visitors.distinct().count() would be the obvious ways, with the first way in distinct you can specify the level of parallelism and also see improvement in the speed. If it is possible to set up visitors as a stream and use D-streams, that would do the count in realtime. You can stream directly from a directory and … Read more
The most simple thing I discovered is to use DISTINCT over jsonb (not json!). (jsonb_build_object creates jsonb objects) SELECT JSON_AGG( DISTINCT jsonb_build_object(‘tag_id’, photo_tag.tag_id, ‘name’, tag.name)) AS tags FROM photo LEFT OUTER JOIN comment ON comment.photo_id = photo.photo_id LEFT OUTER JOIN photo_tag ON photo_tag.photo_id = photo.photo_id LEFT OUTER JOIN tag ON photo_tag.tag_id = tag.tag_id GROUP BY … Read more
LINQ to Objects doesn’t provide this functionality easily in a built-in way, but MoreLINQ has a handy DistinctBy method: messages = messages.DistinctBy(m => new { m.id, m.date }).ToList();
;with cteRowNumber as ( select COMPUTERNAME, SERIALNUMBER, USERNAME, LASTIP, LASTUPDATE, SOURCE, row_number() over(partition by COMPUTERNAME order by LASTUPDATE desc) as RowNum from YourTable ) select COMPUTERNAME, SERIALNUMBER, USERNAME, LASTIP, LASTUPDATE, SOURCE from cteRowNumber where RowNum = 1
A pure XPath 1.0 — one-liner: Use: count(/*/group/user[not(. = ../following-sibling::group/user)])
It’s annoying, certainly. It’s also part of my “MoreLINQ” project which I must pay some attention to at some point 🙂 There are plenty of other operations which make sense when acting on a projection, but returning the original – MaxBy and MinBy spring to mind. As you say, it’s easy to write – although … Read more
Distinct takes an IEqualityComparer as the second argument, so you will need an IEqualityComparer. It’s not too hard to make a generic one that will take a delegate, though. Of course, this has probably already been implemented in some places, such as MoreLINQ suggested in one of the other answers. You could implement it something … Read more
The problem is with your GetHashCode. You should alter it to return the hash code of AllianceName instead. int IEqualityComparer<Village>.GetHashCode(Village obj) { return obj.AllianceName.GetHashCode(); } The thing is, if Equals returns true, the objects should have the same hash code which is not the case for different Village objects with same AllianceName. Since Distinct works … Read more
Use a terms aggregation on the color field. And you need to pay attention to how that field you want to get distinct values on is analyzed, meaning you need to make sure you’re not tokenizing it while indexing, otherwise every entry in the aggregation will be a different term that is part of the … Read more
This question is thoroughly explained on Hibernate FAQ: First, you need to understand SQL and how OUTER JOINs work in SQL. If you do not fully understand and comprehend outer joins in SQL, do not continue reading this FAQ item but consult a SQL manual or tutorial. Otherwise you will not understand the following explanation … Read more