and Dave. - Carol follows Dave and Alice. I want to fetch Alice's first-degree friends (Bob and Carol) and each of their first-degree friends ((Carol and Dave) for Bob, and (Dave and Alice) for Carol). This example is easy enough: START zero=(Alice) MATCH (zero) -[:FOLLOWS]- (first) -[:FOLLOWS]- (second) RETURN first, second I expect to get back results like: - Bob, Carol - Bob, Dave - Carol, Dave - Carol, Alice This is great because I can get Alice's first-degree friends by unique()'ing the first column, for each person, I know their second-degree friends via the second column. But this doesn't work if any of my first-degree friends don't follow anyone. For that, it would be great if I could specify that the second part of that match was optional, and I'd get back a row like this in that case: - Elizabeth, null If I'm returning nodes, it also duplicates a ton of info: each first-degree friend's info is returned in full for each of their friends. Maybe there's no way around that, but this is also where I would find it convenient to be able to return IDs somehow for all but one of the results. Just thinking out loud here, but I'd greatly appreciate any feedback or ideas for this scenario. Thanks much! Aseem ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
Re: [Neo4j] Cypher: how would you query first- and second-degree friends?
Good and valid requests, Aseem. Consider them heard. We'll see how soon we can get it to you. Andrés On Thu, Aug 25, 2011 at 12:57 PM, Aseem Kishore aseem.kish...@gmail.comwrote: Man, I am loving Cypher. Thanks so much guys for introducing it. I'm a bit stuck on one query, though, and I wanted to ask for help. I think the reasons I'm stuck are related to the two feature requests I made yesterday (optional matches, and returning IDs). I want to fetch first- and second-degree friends in one query, in a way that preserves the first-degree friend(s) for each second-degree friends. Assume an asymmetrical friendship model, like Twitter's following, because that makes things a bit easier. Here's a simple transitive example: - Alice follows Bob and Carol. - Bob follows Carol and Dave. - Carol follows Dave and Alice. I want to fetch Alice's first-degree friends (Bob and Carol) and each of their first-degree friends ((Carol and Dave) for Bob, and (Dave and Alice) for Carol). This example is easy enough: START zero=(Alice) MATCH (zero) -[:FOLLOWS]- (first) -[:FOLLOWS]- (second) RETURN first, second I expect to get back results like: - Bob, Carol - Bob, Dave - Carol, Dave - Carol, Alice This is great because I can get Alice's first-degree friends by unique()'ing the first column, for each person, I know their second-degree friends via the second column. But this doesn't work if any of my first-degree friends don't follow anyone. For that, it would be great if I could specify that the second part of that match was optional, and I'd get back a row like this in that case: - Elizabeth, null If I'm returning nodes, it also duplicates a ton of info: each first-degree friend's info is returned in full for each of their friends. Maybe there's no way around that, but this is also where I would find it convenient to be able to return IDs somehow for all but one of the results. Just thinking out loud here, but I'd greatly appreciate any feedback or ideas for this scenario. Thanks much! Aseem ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
Re: [Neo4j] Cypher: how would you query first- and second-degree friends?
Thanks Andrés! Couple of random extensions to the thoughts above: - To be consistent, perhaps for optional matches, the results could always include a row where that optional match was ignored. That way, Elizabeth, null wouldn't be an inconsistency; *every* first-degree friend would have an [friend], null row. - And perhaps *that* row could be the one that has the full node data, and every subsequent row for the second optional match would just reference the URL/ID of the first-degree node. After chewing on this a lot more, though, I've realized that what I really want in this scenario is not tabular data, but a real subgraph. Hence my email just now. =) Aseem On Thu, Aug 25, 2011 at 1:34 PM, Andres Taylor andres.tay...@neotechnology.com wrote: Good and valid requests, Aseem. Consider them heard. We'll see how soon we can get it to you. Andrés On Thu, Aug 25, 2011 at 12:57 PM, Aseem Kishore aseem.kish...@gmail.com wrote: Man, I am loving Cypher. Thanks so much guys for introducing it. I'm a bit stuck on one query, though, and I wanted to ask for help. I think the reasons I'm stuck are related to the two feature requests I made yesterday (optional matches, and returning IDs). I want to fetch first- and second-degree friends in one query, in a way that preserves the first-degree friend(s) for each second-degree friends. Assume an asymmetrical friendship model, like Twitter's following, because that makes things a bit easier. Here's a simple transitive example: - Alice follows Bob and Carol. - Bob follows Carol and Dave. - Carol follows Dave and Alice. I want to fetch Alice's first-degree friends (Bob and Carol) and each of their first-degree friends ((Carol and Dave) for Bob, and (Dave and Alice) for Carol). This example is easy enough: START zero=(Alice) MATCH (zero) -[:FOLLOWS]- (first) -[:FOLLOWS]- (second) RETURN first, second I expect to get back results like: - Bob, Carol - Bob, Dave - Carol, Dave - Carol, Alice This is great because I can get Alice's first-degree friends by unique()'ing the first column, for each person, I know their second-degree friends via the second column. But this doesn't work if any of my first-degree friends don't follow anyone. For that, it would be great if I could specify that the second part of that match was optional, and I'd get back a row like this in that case: - Elizabeth, null If I'm returning nodes, it also duplicates a ton of info: each first-degree friend's info is returned in full for each of their friends. Maybe there's no way around that, but this is also where I would find it convenient to be able to return IDs somehow for all but one of the results. Just thinking out loud here, but I'd greatly appreciate any feedback or ideas for this scenario. Thanks much! Aseem ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
