我有案例类列表。输出需要对案例类的不同参数进行聚合。寻找更优化的方法来做到这一点。
示例：
case class Students(city: String, college: String, group: String,
name: String, fee: Int, age: Int)
object GroupByStudents {
val studentsList= List(
Students("Mumbai","College1","Science","Jony",100,30),
Students("Mumbai","College1","Science","Tony", 200, 25),
Students("Mumbai","College1","Social","Bony",250,30),
Students("Mumbai","College2","Science","Gony", 240, 28),
Students("Bangalore","College3","Science","Hony", 270, 28))
}
现在要了解一个城市学生的详细信息，我需要首先按城市汇总，然后根据大学明智地分解这些细节，然后分组。
输出是以下格式的案例类列表。
Students(Mumbai,,,,790,0) -- aggregate city wise
Students(Mumbai,College1,,,550,0) -- aggregate college wise
Students(Mumbai,College1,Social,,250,0)
Students(Mumbai,College1,Science,,300,0)
Students(Mumbai,College2,,,240,0)
Students(Mumbai,College2,Science,,240,0)
Students(Bangalore,,,,270,0)
Students(Bangalore,College3,,,270,0)
Students(Bangalore,College3,Science,,270,0)
实现此目的的两种方法：
1）循环所有列表，为每个组合创建一个地图（以上情况3组合 ），聚合数据并创建新的结果列表并将数据附加到其中。
2）使用foldLeft选项
studentsList.groupBy(d=>(d.city))
.mapValues(_.foldLeft(Students("","","","",0,0))
((r,c) => Students(c.city,"","","",r.fee+c.fee,0)))
studentsList.groupBy(d=>(d.city,d.college))
.mapValues(_.foldLeft(Students("","","","",0,0))
((r,c) => Students(c.city,c.college,"","",r.fee+c.fee,0)))
studentsList.groupBy(d=>(d.city,d.college,d.group))
.mapValues(_.foldLeft(Students("","","","",0,0))
((r,c) => Students(c.city,c.college,c.group,"",r.fee+c.fee,0)))
在这两种情况下，不止一次循环列表。有没有办法通过单程和优化的方式实现这一目标。
答案 0 :(得分：2)
代码看起来更好一些，但我认为它并不快。使用groupby，你总是有2个“循环”
studentsList.groupBy(d=>(d.city)).map { case (k,v) =>
Students(v.head.city,"","","",v.map(_.fee).sum, 0)
}
studentsList.groupBy(d=>(d.city,d.college)).map { case (k,v) =>
Students(v.head.city,v.head.college,"","",v.map(_.fee).sum, 0)
}
studentsList.groupBy(d=>(d.city,d.college,d.group)).map { case (k,v) =>
Students(v.head.city,v.head.college,v.head.group,"",v.map(_.fee).sum, 0)
}
你会得到像这样的东西
List(Students(Bangalore,College3,Science,Hony,270,0),
Students(Mumbai,College1,Science,Jony,790,0))
List(Students(Mumbai,College2,,,240,0),
Students(Bangalore,College3,,,270,0),
Students(Mumbai,College1,,,550,0))
List(Students(Bangalore,College3,Science,,270,0),
Students(Mumbai,College2,Science,,240,0),
Students(Mumbai,College1,Social,,250,0),
Students(Mumbai,College1,Science,,300,0))
它与您的示例中的输出不完全相同，但它是所需的输出：案例类学生列表。
如果你自己分组，你可以避免这种循环。只有城市的例子，另一个是直接的。
var m = Map[String, Students]()
for (v <- studentsList) {
m += v.city -> Students(v.city,"","","",v.fee + m.getOrElse(v.city, Students("","","","",0,0)).asInstanceOf[Students].fee, 0)
}
m
<强>输出
与studenList相同的输出，但我只为每个Map[String,Students]输出循环一次。
Map(Mumbai -> Students(Mumbai,,,,790,0), Bangalore -> Students(Bangalore,,,,270,0))
只需在完整列表中循环一遍。
val emptyStudent = Students("","","","",0,0);
studentsList.foldLeft(Map[String, Students]()) { case (m, v) =>
m + (v.city -> Students(v.city,"","","",
v.fee + m.getOrElse(v.city, emptyStudent).fee, 0))
}
studentsList.foldLeft(Map[(String,String), Students]()) { case (m, v) =>
m + ((v.city,v.college) -> Students(v.city,v.college,"","",
v.fee + m.getOrElse((v.city,v.college), emptyStudent).fee, 0))
}
studentsList.foldLeft(Map[(String,String,String), Students]()) { case (m, v) =>
m + ((v.city,v.college,v.group) -> Students(v.city,v.college,v.group,"",
v.fee + m.getOrElse((v.city,v.college,v.group), emptyStudent).fee, 0))
}
<强>输出
与studenList相同的输出，但我只为每个Map[String,Students]输出循环一次。
Map(Mumbai -> Students(Mumbai,,,,790,0),
Bangalore -> Students(Bangalore,,,,270,0))
Map((Mumbai,College1) -> Students(Mumbai,College1,,,550,0),
(Mumbai,College2) -> Students(Mumbai,College2,,,240,0),
(Bangalore,College3) -> Students(Bangalore,College3,,,270,0))
Map((Mumbai,College1,Science) -> Students(Mumbai,College1,Science,,300,0),
(Mumbai,College1,Social) -> Students(Mumbai,College1,Social,,250,0),
(Mumbai,College2,Science) -> Students(Mumbai,College2,Science,,240,0),
