How to groupBy groupBy? - scala

Question

I need to map through a List[(A,B,C)] to produce an html report. Specifically, a
List[(Schedule,GameResult,Team)]
Schedule contains a gameDate property that I need to group by on to get a
Map[JodaTime, List(Schedule,GameResult,Team)]
which I use to display gameDate table row headers. Easy enough:
val data = repo.games.findAllByDate(fooDate).groupBy(_._1.gameDate)
Now the tricky bit (for me) is, how to further refine the grouping in order to enable mapping through the game results as pairs? To clarify, each GameResult consists of a team's "version" of the game (i.e. score, location, etc.), sharing a common Schedule gameID with the opponent team.
Basically, I need to display a game result outcome on one row as:
3 London Dragons vs. Paris Frogs 2
Grouping on gameDate let's me do something like:
data.map{case(date,games) =>
// game date row headers
<tr><td>{date.toString("MMMM dd, yyyy")}</td></tr>
// print out game result data rows
games.map{case(schedule,result, team)=>
...
// BUT (result,team) slice is ungrouped, need grouped by Schedule gameID
}
}
In the old version of the existing application (PHP) I used to
for($x = 0; $x < $this->gameCnt; $x = $x + 2) {...}
but I'd prefer to refer to variable names and not the come-back-later-wtf-is-that-inducing:
games._._2(rowCnt).total games._._3(rowCnt).name games._._1(rowCnt).location games._._2(rowCnt+1).total games._._3(rowCnt+1).name
maybe zip or double up for(t1 <- data; t2 <- data) yield(?) or something else entirely will do the trick. Regardless, there's a concise solution, just not coming to me right now...

Answer 1

Maybe I'm misunderstanding your requirements, but it seems to me that all you need is an additional groupBy:
repo.games.findAllByDate(fooDate).groupBy(_._1.gameDate).mapValues(_.groupBy(_._1.gameID))
The result will be of type:
Map[JodaTime, Map[GameId, List[(Schedule,GameResult,Team)]]]
(where GameId is the type of the return type of Schedule.gameId)
Update: if you want the results as pairs, then pattern matching is your friend, as shown by Arjan. This would give us:
val byDate = repo.games.findAllByDate(fooDate).groupBy(_._1.gameDate)
val data = byDate.mapValues(_.groupBy(_._1.gameID).mapValues{ case List((sa, ra, ta), (sb, rb, tb)) => (sa, (ta, ra), (tb, rb)))
This time the result is of type:
Map[JodaTime, Iterable[ (Schedule,(Team,GameResult),(Team,GameResult))]]
Note that this will throw a MatchError if there are not exactly 2 entries with the same gameId. In real code you will definitely want to check for this case.

Answer 2

Ok a soultion from Régis Jean-Gilles:
val data = repo.games.findAllByDate(fooDate).groupBy(_._1.gameDate).mapValues(_.groupBy(_._1.gameID))
You said it was not correct, maybe you just didnt use it the right way?
Every List in the result is a pair of games with the same GameId.
You could pruduce html like that:
data.map{case(date,games) =>
// game date row headers
<tr><td>{date.toString("MMMM dd, yyyy")}</td></tr>
// print out game result data rows
games.map{case (gameId, List((schedule, result, team), (schedule, result, team))) =>
...
}
}
And since you dont need a gameId, you can return just the paired games:
val data = repo.games.findAllByDate(fooDate).groupBy(_._1.gameDate).mapValues(_.groupBy(_._1.gameID).values)
Tipe of result is now:
Map[JodaTime, Iterable[List[(Schedule,GameResult,Team)]]]
Every list again a pair of two games with the same GameId