(Bangalore,College3,Science) -> Students(Bangalore,College3,Science,,270,0))
您可以使用所有列表生成一个大地图。
val emptyStudent = Students("","","","",0,0);
studentsList.foldLeft(Map[(String,String,String), Students]()) { case (m, v) =>
{
var t = m + ((v.city,"","") -> Students(v.city,"","","",
v.fee + m.getOrElse((v.city,"",""), emptyStudent).fee, 0))
t = t + ((v.city,v.college,"") -> Students(v.city,v.college,"","",
v.fee + m.getOrElse((v.city,v.college,""), emptyStudent).fee, 0))
t + ((v.city,v.college,v.group) -> Students(v.city,v.college,v.group,"",
v.fee + m.getOrElse((v.city,v.college,v.group), emptyStudent).fee, 0))
}
}
<强>输出
在这种情况下，您循环一次并返回所有聚合的结果，但仅在一个Map中。这也适用于理解。
Map((Mumbai,College1,Science) -> Students(Mumbai,College1,Science,,300,0),
(Bangalore,,) -> Students(Bangalore,,,,270,0),
(Mumbai,College2,Science) -> Students(Mumbai,College2,Science,,240,0),
(Mumbai,College2,) -> Students(Mumbai,College2,,,240,0),
(Mumbai,College1,Social) -> Students(Mumbai,College1,Social,,250,0),
(Mumbai,,) -> Students(Mumbai,,,,790,0),
(Bangalore,College3,) -> Students(Bangalore,College3,,,270,0),
(Mumbai,College1,) -> Students(Mumbai,College1,,,550,0),
(Bangalore,College3,Science) -> Students(Bangalore,College3,Science,,270,0))
始终会复制地图，因此可能会出现一些性能和内存问题。要解决这个问题，请使用“理解”
这将生成一个包含3种聚合类型的Map。
val emptyStudent = Students("","","","",0,0);
var m = Map[(String,String,String), Students]()
for (v <- studentsList) {
m += ((v.city,"","") -> Students(v.city,"","","", v.fee + m.getOrElse((v.city,"",""), emptyStudent).fee, 0))
m += ((v.city,v.college,"") -> Students(v.city,v.college,"","", v.fee + m.getOrElse((v.city,v.college,""), emptyStudent).fee, 0))
m += ((v.city,v.college,v.group) -> Students(v.city,v.college,v.group,"", v.fee + m.getOrElse((v.city,v.college,v.group), emptyStudent).fee, 0))
}
m
这在内存消耗方面应该更好，因为你不像foldLeft示例中那样复制地图
<强>输出
Map((Mumbai,College1,Science) -> Students(Mumbai,College1,Science,,300,0),
(Bangalore,,) -> Students(Bangalore,,,,270,0),
(Mumbai,College2,Science) -> Students(Mumbai,College2,Science,,240,0),
(Mumbai,College2,) -> Students(Mumbai,College2,,,240,0),
(Mumbai,College1,Social) -> Students(Mumbai,College1,Social,,250,0),
(Mumbai,,) -> Students(Mumbai,,,,790,0), (Bangalore,College3,) -> Students(Bangalore,College3,,,270,0),
(Mumbai,College1,) -> Students(Mumbai,College1,,,550,0),
(Bangalore,College3,Science) -> Students(Bangalore,College3,Science,,270,0))
在所有情况下，如果您在案例类学生中将参数设置为可选，则可以减少代码，因为您可以在分组期间执行类似Students(city=v.city,fee=v.fee+m.getOrElse(v.city,emptyStudent).fee的操作
答案 1 :(得分：1)
使用foldLeft
首先，让我们定义一些类型别名以使语法更容易
object GroupByStudents {
type City = String
type College = String
type Group = String
type Name = String
type Aggregate = Map[City, Map[College, Map[Group, List[Students]]]]
def emptyAggregate: Aggregate = Map.empty
case class Students(city: City, college: College, group: Group,
name: Name, fee: Int, age: Int)
}
您可以将学生列表汇总到一个Aggregate
foldLeft地图中
object Test {
import GroupByStudents._
def main(args: Array[String]) {
val studentsList = List(
Students("Mumbai","College1","Science","Jony",100,30),
Students("Mumbai","College1","Science","Tony", 200, 25),
Students("Mumbai","College1","Social","Bony",250,30),
Students("Mumbai","College2","Science","Gony", 240, 28),
Students("Bangalore","College3","Science","Hony", 270, 28))
val aggregated = studentsList.foldLeft(emptyAggregate){(agg, students) =>
val cityBin = agg.getOrElse(students.city, Map.empty)
val collegeBin = cityBin.getOrElse(students.college, Map.empty)
val groupBin = collegeBin.getOrElse(students.group, List.empty)
val nextGroupBin = students :: groupBin
val nextCollegeBin= collegeBin + (students.group -> nextGroupBin)
val nextCityBin = cityBin + (students.college -> nextCollegeBin)
agg + (students.city -> nextCityBin)
}
}
}
aggregated来计算费用。
如果你真的想要，你可以在foldLeft本身计算费用，但这会使代码更难阅读。
请注意，您也可以尝试monocle's lenses将学生的价值放在汇总结构中。